Let the real functions $ f\circ g$ and $ g$ defined as:

$ f\circ g(x) \equal{} \begin{cases}4x^2 \minus{} 6x \minus{} 1;x\geq 1 \\
4x \plus{} 3;x< 1\end{cases}$

And $ g(x) \equal{} 2x \minus{} 3$

Determine $ f\circ f(x)$

[hide="what I got so far"]
$ f(x) \equal{} \begin{cases}2x \plus{} 3;x < 1 \\
x^2 \plus{} 3x \plus{} 1;x\geq 1\end{cases}$

I am lost, however, on all the conditions that come afterwards on $ f(f(x))$[/hide]

Actually,

$ f(x) \equal{}\begin{cases}2x\plus{}9;x < 1\\ x^{2}\plus{}3x\minus{}1;x\geq 1\end{cases}$

[quote="modularmarc101"]Actually,

$ f(x) \equal{} \begin{cases}2x \plus{} 9;x < 1 \\
x^{2} \plus{} 3x \minus{} 1;x\geq 1\end{cases}$[/quote]
No that is not correct too
$ f(x) \equal{} \begin{cases}2x \plus{} 9;x < \minus{}1 \\
x^{2} \plus{} 3x \minus{} 1;x\geq \minus{}1\end{cases}$[/quote] :)
since the condition for f is the range of 2x-3


Is there a special formula or formulae for finding
 $2^{2^{2^{...}}}$ or  $n^{n^{n^{...}}}$
 :|

${f(x)=n^{f(x-1)}}$
${f(0)=n}$

I forgot the name of this function........

shift!!!

[quote="DaReaper"]so..$n=0$ ?[/quote]

What do you mean?

If $f(n)=n^{n^{n^{n\cdots}}}$, then $f(n)=n^{f(n)}$. This isn't very easy to solve for. I think think this involves Lambert's W function (way above Classroom Math. :( )

EDIT: For those who want the answer, here is the answer: $f(n)=\frac{W(\log n)}{\log n}$, where $W(x)$ is Lambert's W-function. I'm sorry if this is confusing.

oh.. yes... the lagrange again..

but the solution is

${a = \frac{W(-\log(n))}{-\log(n)}}$

not  ${f(n)}$

because the solution is for ${x}$ going to infinity
SEe

http://en.wikipedia.org/wiki/Lambert's_W_function


A bowl contains 4 red jelly beans and 5 black jelly beans. Four jelly beans are selected at random without replacement from the bowl. What is the probability that there will be more black jelly beans than red jelly beans in the selection? Express your answer as a common fraction.

[hide] There are 9 choose 4=126 ways to randomly pick 4 out of 9 balls.  Now, we can either have 3 black and 1 red or 4 black and 0 red.  There are (5 choose 3)(4 choose 1)=40 ways to choose 3 blacks and 1 red.  There are (5 choose 4)(4 choose 0)=5 ways to choose 4 blacks and 0 red.  Therefore, you add them up to get 45.  Our probability is thus 45/126=$\frac{5}{14}$.  [/hide]

[hide="i think..."]$\frac{5C3*4C1+5C4*4C0}{9C4}$, watever that value is...[/hide]

What is the formula for Combinations? and Permutations?

[quote="mchoi815"]n C r (combinations) is
n! / (n-r)!(r!)
n P r (permutations) is
n! / (n-r)!

i think[/quote]
yea thats right.

[quote="mchoi815"]n C r (combinations) is
n! / (n-r)!(r!)
n P r (permutations) is
n! / (n-r)!

i think[/quote]
Thanks.
Another question, does $\binom{9}{4}$ mean $(9C4)$ or $(9P4)$?

$\binom94=C(9,4)$, because it literally means "Nine Choose 4."


geometry
circumcircle
Euler
incenter
power of a point
radical axis

Από το φίλο μας (του forum) April(Vietnam) έλαβα κατόπιν παράκλησης τις παρακάτω ασκήσεις (του ζήτησα μερικές για το ριζικό άξονα κλπ).Αν βρείτε χρόνο, ρίξτε μια ματιά. Τις λύσεις θα τις συμπεριλάβω επώνυμα σε μια προσπάθεια που κάνω σε μια γεωμετρική  ενότητα(θα σας ενημερώσω την κατάλληλη  στιγμή).

 ΑΣΚΗΣΕΙΣ

1. Let AB and CD are two diagonals of a given circle (O). The tangent line to (O) in B meets AC at E, let F the second intersection of DE with the circle (O). Prove that BC, AF and OE are concurrent.

2. Let be given a triangle ABC. Let O, H be respectively the circumcenter, orthocenter of triangle ABC. Denote by M, N be the middle points of the segments AB, AC, respectively, and E, F be the feet of B-altitude, C-altitude of triangle ABC. Prove that OH is perpendicular to AK, where K is the intersection of MN and EF.

3. Given two non-intersecting circles (O) and (O'). Let MN, PQ be their external common tangent, respectively their internal common tangent. Prove that MP, NQ and OO' concur.

 Μπάμπης

[quote="bilot4"]Η πρώτη δεν μου φαίνεται σωστή (συγγνώμη αν κάνω λάθος :oops: ). Φαίνεται πάντως πως όλα τα σημεία εκτός του F είναι ανεξάρτητα του D. Ετσι η τομή BC, OE είναι ανεξάρτητη του D. Αρα  η AF δεν περνάει πάντα απ' το ίδιο σημείο (το οποίο είναι σταθερό), αλλά αντίθετα αλλάζει "μετακινώντας" το D.

Και στην τρίτη κάτι δεν πάει καλά... κάτι άλλο μάλλον λέει...[/quote]

 ΔΗΜΗΤΡΗ, δες μήπως οι παρατηρήσεις βοηθάνε:

I'm sorry, jn the fjrst problem, AB and CD are djameters of the gjven
cjrcle. In the thjrd problem two cjrcles are outsjde to eachother, and
M, P lje on (O), N, Q lje on (O').

Μπάμπης

Για το [b][size=100](1)[/size][/b] : [hide]- Αρκεί να αποδείξουμε ότι τα σημεία $ B,\ M\equiv AF\cap CD,\ N\equiv AD\cap CF,$ είναι συνευθειακά και έτσι προκύπτει το ζητούμενο από τα προοπτικά τρίγωνα $ \bigtriangleup PCF,\ \bigtriangleup OAD,$ όπου $ P\equiv AF\cap CD.$

Η ευθεία $ MX\perp NO,$ πολική του $ N$ ως προς τον κύκλο $ (O),$ τέμνει την ευθεία $ NAD$ στο σημείο έστω $ Z,$ αρμονικό συζυγές του $ N,$ ως προς τα $ A,\ D.$

Αλλά τότε, στο πλήρες τετράπλευρο $ CEFMAD,$ η ευθεία $ MZ$ περνάει από το $ E$ και άρα η ευθεία $ BM,$ η πολική του $ E$ ως προς τον κύκλο $ (O),$ περνάει από το σημείο $ N.$ [/hide]
Για το [b][size=100](2)[/size][/b] : [hide]– Αποδεικνύεται με το [b][size=100]Θεώρημα Πάππου[/size][/b],  ότι το σημείο $ P\equiv ME\cap NF,$ ανήκει στην  [b][size=100]Ευθεία Euler[/size][/b] $ OH$ του τριγώνου $ \bigtriangleup ABC.$

Έτσι στο εγγεγραμμένο τετράπλευρο $ EFMN,$ έχουμε $ OH\equiv O'P\perp AK,$ γνωστό αποτέλεσμα, που αποδεικνύεται άμεσα με πολικές, αλλά και με στοιχειώδη τρόπο. http://www.mathlinks.ro/Forum/viewtopic.php?t=110887[/hide]
Για το [b][size=100](3)[/size][/b] : [hide]– Έχει προταθεί από τον Σιλουανό στο αγγλικό φόρουμ.

http://www.Mathlinks.ro/Forum/viewtopic.php?t=223789 .

Επίσης υπάρχει ως προτεινόμενη άσκηση ( με ακριβέστερη εκφώνηση ), στο βιβλίο [b][size=100]Γεωμετρία – Μαθηματικές Ολυμπιάδες[/size][/b] του [b][size=100]Δ. Κοντογιάννη[/size][/b] ( σελίδα 204, άσκηση 555 ).[/hide]

Κώστας Βήττας.

In this message, $ (XYZ)$ means the circumcircle of triangle $ XYZ$, and $ (XY)$ the circle with diameter $ XY$. I would like to present my solutions using radical axis for these problems here.  :D

1. We have $ BC$ is the radical axis of $ (O)$ and $ (BCE)$, $ AF$ is the radical axis of $ (O)$ and $ (AEF)$. So we have to prove that $ OE$ is the radical axis of $ (BCE)$ and $ (AEF)$.

Indeed, we have $ \angle OAF\equal{}\angle BDF\equal{}\angle AEF$, so $ OA$ is tangent to $ (AEF)$ $ \Longrightarrow \mathcal{P}_{O/(AEF)}\equal{}OA^2\equal{}R^2$ $ (1)$, where $ \mathcal{P}_{X/(MNP)}$ is the power of point $ X$ w.r.t. $ (MNP)$, and $ R$ the radius of circle $ (O)$.

Similarly, $ \angle OBC\equal{}\angle BEC$, so $ OB$ is tangent to $ (BCE)$ $ \Longrightarrow \mathcal{P}_{O/(BCE)}\equal{}OB^2\equal{}R^2$ $ (2)$

From $ (1)$ and $ (2)$ we conclude that $ O$ lies on the radical axis of $ (BCE)$ and $ (AEF)$. Our solution is completed.

2. We have $ \bullet$ $ \mathcal{P}_{A/(OA)}\equal{}0\equal{}\mathcal{P}_{A/(HA)}$ $ (1)$

$ \bullet$ $ \angle MFE\equal{}\angle BCA\equal{}\angle MNA$, it implies that the quadrilateral $ MNEF$ is cyclic. Therefore $ KM\cdot KN\equal{}KF\cdot KE$, i.e., $ \mathcal{P}_{K/(OA)}\equal{}\mathcal{P}_{K/(HA)}$ $ (2)$

Combining $ (1)$ and $ (2)$, we get $ AK$ is the radical axis of $ (OA)$ and $ (HA)$, in conclusion $ OH$ is perpendicular to $ AK$.

I'll post my solution for problem 3 here later.

Denote $ S\equal{}MN\cap PQ$ and $ T\equal{}MP\cap NQ$.

We have $ \angle MSO\equal{}\frac{1}{2}\cdot\angle MSP\equal{}\frac{1}{2}\cdot 2\angle SNQ\equal{}\angle SNQ$, it implies that $ SO$ is parallel to $ NQ$, i.e., $ NQ$ is perpendicular to $ MP$. So, we conclude that $ T$ lies on the radical axis of $ (MN)$ and $ (PQ)$ $ (1)$.

On the other hand, we have $ \mathcal{P}_{O/(MN)}\equal{}OM^2\equal{}OP^2\equal{}\mathcal{P}_{O/(PQ)}$ and $ \mathcal{P}_{O'/(MN)}\equal{}O'N^2\equal{}O'Q^2\equal{}\mathcal{P}_{O'/(PQ)}$. Therefore $ OO'$ is the radical axis of $ (MN)$ and $ (PQ)$ $ (2)$.

From $ (1)$ and $ (2)$ we conclude that $ MP$, $ NQ$ and $ OO'$ are concurrent.

[b]AN APPLICATION.[/b] Let be given a triangle $ ABC$ and an arbitrary point $ M$ on its side $ BC$. Circles $ (O_1)$, $ (O_2)$ are all tangent internally to $ (O)$, also tangent $ MA$ and respectively tangent to $ MB$, $ MC$. Prove that $ O_1O_2$ passes through the incenter $ I$ of triangle $ ABC$.

[quote="April"]Denote $ S \equal{} MN\cap PQ$ and $ T \equal{} MP\cap NQ$.

We have $ \angle MSO \equal{} \frac {1}{2}\cdot\angle MSP \equal{} \frac {1}{2}\cdot 2\angle SNQ \equal{} \angle SNQ$, it implies that $ SO$ is parallel to $ NQ$, i.e., $ NQ$ is perpendicular to $ MP$. So, we conclude that $ T$ lies on the radical axis of $ (MN)$ and $ (PQ)$ $ (1)$.

On the other hand, we have $ \mathcal{P}_{O/(MN)} \equal{} OM^2 \equal{} OP^2 \equal{} \mathcal{P}_{O/(PQ)}$ and $ \mathcal{P}_{O'/(MN)} \equal{} O'N^2 \equal{} O'Q^2 \equal{} \mathcal{P}_{O'/(PQ)}$. Therefore $ OO'$ is the radical axis of $ (MN)$ and $ (PQ)$ $ (2)$.

From $ (1)$ and $ (2)$ we conclude that $ MP$, $ NQ$ and $ OO'$ are concurrent.

[b]AN APPLICATION.[/b] Let be given a triangle $ ABC$ and an arbitrary point $ M$ on its side $ BC$. Circles $ (O_1)$, $ (O_2)$ are all tangent internally to $ (O)$, also tangent $ MA$ and respectively tangent to $ MB$, $ MC$. Prove that $ O_1O_2$ passes through the incenter $ I$ of triangle $ ABC$.[/quote]

Dear April,
 many thanks from all members in the greek community :) . TomorrowI will print and I will read carefully your problems and solutions.It is too late for me now.

 Have a nice day- Babis


[quote="N.T.TUAN"]If $a=0$ then done else
\[\frac{y_{n}}{a^{n}}=\frac{x_{n}}{a^{n}}-\frac{x_{n-1}}{a^{n-1}}(n=1,2,...) \]
Therefore
\[\sum_{k=1}^{n}\frac{y_{k}}{a^{k}}=\frac{x_{n}}{a^{n}}-x_{0}(n=1,2,...) \]
also
\[x_{n}=\sum_{k=0}^{n}a^{n-k}y_{k}(n=0,1,2,...) \]
But at this time , that is all what I know!  :([/quote]
thanks for your hint. i have solved it :)
let $L$ be the upper bound of $\{y_{n}\}$
$\forall\varepsilon>0,\exists N(N=\max\{1,log_{|a|}\frac{(1-|a|)\varepsilon}{L}\}),\forall n>m>M$
$|x_{n}-x_{m}|\le L(|a|^{m}+...+|a|^{n})\le\varepsilon$
so $\{x_{n}\}$ is convergent


1. Find the period of $ a_{n\plus{}1} \equal{} \frac{1}{2}\left(a_n \minus{} \frac{1}{a_n}\right)$ for starting value $ a_1 \equal{} \frac{1}{2}$, or prove that it does not repeat.
2. Describe the period of the sequence for any $ a_1 \in \mathbb{Q}$.
3. Prove that for $ a_1 \in \mathbb{R} (x \notin \mathbb{Q})$, $ \{a_1, a_2, a_3, ...\} \subsetneq \mathbb{R}$

[quote="Aristocrat"]1. Find the period of $ a_{n \plus{} 1} \equal{} \frac {1}{2}\left(a_n \minus{} \frac {1}{a_n}\right)$ for starting value $ a_1 \equal{} \frac {1}{2}$, or prove that it does not repeat.
2. Describe the period of the sequence for any $ a_1 \in \mathbb{Q}$.
3. Prove that for $ a_1 \in \mathbb{R} (x \notin \mathbb{Q})$, $ \{a_1, a_2, a_3, ...\} \subsetneq \mathbb{R}$[/quote]
[hide="Intermediate result"]
Let $ a_n \equal{} \cot(u_n)$. Then $ a_{n \plus{} 1} \equal{} \cot(u_{n \plus{} 1})$ $ \equal{} \frac 12\left(\frac {\cos(u_n)}{\sin(u_n)} \minus{} \frac {\sin(u_n)}{\cos(u_n)}\right)$ $ \equal{} \frac {\cos(u_n)^2 \minus{} \sin(u_n)^2}{2\sin(u_n)\cos(u_n)}$ $ \equal{} \cot(2u_n)$

And so $ a_n \equal{} \cot(2^{n \minus{} 1}u_1)$

So $ a_n$ is periodic $ \iff$ $ u_1 \equal{} \frac {k\pi}{2^p \minus{} 1}$ (where $ k\in\mathbb Z$ , $ p\in\mathbb N$ is the period and $ \frac {k}{2^p \minus{} 1}\notin\mathbb Z$)

Notice that we can also have $ a_n$ periodic from a given starting point $ a_s$ if $ u_1 \equal{} \frac {k\pi}{2^{s \minus{} 1}(2^p \minus{} 1)}$. We'll from now consider that the problem involves uniquely sequence periodic from the beginning $ a_1$

So, $ a_n$ is periodic $ \iff$ $ a_1 \equal{} \cot(\frac {k\pi}{2^p \minus{} 1})$ with $ p\in\mathbb N$, $ k\in\mathbb Z$ and $ \frac {k}{2^p \minus{} 1}\notin\mathbb Z$

So the problem is linked to the study of the rationals values of $ \tan(\frac {k\pi}{2^p \minus{} 1})$ (for example, the first question implies to answer to "does it exist $ k,p$ such that $ \tan(\frac {k\pi}{2^p \minus{} 1}) \equal{} 2$ ??") ...

I'm sure the answer is no but I'm blocked here for the moment ...
[/hide]

Thanks! I wasn't aware that the periodicity of performing Newton's method on $ x^2 \plus{} 1$ in the reals was explainable by trig. Maybe this warrants a little more investigation, but right now I'm fairly lost in the math :(. If anyone else wants to contribute their two cents, please do; it'll be of great help!


AMC AIME USA(J)MO USAMO trigonometry geometry probability

Do u guys think I can make it to USAMO with a 111 on AMC 10A? I know it's a very bad score and I'm not quite optimistic this time. :(

But anyways, what should I review for AIME? Like trigonometry and stuff? I heard that there should be a lot of geometry and probability and I suck at those.

Any advice? :?:

How did you get a 111 on the 10A and make AIME?

[quote="Art of Owna"]Review the number 2007..seriously..do some stuff with it..

and yea.. theres always like 5+ geometry/trigonometry each year, as geometry/trigonometry usually(maybe always) is the majority of the test..compared to combinatorics,number theory, functions+polynomials, and sequences

and yea..you probably need atleast 10 for usamo[/quote]

$2007=3^{2}*223$, and there isn't too many interesting things you can do with it (other than the fact that it's a multiple of 9)

The prime (as in best, not two divisors) years for year-related problems were definately $2000$ and $2002$.

[quote="Art of Owna"]I think he meant amc12..since hes implying that hes adding taht score with his AIME score....since I qualified through the 10 (9th grader), my amc score doesnt really matter, so it would of no use to post ur 10 score when you are also talking about usamo[/quote]

No. Really. It's AMC 10A. I was confused, too. I heard that the cutoff was 117 or something and I'm definitely not in the first 1%. I dont know how I'm qualified for the AIME.

And yea, u guys have any advice about prep books or how to study for the last 2 weeks?

oh well, maybe i have to check my score again. the teacher's probably confused with some1 else. She didnt even look at me when i asks for my score. :P


[quote="ColinA"]Please find the patterns for following 2 sequences :

72, 91, 103, ?, 89, ?, 96, ?, 72

57, 13, 60, 13, ?, 16, ?, 25, ?, 29

Please send solution(s) to colinaye@hotmail.com. Payment will be available for the satisfactory solution(s).

Thanks[/quote]

Firstly, this is not an [b]unsolved olympiad problem in Combinatorics[/b].
Secondly, if payment is available for satisfactory solutions, then why do you need the solutions?
Thirdly, this forum section is about proofs or theorems or other mathematical ideas, [i]not[/i] pattern matching. Why? Because there are infinitely many solutions to any "pattern". We might as well give a sequence of randomly generated numbers and ask for the next one.


Can you help me prove?

$cotx-2cot2x=tanx$

...using double angle forumlas

$\cot{2\theta} = \frac{1 - \tan^2{\theta}}{2\tan{\theta}}$

going to your problem,

$\frac{1}{\tan{\theta}} - 2(\frac{1 - \tan^2{\theta}}{2\tan{\theta}}) = \frac{1}{\tan{\theta}} - \frac{1-\tan^2{\theta}}{\tan{\theta}} = \frac{\tan^2{\theta}}{\tan{\theta}} = \tan{\theta}$

there you go.

thanks, but im still confused about the inbetween steps u left out  :blush:

Edit: Ah OK I get it. I didn't know you could take the 2 out of the bottom without doing something to the top since the two you took out was in the parentheses

~thanks.


Acute triangle ABC has circumcenter O. Let the feet of the perpendiculars from O to BC, AC, and AB be $ O_A$, $ O_B$, and $ O_C$ respectively. Prove that $ r\plus{}R\equal{}OO_A\plus{}OO_B\plus{}OO_C$.

See the attached diagram. Let $ AA', BB', CC'$ be the altitudes of the triangle.

Since $ \angle COB \equal{} 2\alpha\implies \angle COO_A \equal{} \alpha$, we have three similar right triangles: $ \triangle ACC'\sim\triangle ABB'\sim OCO_A$. Therefore $ {AC'\over b} \equal{} {AB'\over c} \equal{} {OO_A\over R}$. This yields $ R\cdot AC' \equal{} b\cdot OO_A\land R\cdot AB' \equal{} c\cdot OO_A$. Adding those two up, we get $ R(AC' \plus{} AB') \equal{} OO_A(b \plus{} c)$

In the similar manner, we find $ R(BA' \plus{} BC') \equal{} OO_B(a \plus{} c)$ and $ R(CA' \plus{} CB') \equal{} OO_C(a \plus{} b)$. Summing the three equalities, we get $ R(a \plus{} b \plus{} c) \equal{} OO_A(b \plus{} c) \plus{} OO_B(a \plus{} c) \plus{} OO_C(a \plus{} b)\quad(1)$.

We also know that $ [ABC] \equal{} [OAB] \plus{} [OBC] \plus{} [OCA]$, hence $ r(a \plus{} b \plus{} c) \equal{} aOO_A \plus{} bOO_B \plus{} cOO_C\quad(2)$

Adding $ (1)$ and $ (2)$ and dividing by $ a \plus{} b \plus{} c$, we get $ OO_A \plus{} OO_B \plus{} OO_C \equal{} R \plus{} r$, as desired.

NOTE: As $ OO_A\equal{}R\cos\alpha$ (see the picture) etc. cyclically, as a direct consequence we get $ \cos\alpha\plus{}\cos\beta\plus{}\cos\gamma\equal{}1\plus{}{r\over R}$


I was just wondering if anyone could give me some advice on how to prepare for AP exams. I plan to just drill myself for calculus, but I'm concerned about the AP Biology test. Our class has gone rather slow this year and I've tried to stay ahead..but there was a brief hiatus in the February/March time when math comps were imminent. I plan to finish evolution soon, but I still have animal evolution, human body systems, and ecology left and a lot of AP books to work through. Any veterans out there with some good advice as to how to most wisely use my time?

Well, You are taking the AB exam which is like a sinch. And second of all about 20% of people get fives. And most people here on AOPS are very gifted so you will definitely get a 5.

AP Biology is heck. This is because NO MATTER HOW SMART YOU ARE, YOU STILL HAVE TO WORK YOUR BUTT OFF :(.

If I were you, I'd hurry up. You've got a month and a week or two left, and you should be done in two weeks. Instead, you've still got the ABSOLUTE hardest part of AP Biology ahead of you (human body systems). Evolution and ecology are a breeze though.

Do not underestimate human body systems. Collectively, it's about as hard as the rest of AP Biology put together :(.

LOOP OF HENLE! CIRCULATORY SYSTEMS! ACTIN AND MYOSIN! They still give me nightmares =/.

I got a 5 though, so I guess it was worth it :). AP Bio builds character in terms of work habits.

Oh yeah, advice wise, there's not much to give. Just read, and read a lot. What text book are you using? I used the Campbell-Reece-Mitchell book, and it's good. I'd try to finish the rest of biology before the last week of April, and then try reviewing for about two weeks.

Yeah. I took physiology this year hoping that it would help my ap class..but I learned nothing. Absolutely nothing. Yeah. I think I can breeze through evolution and ecology pretty quickly, it's human body systems and plants that are stressing me. Two weeks to review? That's all? And should I just keep working through books and practice tests?

The AP Biology teacher here works your butt off. Half the students get 5. Because his tests are so dang hard. And everybody studies like crazy. Last year three stupid people opted to not take the AP exam and instead had to take the teacher's final exam. they all got Fs hehe!

I'm worried about my BC exam (not exactly its more like getting ready for the calc 3 class i'm going to take). Is it a good idea to go through every odd problem in a really good college Calculus book? (because I have already started and am 1/3 of the way done).

I'm not even going to really study for BC. If I can wing it completely and get a 5 while I'm not as good at the six other tests I'm taking, I don't see the point.

But good luck to you :). I use Leithold's calculus textbook. The problems range from easy/basic to pretty difficult proofs at the end. You can give it a shot if you want. If you want a "preview," PM me and I'd be happy to type out a few of the harder questions at the end to see if you like them.

rlee, you are taking 6 AP exams?! Did you study independently for some of them? At my school, we can only fit in at most 4 AP classes per year because of the way we do our schedules? What exams are you taking-- 6 full year exams or some half year exams?

I take all the courses at school. I go to school an extra hour per day (zero period, go to school at 7:00 instead of 8:00).

I am taking:
Computer Science AB
Calculus BC
Physics C Mechanics
Physics C E&M
Language and Composition
Art History
U.S. History

So yeah, they are all year-long courses. In addition to the six classes I take Mandarin 3 :).

My school only offers two of the classes on your list, and I am not allowed to take AP English until senior year.


Να βρεθεί πόσα χρήματα κερδίζει κάποιος σε μια εβδομάδα δουλεύοντας μέρα και νύχτα ,αν κερδίζει 5 ευρώ κάθε φορά που οι δείκτες του ρολογιού σχηματίζουν γωνία 90 μοιρών ?

Από ορισμό μέτρου γωνιακής ταχύτητας στην ομαλή κυκλική κίνηση παίρνουμε: ω = φ/t ή φ=ωt, αλλά ω=2π/Τ οπότε φ=2πt/Τ . Αν μετά από χρόνο t ο ωροδείκτης έχει διαγράψει γωνία φ1 και ο λεπτοδείκτης γωνία φ2, τότε θα ισχύει φ1=2πt/Τω και φ2 = 2πt/Τλ . O ωροδείκτης εκτελεί πλήρη περιστροφή σε Τω = 12h, άρα έχει περίοδο 12h, ενώ ο λεπτοδείκτης εκτελεί μία πλήρη περιστροφή σε χρόνο 1h, άρα έχει περίοδο Tλ = 1h. Αν φ=π/2 θα έχουμε: π/2=φ2 - φ1 ή π/2 = (2π/Τλ - 2π/Τω)t ή π/2 = 2π[(Τω - Τλ)/Τλ*Τω]t ή t = (π/2*Τλ*Τω/2π(Τω-Τλ) = ... = 3/11h . Έτσι για 3/11h κερδίζει 5 euro. Με την ίδια λογική βρίσκουμε για όλοκληρη τη μέρα και στη συνέχεια για όλη την εβδομάδα..

Ωραία mostel

 αλλά παρέλειψες τη περίπτωση φ = 3π/2  !

 Γίνε λίγο πιο σαφής για ολόκληρη τη μέρα .

φυσική λυκείου εγώ θα το έλεγα... έχω συναντήσει κάτι παρόμοιο στο σχολικό βιβλίο

ποτέ δε  μου άρεσαν αυτές οι ασκήσεις με τα αναλογικά ρολόγια... μια φορά έγινα ρεζίλι μπροστά σε όλη την τάξη γιατί θεώρησα ότι ο ωροδείκτης κάνε μια πλήρη περιστροφή κάθε 24 ώρες :blush:  τι να κάνω, προτιμώ τα ψηφιακά!!

Σε χρόνο $t$ ωρών, ο ωροδείκτης διαγράφει γωνία $30t$ (σε μοίρες), ενώ ο λεπτοδείκτης γωνία $360t$. Εμείς θέλουμε να βρούμε τις χρονικές στιγμές $t$ για τις οποίες ισχύει $360t-30t \equiv 90 \pmod{360}\Rightarrow 11t \equiv 3 \pmod{12}\Rightarrow t=\frac{12k+3}{11}$ για κάποιο $k \in \mathbb N$ και επειδή η εβδομάδα έχει $7\cdot{24}=168$ ώρες θα ισχύει επίσης και $0\leq t\leq 168 \Rightarrow 0\leq \frac{12k+3}{11}\leq 168 \Rightarrow-\frac{1}{4}\leq k\leq \frac{615}{4}\Rightarrow^{k \in \mathbb N}0\leq k\leq 153$. Άρα, οι ζητούμενες χρονικές στιγμές μέσα στην εβδομάδα είναι στο πλήθος $154$.
Συνεπώς, ο τύπος του προβλήματος θα εισπράξει συνολικά $154\cdot5=770$€.

Στέλιος...

P.S. Κι'αυτή σε βιβλίο δημοτικού τη βρήκες;; :huh:  Μ'αρέσει πάντως αυτή η ξαφνική στροφή του forum προς τις ασκήσεις του δημοτικού... :P !

Τέλεια Στέλιο

 αλλά σου "έφυγε" η περίπτωση

                   360t - 30t = 270 (mod360)

 οπότε του κλέβεις τα μισά λεφτά του εργαζόμενου !!!!!

Σωστός..., ελπίζω να με συγχωρήσει ο εργαζόμενος  :P , ο οποίος τελικά θα πάρει 1540€ (και λίγα είναι αν λάβουμε υπόψιν μας ότι δουλεύει 24 ώρες τη μέρα επί μια εβδομάδα...  :huh: ).

[quote="stedes"]τελικά θα πάρει 1540€ (και λίγα είναι αν λάβουμε υπόψιν μας ότι δουλεύει 24 ώρες τη μέρα επί μια εβδομάδα...  :huh: ).[/quote]

24 ώρες * 7 ημέρες= 168 ώρες
168/8= 21 ημέρες=21/5= (περίπου) 4 εβδομάδες= 1 μήνας

άρα το ποσό που θα πάρει αντιστοιχεί στο ποσό που θα έπαιρνε αν δούλευε 8ώρο για ένα μήνα

τώρα το αν αυτό είναι λίγο... προϋποθέτει ορισμό του λίγου

φτάσαμε στο νηπιαγωγείο?

[u]Συνοδική περίοδος[/u] δύο πλανητών ονομάζεται το χρονικό διάστημα που μεσολαβεί μεταξύ δύο συνόδων των πλανητών.
Λέμε ότι έχουμε [u]σύνοδο[/u] δύο πλανητών όταν οι δύο πλανήτες και ο ήλιος βρίσκονται σε ευθεία γραμμή.

Ισχύει $\frac{1}{S}= \frac{1}{E}-\frac{1}{P}$, όπου:

$E,P:$ Οι περίοδοι περιφοράς των δύο πλανητών όπου $E<P$.
$S:$ Η συνοδική περίοδος

http://elvis.rowan.edu/astronomy/observe/cop.htm (στο τέλος της σελίδας)

Ο λεπτοδείκτης και ο ωροδείκτης μπορούν να αντικατασταθούν από δύο πλανήτες με περιόδους $E=1h$ και $P=12h$ αντίστοιχα.

Τότε: $\frac{1}{S}= \frac{1}{1}-\frac{1}{12}= \frac{11}{12}$

Επομένως η περίοδος συνάντησης θα είναι $\frac{12}{11}h$. Μέσα σε 12 ώρες θα έχουμε 11 συναντήσεις.

Η διαφορά είναι ότι θέλουμε να βρούμε πόσες φορές γίνονται κάθετοι.
Μ' άλλα λόγια, αν φανταστούμε τους δύο (νοητούς) δείκτες οι οποίοι είναι συνεχώς κάθετοι στον ωροδείκτη, ψάχνουμε να βρούμε πόσες φορές αυτοί οι δύο δείκτες θα συναντηθούν με το λεπτοδείκτη. Επειδή οι δύο νοητοί δείκτες έχουν κι αυτοί περίοδο 12 ώρες, προκύπτει ότι μέσα σε 12 ώρες ο λεπτοδείκτης θα συναντήσει 11 ώρες τον καθένα, το σύνολο 22, άρα 308 σε μια εβδομάδα, που σημαίνει 1540€

Einai dinaton oi $a, b$ na einai aritoi  kai o $a^{b}$ na einai
ritos

Ναι γίνεται ........ :wink:
Έστω $a$  ένας φυσικός όχι τέλειο τετράγωνο .

Αν ο $\sqrt{a}^{\sqrt{a}}$  είναι ρητός τελειώσαμε .
Αλλιώς $\sqrt{a}^{\sqrt{a}^{\sqrt{a}}}=a$     :wink:   ρητός.

Σιλουανέ θα ήθελα να μου δείξεις τις πράξεις με τις οποίες έβγαλες τα 2 αυτά συμπεράσματα


There is an odd number of points in 3d-space. Prove that there exists a plane passing  through one of them which divides  the space in two parts both containing the other points in equal amounts.

[hide="i think..."]We can turn this into a 2d problem. Then we must have the initial condition that no three of the points lie on the same line. Choose a point $ A_{1}$. Start with the line that passes through two of the points, say $ A_{1},A_{2}$, in such a way that all other points of the set lie on one side of $ A_{1}A_{2}$, where $ A_{2}$ is a point such that no $ A_{i}\ (i\neq 1,2)$ lies on $ A_{1}A_{2}$. Then move the line so that it has $ A_{2}$ on one side and all other points on another side, but the line always goes through $ A_{1}$.Then move again so that it has $ A_{2}$ and $ A_{3}$ on one side and all other points on the other side. Continue this process until we reach the situation when half of the points lie on one side of the line and the other half lies on the other side.[/hide]

EDIT: some gaps filled in. sorry i was in a rush.

we take $ S$ the set of these points with $ |S|=2n+1$
there exists a plan $ P$ where we can project the 2n+1 points orthogonally knowing that two points do not have the same projection.
now we draws a line $ L$ which is not parrallel with any line of the lines of $ S$ (the lines passer by 2 points of $ S$) such that all the points of $ S$ are on the left of the line $ L$, we move the line $ L$ from left to right with uniform speed, and we calls $ t_{i}$ the $ i^{th}$ moment when a point of $ S$ will be on $ L$.
there will be $ 2n+1$ moment because $ L$ cannot contain 2 point of $ S$ at the same time.at the moment $ t_{n+1}$ there we stoppe $ L$. there will be $ n$ points on the right and $ n$ points on the left and one point in $ L$. and our plan which checks the condition of the exercise the plan orthogonal with P passing by $ L$ stopped at the moment $ t_{n+1}$.

sorry for my bad english.

[quote="aviateurpilot"]...
sorry for my bad english.[/quote]

Yes,  this is almost the same as my own.
Firstly reduce the problem to 2dim. Drow lines through any 2 points of the set. As there exist only finite number such lines there exist a line which is parallel to none of them. Take a plane P which is orthogonal to this chosen line and project all the points A to A' belonging to P. The projections will be all distinct.
Now chose a line in the P which is parallel to none of line passing through 2 projection A' and B' of the points and chose an orthogonal line L. The projections A'' of projections A' of points A on line L will be distinct. As all A'' belong to L and they consist odd amount we can take those point B'', which separates all the other points A'' into 2 equal parts. Now the plane passing through B'' and orthogonal to L is those we are seeking.
It is evident by induction that the problem is true for arbitrary dimension.

P.S. I was ispired by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=155321]extremal geometry[/url]


let a,b,c>0:abc=1.prove that
\sum\frac\{a^3c}{(b+c)(c+a)}\geq\frac\{3}{4}

You are a spamer? :P
[quote="rem"]After clearing denominators we obtain:
$ 4\sum_{cyc}(a^4c \plus{} a^3cb)\ge 3(\sum_{cyc}(a^2b) \plus{} \sum_{cyc}(a^2c) \plus{} 2abc)$ multiply $ LS$ by $ 1 \equal{} a^{\frac {2}{3}}b^{\frac {2}{3}}c^{\frac {2}{3}}$
$ \Leftrightarrow 4\sum_{cyc}(a^4c) \plus{} 4\sum_{cyc}(a^3cb)\ge 3\sum_{cyc}(a^{\frac {8}{3}}b^{\frac {5}{3}}c^{\frac {2}{3}}) \plus{} 3\sum_{cyc}(a^{\frac {5}{3}}c^{\frac {8}{3}}b^{\frac {2}{3}}) \plus{} 6a^{\frac {5}{3}}b^{\frac {5}{3}}c^{\frac {5}{3}}$
$ \Leftrightarrow \sum_{cyc}(a^4c \plus{} a^3bc \plus{} b^4a \minus{} 3a^{\frac {8}{3}}b^{\frac {5}{3}}c^{\frac {2}{3}}) \plus{} \sum_{cyc}(a^4c \plus{} a^3bc \plus{} c^3ab \minus{} 3a^{\frac {8}{3}}b^{\frac {2}{3}}c^{\frac {5}{3}}) \plus{} (a^4c \plus{} b^4a \plus{} c^4b \minus{} 3a^{\frac {5}{3}}b^{\frac {5}{3}}c^{\frac {5}{3}}) \plus{} (a^3bc \plus{} b^3ca \plus{} c^3ab \minus{} a^{\frac {5}{3}}b^{\frac {5}{3}}c^{\frac {5}{3}})\ge 0$  (*)
Each of the expressions in the brackets is greater than or equal to $ 0$ by AM-GM and therefore inequality (*) is true, hence the original inequality is true.[/quote]


Prove that:
$\tan 5x < \tan 4x + 2\tan 3x,\forall x \in \left( {0,\frac{\pi }{{12}}} \right)$

I am pretty sure that this isin't "Getting Started" material  ;)

[hide="hint"]
tan (x+y) = (tan x + tan y) / (1 - tan x tan y)
[/hide]

I'm not sure how hard this is so I'm moving this to Intermediate.

Hmm...he posted this twice...once in getting started and once in intermediate.  I'm deleting the other post in here.


ratio
geometry
cyclic quadrilateral
geometry unsolved

Given $\triangle ABC$ inscribed $(O)$ find locus $M$ such that
$a\frac{MA}{MA'}+b\frac{MB}{MB'}+c\frac{MC}{MC'}=a+b+c$ with $A'$ is intersection of $MA$ and $(O)$ $B',C'$ similar!

We have that $MA\cdot MA'=\rho(M)=R^{2}-OM^{2}$.So we have to find the locus such that $aMA^{2}+bMB^{2}+cMC^{2}=(a+b+c)(R^{2}-OM^{2})$.Now we have $\sum aMA^{2}=(a+b+c)MI^{2}+abc$ so we have $MI^{2}+MO^{2}=constant$ so the geometrical locus is a circle centered at the midpoint of OI.Noting that O is also in the geometrical locus it is the circle of diameter OI :)[/list]

yes thank you for your proof  :) maybe you are interested in this problem related it, with suppose $M$ belong to circular disk with diameter $OI$
prove that
$S_{A'B'C'}\ge \frac{27Rr^{2}}{a+b+c}$

It should be $ \frac{27Rr^{2}}{ (a+b+c)}$ :D

ok you are right mateivld  :D I have fixed it can you post your proof  :D

We have from your first problem $\rho(M)(a+b+c)=\sum aMA^{2}$.it is easy to prove $\frac{S_{A'B'C'}}{S_{ABC}}=\frac{\rho(M)^{3}}{MA^{2}MB^{2}MC^{2}}$.Now we have $(\sum aMA^{2})^{3}\geq 27abc MA^{2}MB^{2}MC^{2}$.Now we are done.(QED) :lol:

[color=darkblue]The initial problem is well-known generally : [b]Let[/b] $m,n,p$[b] be three fixed real numbers for which[/b] $m+n+p=1$ [b]and let[/b] $ABC$ [b]be a given triangle inscribed in the circle[/b] $w$. [b]For a mobile point[/b] $L$ [b]define the second intersections[/b] $A',B',C'$ [b]between the circle[/b] $w$ [b]and the lines[/b] $LA$, $LB$, $LC$ [b]respectively. Ascertain the geometrical locus of the point[/b] $L$ [b]for which[/b] $m\cdot\frac{LA}{LA'}+n\cdot \frac{LB}{LB'}+p\cdot\frac{LC}{LC'}=1\ .$

[b]Question.[/b] Can you extend this problem to a given cyclic quadrilateral $ABCD$ ? A particular case is at http://www.mathlinks.ro/Forum/viewtopic.php?t=71732 .[/color]

Dear Virgil Nicula mateivld! I think we can extend this problem!
From your first problem, if these ratios $\frac{LA}{LA'},\frac{LB}{LB'},\frac{LC}{LC'}$ is ratio of geometric length($>0$ then $m,n,p$ should positive,if they are algebraic length then we can do with any $m,n,p \in R$ and our problem is finding locus of $m\frac{\overline{LA}}{\overline{LA'}}+n\frac{\overline{LB}}{\overline{LB'}}+p\frac{\overline{LC}}{\overline{LC'}}=1$ then result is $L\in$ circle with diameter $OI$ with $I$ is point such that barycentric is $(m,n,p)$

For your general problem:
Given circle (O) and $n$ point $A_{i}, i=\overline{1,n}$ on it find locus $\sum_{i=1}^{n}a_{i}\frac{\overline{LA_{i}}}{\overline{LA_{i}'}}=1$ with $\sum_{i=1}^{n}a_{i}=1,a_{i}\in R$we use $\overline{AB}$ to show algebraci length os segment $AB$ then locus is circle with diameter $OI$ with $I$ is the fix point such that
$\sum_{i=1}^{n}a_{i}\vec{IA_{i}}=\vec{0}$ here we can using this formula like leibnitz formula
$\sum_{i=1}^{n}a_{i}MA_{i}^{2}=(\sum_{i=1}^{n}a_{i})MI^{2}+\sum_{i=1}^{n}a_{i}IA_{i}^{2}$  and $\overline{LA}.\overline{LA'}=\rho (L)/(O)$:)  :)


Let there be $ x,y,z$ positive real numbers such that $ x\plus{}y\plus{}z\equal{}1$.Prove that:
\[ \sqrt{1\plus{}\frac{x}{yz}}\plus{}\sqrt{1\plus{}\frac{y}{zx}}\plus{}\sqrt{1\plus{}\frac{z}{xy}} \geq 6\]

[quote="Vrajitoarea Andrei"]Let there be $ x,y,z$ positive real numbers such that $ x \plus{} y \plus{} z \equal{} 1$.Prove that:
\[ \sqrt {1 \plus{} \frac {x}{yz}} \plus{} \sqrt {1 \plus{} \frac {y}{zx}} \plus{} \sqrt {1 \plus{} \frac {z}{xy}} \geq 6
\]
[/quote]

[b]First proof:[/b]
By Minkowski inequality, we have:

$ \sqrt {1 \plus{} \frac {x}{yz}} \plus{} \sqrt {1 \plus{} \frac {y}{zx}} \plus{} \sqrt {1 \plus{} \frac {z}{xy}} \geq \sqrt{3^2\plus{}\left(\sum \sqrt{\frac{x}{yz}}\right)^2}$ (1)

By AM-GM inequality:

$ \sum \sqrt{\frac{x}{yz}} \ge 3\sqrt[6]{\frac{1}{xyz}} \ge 3\sqrt[6]{\frac{27}{(x\plus{}y\plus{}z)^3}} \equal{}3\sqrt{3}$ (2)

From (1) and (2), we are done.

[b]Second proof:[/b]

Using AM-GM inequality:

$ 1\plus{}\frac{x}{yz} \ge 4\sqrt[4]{\frac{x^3}{(3yz)^3}}$

So:

$ \sqrt {1 \plus{} \frac {x}{yz}} \plus{} \sqrt {1 \plus{} \frac {y}{zx}} \plus{} \sqrt {1 \plus{} \frac {z}{xy}} \ge 2\sum \sqrt[8]{\frac{x^3}{3^3y^3z^3}} \ge 6\sqrt[24]{\frac{1}{3^9x^3y^3z^3}} \ge 6$

[b]Third solution:[/b]

From the hypothesis and Cauchy-Schwarz inequality, we get that:

$ 1\plus{}\frac{x}{yz} \equal{}1 \plus{} \frac{x(x\plus{}y\plus{}z)}{yz} \equal{} \left(\frac{x}{y}\plus{}1\right)\left(1\plus{}\frac{x}{z}\right) \ge \left(\sqrt{\frac{x}{y}}\plus{}\sqrt{\frac{x}{z}}\right)^2$

So $ \sqrt {1 \plus{} \frac {x}{yz}} \plus{} \sqrt {1 \plus{} \frac {y}{zx}} \plus{} \sqrt {1 \plus{} \frac {z}{xy}} \ge \sum \left(\sqrt{\frac{x}{y}}\plus{}\sqrt{\frac{x}{z}}\right) \ge 6$ (by AM-GM inequality)


Set $ A$ contains all of the factors of 45, and set $ B$ contains all of the factors of 36. What is the product of the elements in $ A\cap B$?

The factors of 45 are 1, 3, 5, 9, 15, and 45. The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. The numbers in common are 1, 3, and 9. These multiplied together gives you $ \boxed{27}$


In acute triangle $ABC$, let $D$, $E$, and $F$ be the feet of the altitudes.  Let $P$ and $Q$ be points on $DF$ and $EF$, respectively, such that such that $\angle PAQ = \angle DAC$.  Prove that $AP$ bisects $\angle FPQ$.

Well this solution is not short so let's start: First notation:

Triangle $ABC$ foot of altitude from $A$ is $D$ from $B$ is $E$ and from $C$ is $F$.
Orthocenter of triangle $ABC$ is $H$ Let $AP \cap FE=M$,$AQ \cap CF=N$ and $AP\cap
CF=X$, and last but not the least $QX\cap AB=R$. I will show that $X$ is incentre of
$\triangle FPQ$ and that will be enough. So first

$Lemma$ Orthocenter $H$ is incenter for triangle $DEF$

Proof: This lemma is trivial, we have $\angle FDH=\angle FBH\equiv \angle FBE=\angle
FCE \equiv \angle HCE= \angle HDE$. It's the same thing for all other angles.

$\angle PAQ=\angle DAC$ so from that $\angle PAD=\angle QAC\equiv \angle NAC$ and
$\angle FDA\equiv PDA=\angle FCA\equiv \angle NCA$ thus $\angle FPA=\angle FNA$. So
from that we get $PNAF$ is cyclic quadraliteral so $\angle PNF= \angle PAF$ and
$\angle NPA=\angle NFA=90^o$.
\newline
$\angle XFQ=\angle HFQ=\angle HFD=$ (from lemma) $=\angle CFD=\angle DAC=\angle
PAQ=\angle XAQ$ so $\angle XFQ=\angle XAQ$ thus quadraliteral $XQAF$ is cyclic, so
$\angle FAX=\angle FQX$, and $\angle XQA=180^o-\angle XFA=90^o$. Now we have $\angle
NPX=90^o=\angle XQA$ so quadraliteral $NQXP$ is cyclic and from that $\angle PQX=
\angle PNX=\angle PAF=\angle XQF$ so $\angle PQX=\angle XQF$ and $\angle PFX=\angle
XFQ$ because of lemma so $X$ is incentre of triangle $PQF$ thus $AP$ is bisector of
$\angle FPQ$

That's it, ask if something is not clear or I made some mistakes....

$P.S.$ Interesting note is that points $N,P$ and $R$ (last but not the least) are collinear!!

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=1170 .

  darij

Well from my solution you can get a lot of other stuf like already told $R,P,N$ collinear, then $XQ||BE$ and I can't remember the others at the moment but I know that I notice really a lot of interesting sutff.
So mine is better :D  :D :D

[i]Well ,
I think I have solved this problem between 1980 and 1990. It was an IMO - proposed problem from Russia(short or long list).
After so many years I tried to solve the problem again.I think that what follows , is a simple (not easy)and clear solution.(please make a good figure)[/i]

 Let $AP$ and $CF$ inersect at $M$ and the line $AQ$ inersects $CF$ at $N$( In my figure I have $AB<AC$).Since
         $\angle PAQ = \angle DAC = \angle EFC = \angle CFD$ and $\angle AFE= \angle DFB$, we' ll have

    [color=red]a)[/color] $AFMQ$ is insribed so $\angle AFQ = \angle AMQ$

    [color=red]b)[/color] $AFPN$is inscribed , so $\angle ANP = \angle PFB = \angle QFA = \angle QMA$  , which means that $NPMQ$ is also inscribed. Now we are finished , because

      $\angle MPQ = \angle MNQ = \angle FNA = \angle FPA$

[u]   Babis[/u]


I was wondering, which background do you use?

Currently, I am using Mathlinks, but for over a year, I used AoPS (primarily because I did not know you could change it to Mathlinks, and I signed up on AoPS).  Also, the MathLinks background currently has a calendar that does not work properly, so it is better  :D

Please state which you use and explain WHY you use which ever one you use.

I use AOPS. I find it nice. more colorful...
The main reason why I use it is because I have not tried using Mathlinks yet...

I use MathLinks, because what I use of AoPS is the community part, not he other things that buying books or that, so it's more directly to me to use the MathLinks one

I just use AoPS, because I am use to it and I am taking a class, so it is much easier to access it through AoPS. I still think that MathLinks is cool though.

i use AOPS because it makes me happy.  :D   :D

MathLinks.  It's easier to read, loads faster (no Flash Proof-Without-Words : I have a [i]very[/i] slow connection), and has a nice grey color-scheme.  I like it.  :)

Mathlinks. It seems more user friendly.  Also, the new web server thing looks better on mathlinks.

I use the AoPs one because that was the first website I saw.

I use AoPS because it was the one I first started using, and I've never bothered to change...

I use artofproblemsolving because it was the one I was originally introduced to. It's the one I've gotten used to.

Yeah, same with me. I'm just more used to AoPS and am too lazy to try MathLinks. Besides is there really much of a difference anyway?

I use AoPS because it's magically delicious

Actually, it is a lot different.  I forgot to add the option both, so I will just vote for Mathlinks right now (unless I can edit and add an option).

I use AoPS..........I'm rather addicted to those flash things on the left explaining some concepts in a really nice way.... :D

Yeah, I love those little movies.  Now, I switched back to AoPS because you can't access AoPSWiki from Mathlinks, and that is a nice feature.

Off topic, but thanks for telling me about AoPS wiki.......... Here of course, I guess the accuracy of things will be much better than at wikipedia   :lol:


Here is the IMO shortlist problems 1995, I got it from internet.


function
linear algebra
linear algebra unsolved

Find all $f$:$M(n,C)\rightarrow M(n,C)$ such that for any $A,B \in M(n,C)$
wich commute $AB=BA$ we have $f(A+B)=f(A)f(B)$

Moubinool, do you have a solution for this one or is it just a question?

There are monsters lurking in the answer of this one; there's a whole lot more than just the exponential here.

Exactly, this is why I asked, since I found a very large class of functions and I wonder if one can characterize them all.


[hide]Observing this in $ \mod 4$, we see that

$ x \equal{} 4n \plus{} 1$

$ y \equal{} 4m \plus{} 3$

Don't know if it helps...
[/hide]

:huh: In the thread xpmath pointed out, I find it somewhat funny that neither t0r or any of the other beasts noticed that there is another integer solution so its impossible to prove! (the other integer solution is (3,-5)) :rotfl:

Solve the equation
http://www.mathlinks.ro/viewtopic.php?t=247252


If $\sin{(\cos x)} = \cos{(\sin x)}$, then $\sin^2{(\cos x)} = \cos^2{(\sin x)}$,

$\frac{1 - \cos{(2\cos x)}}{2} = \frac{1 + \cos{(2\sin x)}}{2}$

$- \cos{(2\cos x)} = \cos{(2\sin x)}$

$2\cos x = \pi - 2\sin x$

$\sin x + \cos x = \frac \pi 2$

$\sin x \cos{\frac \pi 4} + \cos x \sin{\frac \pi 4} = \frac{\pi \sqrt 2}{4}$

$\sin{\left(x + \frac \pi 4\right)} = \frac{\pi \sqrt 2}{4} > \frac 1 4 \cdot \frac{25}{8} \cdot \frac{14}{10} = \frac {35}{32} > 1$

which is impossible.

Here is a slight extension of this problem:

Prove that Cos(SinX) >Sin(CosX) for all X

I think we can see that from Psycho's proof of ur first question. obviously since the lower bound of $\cos(\sin{x})$ is greater than the upper bound of $\sin(\cos{x})$, $\cos(\sin{x}) > \sin(\cos{x})$.

[quote="Michael Niland"]Prove that the equation Sin(CosX) = Cos(SinX) has no real solutions.[/quote]

does it have any solutions?? (complex, i mean)

[quote="Psycho"]$-1 \leq \cos {x} \leq 1$
$\Rightarrow -\frac {\pi}{2} < \cos {x} < \frac {\pi}{2}$
$\Rightarrow -\frac {\sqrt2}{2} < \sin ({\cos {x}}) < \frac {\sqrt2}{2}$

$-1 \leq \sin {x} \leq 1$
$\Rightarrow -\frac {\pi}{2} < \sin {x} < \frac {\pi}{2}$
$\Rightarrow \frac {\sqrt2}{2} < \cos ({\sin {x}})$

$\therefore \cos ({\sin {x}}) = \sin ({\cos {x}})$ is impossible[/quote]I don't agree with this. $\sin\cos0>\frac{\sqrt{2}}{2}$ and $\cos\sin\frac{\pi}{2}<\frac{\sqrt{2}}{2}$. We can say $-\sin{1}\le\sin\cos{x}\le\sin{1}$ and $\cos\sin{x}\ge\cos{1}$ but this doesn't lead to the result immediately as your inequalities would have.

the problem is very easy if i hadn't made any mistake :P

let $\alpha=\sin x$, $\beta=\cos x$.

we have to prove that $\sin \beta=\cos \alpha$ has no real solns for $\alpha^2+\beta^2=1$.

we have ${\cos (\frac{\pi}{2}-\beta})=\cos (\pm \sqrt{1-\beta^2})$

$\frac{\pi}{2} - \beta \mp \sqrt{1-\beta^2}=2n\pi$

=> n is 0 or 1

use $\sin \beta +\cos \beta \leq \sqrt{2}$ (well known) and finish it off nicely

as shown above, the inequality proposed is not true(try plotting)


limit
real analysis
real analysis unsolved

[hide="solution"]Here's a sketch, some details are skipped.

Let $ b_n \equal{} \sqrt {n}a_n$, then $ b_{n \plus{} 1} \equal{} \sqrt {\frac {n \plus{} 1}{n}}b_n\left(1 \minus{} \frac {1}{n}b_n^2\right)$, and clearly $ b_n > 0$ for all $ n$.  The only positive fixed point of the map $ \sqrt {\frac {n \plus{} 1}{n}}x\left(1 \minus{} \frac {1}{n}x^2\right)$ is $ x \equal{} \sqrt {n \minus{} \frac {n\sqrt {n}}{\sqrt {n \plus{} 1}}}$.  As $ n\to \infty$ we have $ \frac {n\sqrt {n}}{\sqrt {n \plus{} 1}} \equal{} n \minus{} \frac {1}{2} \plus{} \mathcal{O}\left(\frac {1}{n}\right)$, and therefore $ b_n\to \frac {1}{\sqrt {2}}$ as $ n\to \infty$.[/hide]

[quote="tom-nowy"]$ 0 < a_{1} < 1 \\
 \\
a_{n \plus{} 1} \equal{} a_{n} ( 1 \minus{} a_{n}^{2} ) \\
 \\
\lim_{n \to \infty} \sqrt {n} a_{n} \equal{} ???$[/quote]

First, we realize that  $ a_{n\plus{}1} \le a_n \  \Longrightarrow \  \lim_{n \to \infty} a_n\equal{}0 \  \Longrightarrow \lim_{n \to \infty} \frac{1}{a_n^2}\equal{}\infty$

This means :

$ \lim_{n \to \infty} n.a_n^2\equal{} \lim_{n \to \infty} \frac{n}{\frac{1}{a_n^2}}\equal{}\lim_{n \to \infty} \frac{1}{\frac{1}{a_{n\plus{}1}^2}\minus{} \frac{1}{a_n^2}}\equal{}\lim_{n \to \infty} \frac{(1\minus{}a_n)^2}{2\minus{}a_n^2} \equal{}\frac{1}{2}$

We get the result : $ \boxed{\lim_{n \to \infty} \sqrt {n} a_{n} \equal{} \frac{\sqrt{2}}{2}}$

This is what Polya and Szego say in their book

exercise num.174, p.38, Problems and Theorems in Analysis, I

Assume that $ 0 < f(x) < x$ and $ f(x) = x - ax^k + bx^l + x^l\varepsilon(x),$ $ \lim_{x\to0}\varepsilon(x) = 0,$ for
$ 0 < x < x_0$ where $ 1 < k < l$ and $ a,b$ both positive. The sequence $ x_n$ defined by $ x_{n + 1} = f(x_n)$ satisfies
$ \de \lim_{n\to\infty} n^{1/(k - 1)} x_n = (a(k - 1))^{ - 1/(k - 1)}$

[quote="tom-nowy"]$ 0 < a_{1} < 1$.... [/quote]


How many different ways can you select a block of five adjacent squares on a 8x8 chessboard?

Two squares are adjacents if they have a common side.

(the answer would be 49 for a 3 x 3 board )

Don't make double posts in different sections!!!


algebra
polynomial
quadratics
number theory unsolved
number theory

Given $a \in \mathbb{Z}$ such that $\forall x \in \mathbb{Z}, \exists (b,c) \in \mathbb{Z}, x^{2}+a=b^{2}+c^{2}$, show that $a$ is a square.

I proposed this problem quite long time on the forum. It also appeared in Mathematical Reflections. I don't think I saw a complete solution until now. It is one of the most particular cases of a theorem of Davenport: if $f(n)$ is a sum of two squares for all $n$ and $f$ is an integer polynomial, then it is the sum of the squares of two polynomials with rational coefficients.

Yeah. Actually I haven't started thinking about it yet because it seemed difficult and I wanted a second opinion. I definitely forget about it for now  :D Thanks for the reference, by the way.

You shouldn't say that, it is not so hard. Actually, it is almost solved on the forum.

It's not that hard if we're allowed to use Dirichlet's Theorem on primes in arithmetic progressions. Using it (and quadratic reciprocity), one can prove that unless $a$ is a square, we can find a prime $p$ of the form $4k+3$ such that $\left(\frac{a}{p}\right)=-1$, or, in other words, $p|f(x)=x^{2}+a$ for some $x$. One of $f(x),f(x+p)$ is not divisible by $p^{2}$, so assume it's $f(x)$. Now the prime $p=4k+3$ appears in the factorization of $f(x)$ with an odd exponent (its exponent is $1$, actually), so by Fermat's Theorem characterizing the numbers which can be written as sums of two squares, $f(x)$ is not such a sum.

Yep, that's it. Anyway, I really doubt about a proof not usind Dirichlet's theorem. Now, who wants to make another step: polynomials of degree 3?


inequalities
LaTeX
Cauchy Inequality
IMO Shortlist
IMO

:blush: I am ususally very glad when I can't solve problem and have possibility to view the solution, but sometimes when I solve problemsform shortlist I see that the solution is veru complicated, and then I think that it would be better if people could see my solution too, what I can do for that???

for example to be sure that I am not joking see solution of 1967 shortlist, question 5(Bulgaria) and you'll see what I mean, solution is very long, right? are you interested how I solved it, here you are:
 adding we get x^2+y^2+z^2=3
 taking ones to enother side and multiplying x*y*z=1 or x=y=z=-1, we check (-1,-1,-1) is a solution and we go on
3=x^2+y^2+z^2>=3*cubicrootfrom 1=3 (cauchy inequality), so x=y=z=1 enother solution (1,1,1) and finished, that's it, we don't need any fighting and timespending to solve it!!!

alright, bye, waiting for reply...

P.S. generally all solutions are 7 points, but I like shorter ones:)

Thanks for your interest. You can easily post your solution in the respective IMO Shortlist threads shown in the resources section. But in order to make me your solution added to the resource section itself I insist on FULL solutions written in $\LaTeX$!

Click on the problem number in the resources/contests section and then, once arrived in the topic of discussion, post a reply with your solution.


trigonometry
inequalities
geometry proposed
geometry

Let $A$, $B$ and $C$ be angles of a plane triangle. Show that:

\[1 < \cos A + \cos B + \cos C \leq \frac{3}{2}.\]

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

Using that $\cos{x}+\cos{y}=2\cos{\frac{x+y}{2}}\cos{\frac{x-y}{2}}$, $\cos{x}=1-2\sin^{2}{\frac{x}{2}}$
we get $ \cos{A} +\cos{B}+ \cos{C}=4\sin{\frac{A}{2}}\sin{\frac{B}{2}}
\sin{\frac{C}{2}}+1$. Putting this into our inequality we obtain $0<\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}\leq\frac{1}{8}$. Left side is obvious.
Now right side. After Am-Gm and then Jensen we get
 ${\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}\leq\left(\frac{\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}}{3}\right)^{3}}\leq\sin^{3}{\left(\frac{A+B+C}{6}\right)}=\frac{1}{8}$.
QED.


My favorite one was definitely that hexagon problem (#27 or so?)

Do you remember the question?, because alanchou and I are not agreeing with each other on the answer. I think he misremembered.

[quote="xpmath"]It was: You have an equiangular hexagon, with sides 1, 5, 4, 2, in that order (I think that was it, not sure if I remembered right).

The answer was 9; you basically extend pairs of sides to form a big equilateral triangle, and then it's easy.[/quote]

It was 1, 7, 4, 2. The way I did it was to draw line segments that intersect the unknown sides from the known sides.

Oh, ok.
Yeah, so that was my favorite problem.

I remember trying to do that in my head since I couldn't write, but now that I can write, I really like the problem.  Here is a picture to help out.

two of the people on my team totally guessed this problem right (by randomly writing down an answer) :lol:

[quote="CatalystOfNostalgia"]Extention: ABCDEF is an equilangular hexagon. Prove that

|AB-DE|=|BC-EF|=|CD-FA|.

This can be done with complex numbers as well as the method outlined above.[/quote]

Everyone who participated in WOOT had a ridiculously unfair advantage on that problem.

YAY!!!

i got that right in a minute seconds w/o woot

I guessed on that question with like 3 secs left. I assumed it would be the sum of the four sides divided by two. I should have gotten 7 but I added wrong and got 8! If only I had added wrong and gotten 9...

The known sides of the equiangular hexagon were 1, 7, 2, 4. in order.  (This was one of my favorite problems, too!)

My observation was that the 'slant' distance between the parallel sides is independent of the lengths of the parallel sides.  (See illustration, below.)

This leads to a new rule, which should be added to everybody's Tips and Tricks file:  "The sum of any two adjacent sides of an equiangular hexagon is equal to the sum of the opposite two sides."

I know this is a tangent, but can that be proven?

For this problem I got it completely wrong... got a fraction with a square root. I had a 0% of getting that right, considering it didn't say anything like " express your answer as a simplified root" or "express your answer as a common fraction"
 :rotfl:


The first $ 2007$ positive integers are each written in base $ 3$. How many of these base-$ 3$ representations are palindromes? (A palindrome is a number that reads the same forward and backward.)

$ \textbf{(A)}\ 100 \qquad \textbf{(B)}\ 101 \qquad \textbf{(C)}\ 102 \qquad \textbf{(D)}\ 103 \qquad \textbf{(E)}\ 104$

\[2007=2202100_{3}\]
Thus we can count the number of $1,2,3,4,5,6$ -digit palindromes without issues.
Clearly, there are $2,2,6,6,18,18$ of each; indeed, if there are $2n+1$ digits then the palindrome is of the form: [u]a[/u] [u]b[/u] [u]c[/u] ... [u]k[/u] [u]l[/u] [u]k[/u] ... [u]c[/u] [u]b[/u] [u]a[/u]
Then there are 2 ways to choose the first digit (it can't be 0) and $3^{n}$ ways to choose the digits in spots $2$ to $n+1$ inclusive, which define all palindromes. So $2\times 3^{n}$.
If there are $2n+2$ digits, then we have a number of the form: [u]a[/u] [u]b[/u] [u]c[/u] ... [u]k[/u] [u]k[/u] ... [u]c[/u] [u]b[/u] [u]a[/u]
hence there are still exactly $2\times 3^{n}$ palindromes, since a bijection can be formed by dropping out one of the two middle numbers of this palindrome to reach the next lowest odd digit palindrome.

Now for the case of $7$ digits:
First note that $2007=2202100_{3}>2202022_{3}$. Thus, for each number from $1000_{3}$ to $2202_{3}$ there is a corresponding palindrome. There are exactly $1210_{3}=48_{10}$ such numbers.
Total: $2+2+6+6+18+18+48=100$

I used a similar method as me@home..
I remember this from one of the NYML.. that $2007=2202100_{3}$
if its $1$ digits, or [u]a[/u] you can have $2$ cases.
$2$ digits, or [u]a[/u] [u]a[/u], you have $2$ cases.
$3$ digits, or [u]a[/u] [u]b[/u] [u]a[/u], where a is not necessary distinct from b, you have $2*3=6$ cases.
$4$ digits, [u]a[/u] [u]b[/u] [u]b[/u] [u]a[/u] $2*3=6$ cases
$5$ digits, [u]a[/u] [u]b[/u] [u]c[/u] [u]b[/u] [u]a[/u] $2*3*3=18$ cases
$6$ digits, [u]a[/u] [u]b[/u] [u]c[/u] [u]c[/u] [u]b[/u] [u]a[/u] $2*3*3=18$ cases
$7$ digits, [u]a[/u] [u]b[/u] [u]c[/u] [u]d[/u] [u]c[/u] [u]b[/u] [u]a[/u] $2*3*3*3=54$ cases

BUT, you overcounted $2210122,2211122,2212122,2220222,2221222,2222222$, which is $6$ cases.

Therefore, tha snwer is $2+2+6+6+18+18+54-6=\boxed{100}$

A!

[quote="me@home"]

Now for the case of $7$ digits:
First note that $2007=2202100_{3}>2202022_{3}$. T[b]hus, for each number from $1000_{3}$ to $2202_{3}$ there is a corresponding palindrome.[/b] There are exactly $1210_{3}=48_{10}$ such numbers.
Total: $2+2+6+6+18+18+48=100$[/quote]

Can you explain that part?
thanks.

The last 3 digits are dependent on the first 4

For example, if I have 2102---, then since it must be a palindrome, the number must be 2102012.

Thus, every 7-digit palindrome can be represented by its right-most 4 digits, or by its left-most 4 digits.

It's more meaningful to look at the leftmost because then you get a sense of what's larger than what.


Let be given three circles $(O_{1}), (O_{2}), (O_{3})$. Denote by $\alpha_{ij}, \beta_{ij}$ the two common internal tangents of the circles $(O_{i}), (O_{j})$.
Prove that: $\alpha_{ij}$ $(i\neq j \in \{1,2,3\})$ are concurrent if and only if $\beta_{ij}$ $(i\neq j \in \{1,2,3\})$ are concurrent

This is the Mannheim theorem, mentioned in http://www.mathlinks.ro/Forum/viewtopic.php?t=3680 and http://www.mathlinks.ro/Forum/viewtopic.php?t=80364 . I'm leaving this topic open for a complete solution to be posted ;)

  darij

Let's call $A_{ij}$ the common point of $\alpha_{ij}$ and $\beta_{ij}$; suppose that the alpha-tangents concurr in $P$ and let $Q$ be the intersection of two of the beta-tangets, e.g. $\beta_{12}$ and $\beta_{13}$.
We want to demonstate that $QA_{23}$ is tangent to $(O_{2})$ and $(O_{3})$; proceeding by contraddiction, suppose that the tangent to $(O_{2})$ through $Q$ is $QB$ with $B$ on $PA_{23}$.
Then $(O_{1})$ is tangent to the lines $A_{13}P$, $A_{12}P$, $A_{13}Q$. $A_{12}Q$; considering the quadrilateral $A_{13}PA_{12}Q$, we see that the following relation holds (similar to the one holding in a circumscriptible quadrilateral) : \[A_{13}P+A_{13}Q=A_{12}P+A_{12}Q\] Also, $(O_{2})$ is tangent to the lines $A_{12}P$, $PB$, $BQ$, $QA_{1 2}$, therefore \[A_{12}P+A_{12}Q=BP+BQ\] Then \[A_{13}P+A_{13}Q=BP+BQ\] and from this relation we can derive the existence of a circle touching the lines forming the quadrilateral $A_{13}PBQ$, but this circle must be $(O_{3})$; in particular $BQ$ is tangent to $(O_{3})$, so is a common internal tangent of $(O_{2})$ and $(O_{3})$, but there exist only two such tangents and they cross in $A_{23}$, therefore $B=A_{23}$.


I think it is assume that $R_{n}<\frac{1}{1000}$
(thats what calc teacher said at least)

This isn't really Pre-Olympiad.  Anyway, the error term is $\frac{ \left( \frac{t^{2}}{2}\right)^{n+1}}{(n+1)!}< \frac{1}{1000}$.  Solving for $n$ is difficult without being given a range of $t$...


One day me and my friend were bored in class, so we started playing tic tac toe. After a while, it got pretty boring, so we moved up to 3 diminsional tic-tac-toe. Since 3d tic-tac-toe is played on a 3x3x3 cube, we turned the cube into 3 3x3 squares. You could get tic-tac-toe normally, and any two cells that would be adjacent in 2d are adjacent in two adjact 3x3 squares. After a couple of games of this, we realized that if we had 3 sets of 3 3x3 squares, we would have 4d tic-tac-toe. Just use the conversion of 2d to 3d for this, and you can see what an awesome game this is. Me and my friend usually use a 4x4x4x4, because with the 3x3x3x3 its too easy to force a win. We have even gone up to 6d tic-tac-toe, which takes forever to play, but involves a lot of strategy. If anyone needs anymore explanation on 4d tic-tac-toe just say so.

3D tic-tac-toe is really stupid. The first player to go can force a win. I mess with people on MSN with that.

I expect 4d to be the same way. The first player can force a win.

Well, in 4-d can't the first player just look at a 3-d 'cross section' and force a win there? if the other player ignores him (i.e. plays on another cross section), the first player wins, and otherwise we already know he can force a win.

If that is the case, wouldn't n-dimensional tic tac toe be a forced win for teh first player for n>2?

That's why we went up to a 4x4x4x4 hypercube instead of a 3x3x3x3.

[quote="k00lperson"]ya, maybe try playing with 3 players :D[/quote]

I just said that....

Another discussion on this topic can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=300186]here[/url].


geometry
inradius
inequalities
area of a triangle
Heron\u0027s formula
triangle inequality
algebra

[quote="April"]The inradius of triangle $ ABC$ is $ 1$ and the side lengths of $ ABC$ are all integers. Prove that triangle $ ABC$ is right-angled.[/quote]

We have that $ S=p.r$ where $ S$ is area, $ p$ is semiperimeter and $ r$ is inradius of a given triangle. From Heron's formula we have:

$ S=\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)}$

Thus

$ 1=r=\frac{S}{p}=\sqrt{\frac{\left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)}{\left(\frac{a+b+c}{2}\right)}}$

or

$ \left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=\frac{a+b+c}{2}$

From triangle inequality we have that $ a+b>c$ but since sides of the triangle are integers, we can write $ a+b=c+x$ where $ x$ is positive integer. Similarly, we get that $ b+c=a+y$ and $ c+a=b+z$, with $ y$ and $ z$ positive integers. Also we have that $ x+y=2b$, $ y+z=2c$ and $ z+x=2a$ thus $ x,y,z$ are all odd or all even.

Form equation above we have:

$ \left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=\frac{a+b+c}{2}$ is equivalent to $ \frac{xyz}{8}=\frac{x+y+z}{2}$

or $ xyz=4(x+y+z)$

Since $ x,y,z$ all cannot be odd hence they are all even. We get equation

$ x_{1}y_{1}z_{1}=x_{1}+y_{1}+z_{1}$

We can easily get (maybe it is well known) that the only solution in positive integers of this equation is $ (x_{1},y_{1},z_{1})=(1;2;3)$ and all possible permutations. Hence we get that $ (x,y,z)=(2,4,6)$ and solving the following system of equations

$ a+b=c+2$
$ b+c=a+4$
$ c+a=b+6$

we get that sides of a triangle are $ (a,b,c)=(3;4;5)$ but these are the sides of a right triangle.

I like this problem and your solution  :wink:

[quote]We can easily get (maybe it is well known) that the only solution in positive integers of this equation...[/quote]

Can you show me your proof please ?

But are you sure isn't

$ \left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=(\frac{a+b+c}{2})^{2}$ ??

[quote="zweig"]
[quote]We can easily get (maybe it is well known) that the only solution in positive integers of this equation...[/quote]

Can you show me your proof please ?
[/quote]

$ a+b+c=abc \Rightarrow (c-1)(ab-1)+(a-1)(b-1)=2$

After checking three cases we get solutions.

[quote="zweig"]
But are you sure isn't

$ \left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=(\frac{a+b+c}{2})^{2}$ ??[/quote]

There shouldn't be a square. After squaring both sides of the equation $ S=r.p$ we get:

$ \left(\frac{a+b+c}{2}\right)\left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=\left(\frac{a+b+c}{2}\right)^{2}$

and after cancelling $ \frac{a+b+c}{2}$ we get the result.

This problem was also proposed to 2006 Catalunya Mathematical Olympiad (problem 3):

http://www.xtec.es/recursos/mates/aqui/olimp42/olimp42.pdf

in english, the problem says

3. Prove that the unique triangle with integer sides and such that area is equal to semiperimeter, is the one with sides 3,4 and 5.

 :)

The double area of this triangle is equal to $ pr \equal{} p \equal{} a\plus{}b\plus{}c$, where a,b,c are the sides. So, S=p. After substituting $ a\plus{}b\minus{}c \equal{} x$, $ b\plus{}c\minus{}a \equal{} y$ and $ a\plus{}c\minus{}b \equal{} z$, by Heron we get: $ x\plus{}y\plus{}z \equal{} xyz$, where $ x,y,z$ are all positive integers. Wlog, $ x\leq y\leq z$. Then, $ xyz \equal{} x\plus{}y\plus{}z\leq 3z$, or $ xy\leq 3$, where $ x,y\geq 1$. From here we get that the sides of this triangle are 3,4 and 5. As $ 3^{2}\plus{}4^{2}\equal{} 5^{2}$, this triangle is right angled,
                                                                               Q.E.D.

The area is equal to the semiperimeter, so the semiperimeter is an integer. By Heron's, this is equivalent to:
$$\sqrt{s(s-a)(s-b)(s-c)}=s\Leftrightarrow (s-a)(s-b)(s-c)=s.$$
Let $x=s-a,y=s-b,z=s-c$.  Note that $\{x,y,z\}\subseteq\mathbb Z$, and $xyz=x+y+z$. WLOG, assume that $x\ge y\ge z$, then $xyz\le3z$, so $yz\le3$.

[b]Case 1:[/b] $yz=1$
Then $x=x+y+z$ which is absurd.
[b]Case 2:[/b] $yz=2$
Then $x=y+z=3$ since $y=2$ and $z=1$ must hold. These together imply $s=x+y+z=6$, so $(a,b,c)=(3,4,5)$ and permutations, which are right-angled.
[b]Case 3:[/b] $yz=3$
Then $2x=y+z=4$, so $x=2$, a contradiction with our ordering assumption.


Find the smallest positive integer $ x$ so that the fraction below represents a fraction whose decimal equivalent terminates. \[ \frac1{10\plus{}x}\]

In order for a base $ 10\equal{}2\cdot5$ fraction to terminate, we must have the number composed entirely of $ 2$'s and $ 5$'s (do you see why?), and the smallest $ x$ occurs when we have $ \frac1{10\plus{}\boxed6}\equal{}\frac1{16}\equal{}0.0625$.


inequalities
inequalities proposed
romania

Let $ a,b$ and $ c$ be three positive real numbers with $ a\plus{}b\plus{}c\equal{}3.$ Prove that:

$ \frac{a\plus{}3}{3a\plus{}bc}\plus{}\frac{b\plus{}3}{3b\plus{}ca}\plus{}\frac{c\plus{}3}{3c\plus{}ab}\geq 3.$

[quote="Maxim Bogdan"]Let $ a,b$ and $ c$ be three positive real numbers with $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:

$ \frac {a \plus{} 3}{3a \plus{} bc} \plus{} \frac {b \plus{} 3}{3b \plus{} ca} \plus{} \frac {c \plus{} 3}{3c \plus{} ab}\geq 3.$[/quote]
$ \sum \frac {a \plus{} 3}{3a \plus{} bc} \equal{} \sum \frac {a \plus{} 3}{(a \plus{} b)(a \plus{} c)}$.
We need to prove: $ \sum \frac {a \plus{} 3}{(a \plus{} b)(a \plus{} c)}\ge 3$.
Indeed, the inequality above is equivalent to:
$ \sum (a \plus{} 3)(b \plus{} c)\ge 3(a \plus{} b)(b \plus{} c)(c \plus{} a)$
$ \Longrightarrow \sum (2a \plus{} b \plus{} c)(b \plus{} c)\ge 3(a \plus{} b)(b \plus{} c)(c \plus{} a)$, $ (*)$.
Denote $ (p,q,r) \equal{} (a \plus{} b \plus{} c,ab \plus{} bc \plus{} ca,abc)$
$ (*)\Rightarrow 2p^2 \plus{} 2q \plus{} 3r\ge 3pq$ or $ 6p^2 \plus{} 6q \plus{} 9r \equal{} 2p^3 \plus{} 9r \plus{} 2pq\ge 9pq$.
But $ 2p^3 \plus{} 9r\ge 7pq$ (this result can be proved easily by [b]Schur[/b] and [b]AM-GM[/b])
Therefore $ 2p^3 \plus{} 9r \plus{} 2pq\ge 7pq \plus{} 2pq \equal{} 9pq\Longrightarrow (*)$ is [b]true[/b], which leads to the result of the problem.
Our proof is completed then :wink:

Another approach (Just [b]AM-GM[/b])
$ LHS \equal{} \sum \frac {2a \plus{} b \plus{} c}{(a \plus{} b)(a \plus{} c)}$
$ \equal{} \sum \frac {(a \plus{} b) \plus{} (a \plus{} c)}{(a \plus{} b)(a \plus{} c)}$
$ \equal{} 2\sum \frac {1}{a \plus{} b}\ge \frac {18}{2(a \plus{} b \plus{} c)} \equal{} \frac {18}{6} \equal{} 3$.
Then our proof is completed :D

Hello Maxim, do you have the other problems from Romanian Junior Balkan TST?
Thx

Let $a,b,c > 0$ and $a+b+c=3.$ Prove that
$$\frac{a+k}{3a+bc}+\frac{b+k}{3b+ca}+\frac{c+k}{3c+ab} \ge  \frac{3(k+1)}{4}$$
Where $k\geq 1.$


Persons A, B, and C are to participate in a three-corner pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. When it is their turn, they are to fire at whomever they choose and they take turns in the order A, B, C cyclically. But a hit man no longer takes turns, and is no longer shot at. It continues until only one man is left unhit. What should A's strategy be?

What happens if you change the order of their turns to A, C, B?

This is an awesome problem. Did you get it from [i]Duelling Idiots and Other Probability Puzzlers[/i] by Paul Nahin?

No. It came from a really cheap dover book called Fifty Challenging Problems in Probability with Solutions by Frederick Mosteller. But I recognize a lot of these problems as being rather famous. It's a neat book though. I got it for like 5 bucks.

hmm.. shouldn't A fire at B? If A hits B, he has a .5 chance of being shot by C. If A hits C, he automatically is out next turn since he's B's only target. If A misses, it doesn't matter who he fired at. But B would presumably fire at C, since he has 100 percent accuracy he only has to survive until his second turn to win, and he would have a better chance of doing so if he eliminated C.

The neat thing about this puzzle is that there are more than two options for person A. I shall say no more right now.

I suppose A could fire at himself and wait for B to eliminate C. But since it would logically make more sense for B to kill C regardless of whoever A fired at, I don't understand how that strategy would be to A's advantage...

I think [hide] he should shoot B first because of 314159265358979323846's reasoning. From there on, we have to assume B would shoot C before A. He could shoot A instead of B as a retaliation, so there's no definite telling of whom he's going to shoot.[/hide]

Yeah, A should shoot at B. I had B and C mixed up.

Well I was imagining of a Wild Wild West shootout, because no one would play a "game" where you can get killed.
It's not like their using water guns. They're using real guns. So I doubt they can shoot at someone else not involved in this "game."

Oops. I messed up my source. It actually came from [i]Fermat's Enigma[/i] by Simon Singh. (Thanks to Mindspa for pointing that out.)

The problem in the form given in the book:

A truel is similar to a duel, except there are three participants rather than two. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a conflict by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting his target on average only one time in three. Mr. Gray is a better shot, hitting his target two times out of three. Mr. White is the best shot, hitting his target every time. To make the truel fairer Mr. Black is allowed to shoot first, followed by Mr. Gray (if he is still alive), followed by Mr. White (if he is still alive), and around again until only one of them is alive. The question is this: Where should Mr. Black aim his first shot?

Solution:

Let us examine Mr. Black's options. First, Mr. Black could aim at Mr. Gray. If he is successful then the next shot will be taken by Mr. White. Mr. White has only one opponent left, Mr. Black, and as Mr. White is a perfect shot then Mr. Black is a dead man.

A better option is for Mr. Black to aim at Mr. White. If he is successful then the next shot will be taken by Mr. Gray. Mr. Gray hits his target only two times out of three and so there is a chance that Mr. Black will survive to fire back at Mr. Gray and possibly win the truel.

It appears that the second option is the strategy that Mr. Black should adopt. However, there is a third and even better option. Mr. Black could aim into the air. Mr. Gray has the next shot and he will aim at Mr. White, because he is the more dangerous opponent. If Mr. White survives then he will aim at Mr. Gray because he is the more dangerous opponent. By aiming into the air, Mr. Black is allowing Mr. Gray to eliminate Mr. White or vice versa.

This is Mr. Black's best strategy. Eventually Mr. Gray or Mr. White will die and then Mr. Black will aim at whoever survives. Mr. Black has manipulated the situation so that instead of having the first shot in the truel, he has first shot in the duel.

I had deduced this line. However, in the form given by gauss202, "Mr. White" takes the second turn, not the third.

Why does that matter though? The same strategy reduces a truel to a duel with Mr. Black having the first shot. It is possible, however, that a duel against Mr. Gray is so much better than a duel against Mr. White that it's worth the risk of giving up the first turn and shooting Mr. White to have a better chance of duelling Mr. Gray.

Gee.

The probability of losing a duel vs. B when you have the first shot is 7/10, whereas the probability of losing a duel vs. C when you have the second shot is the sum of the infinite series 1/2 + 1/2*7/10*1/2 + ... = 10/13 > 7/10. So I guess you really do want a duel with B. A close shave.

You screwed it up! The probabilities are 1/3, 2/3, and 1. Or did you purposefully do that to simplify the problem?

-Isaac

The one I got was from [i]Fermat's Enigma[/i]. Gauss202 got a different version of the problem from somewhere else.


The Chapter Competition in my area took place yesterday (2/7), and I assisted with the grading and certificates, etc. There is an incident that I wish to have clarified; those wanting me to get to the point can skip to the bottom. The middle is background information and details.

I suspected something was up when two things happened with the winning school (which will remain anonymous for obvious reasons): first, all their tests were in a separate pile, and also their scores were unusually high given their performance in the countdown round. However, I reserved any critical judgement since at the time I could not make much sense out of these abnormalities; moreover, the countdown round and the individual rounds are not strongly correlated.

Well, after the competition I was informed that this school had taken the Chapter test before coming to the competition site, so that at the competition they took a "mock test" (else they would be bored!!). They justified it as thus: one of their team members had a religious event to attend to on that Saturday, so he could not come. Therefore, all the tests had to be administered beforehand for him to participate. If he were to take the test prior to the actual competition, all his team members would have to as well to avoid unfair content leaks.

I am confounded by this line of reasoning: firstly, if the person had a conflict on this day, he should not be on the team at all; that's what alternates are for. I gave up my position on the states team 4 years ago despite being the top scorer (the reason is an entirely different matter that I do not wish to elaborate on), and I never requested the test to be taken earlier. Second, how could there be any assurance that all the rules were strictly followed, and in particular that none of the students recieved any help? I am cautious to not falsely accuse anyone, but given the personality of their coach, I would not be surprised if their coach extended the time or helped the students with the problems. Unlikely, but possible.

The final question is: [i]can a school take the test remotely and have their scores count against the rest of the teams who take the test at the competition site? [/i] If this were to be allowed, Mathcounts' credibility would be seriously hurt, as then it potentially becomes a contest between coaches.

Taken from the 2008-2009 School Handbook:

 A student who is unable to attend a competition due to religious observances may take the written portion of the competition up to one week in advance of the scheduled competition.
In addition, all competitors from that school must take the exam at the same time.

That is a strange rule. It leaves a lot of leeway for cheaters.

Wow... that's complicated.. I don't think that is allowed! Taking the test remotely? Doesn't one have to take a test with like 50 people?

About the alternates...what if, in a particular school, there was a certain event that wiped out [i]all of the participants[/i]? Surely you're not going to get some *below exceptional students* to take the test.

In fact, this was the exact case last year for my school, and if we had not been able to do Chapter, that would have been devastating. :wink:

No rules can be enforced perfectly--many things are on the honor code strictly. Plus, in a sense, those early competitors could have a disadvantage: less preparation time, and in the case of official countdown, and automatic forfeit.

[b]jonathanchou711:[/b] They are the [url=http://www.claymath.org/millennium/]Millenium Problems[/url]. :)

[quote="jonathanchou711"]I just have a random question. Sorry guys. I'm trying not to spam, but this is a question about aops. What are the stars by the username? Like the Birch and Swinsadas;dasdfaalskdfj and the P v NP??[/quote]

they are how many posts you have.
the more stars you have the more posts you have.

The team is allowed to take the test up to a week before the competition if anyone has a religious observence on the day of the test.

You can appeal to the national board if you wish, but it won't help for this year.

Also, everyone here please read my thread regarding protests. Read and join us in protesting!

Yep, students can take the test early (and remotely) if they have a religious conflict--but it has to be for that reason.  Coaches can't just randomly decide to give the test early because they feel like it.  I've seen it done many times, as Minnesota has a Jewish school that used to make state on a regular basis; they would always take the state tests early.  To do it, they have to get permission from the competition coordinator, who also arranges to have the test proctored.  In particular, it wasn't just their own coach proctoring the test!

Incidentally, this is part of the reason why official countdown rounds at chapter and state are required to use the ladder format.  The idea is that if you have a conflict and can't attend the countdown, you automatically forfeit all countdown matches.  In a bracket format, that can drop you quite a few places, but in a ladder format, you drop exactly one place.

Do we know if early test-takers are [i]required[/i] to arrange for a proctor, like in the case of the Jewish school that Sly Si mentioned? In particular, in 236factorial's case, did the coach act as a proctor, or was there a separate person who administered the test?

Another question: what is to prevent early test-takers from divulging test information to others who have not yet taken the test? Even mentioning the difficulty of the test is considered divulging information, isn't it?

As I said earlier, a lot depends on people being fair and honest. That's the basis of many things.

[quote="BOGTRO"]The team is allowed to take the test up to a week before the competition if anyone has a religious observence on the day of the test.

You can appeal to the national board if you wish, but it won't help for this year.
[/quote]

First of all thanks to everyone who pointed out the rule to me (I have never read the handbook, as you can guess).

And I cannot tell you any more information than what I have, so as I said, whether or not the team did everything fairly will be completely unknown.

A formal protest would be going a little far, for several reasons: first, I cannot be sure about my suspicions, and second, the team would have won anyway. Besides, I am not too closely connected with any of the teams (I assist with the 2nd place team only) so I would not be the one to be involved. I just wanted clarification of the relevant procedures.

As math154 said, the entire system is rooted in honesty. However, I do not doubt that there are people who take advantage of the rules to advance themselves. Sad, but after reading much literature, you learn that humans are inherently evil..

PS: what religious holiday is on 2/7 anyway?

Teams that take the test early are no different than teams that take chapter on the first or second or third weekends.  They are on their honor not to divulge any info about the test.  You have to trust in the integrity of the people and the coaches that are in such a situation.

[quote="ThinkFlow"]Do we know if early test-takers are [i]required[/i] to arrange for a proctor, like in the case of the Jewish school that Sly Si mentioned? In particular, in 236factorial's case, did the coach act as a proctor, or was there a separate person who administered the test?[/quote]
Sounds to me like arranging to take the test early is entirely up to the coordinator, so he or she would be the one to arrange a proctor.  So the school would have no control over whether the test is proctored, and by whom.

[quote="236factorial"]PS: what religious holiday is on 2/7 anyway?[/quote]
The Sabbath.  Jews who observe it in its strictest form are not allowed to travel by motor vehicle, use electronic devices (such as calculators), or even write.  All of these things are rather crucial to competing at Mathcounts.

Folks, a formal protest about this will in all likelihood go nowhere.  It sounds like the school followed the rules, and there is no reason to suspect them of cheating other than the fact that they won.


ratio
geometric sequence
algebra unsolved
algebra

1) FOr a geometric sequence t(7)=5x+2 and t(10)=x-23. If the common ratio is 2, find the numerical value of t(10).

2) The sum of an artihmatic sequence is -40. If t(9)=16 then determine the common difference.

3) THe sum of the terms of a sequence is represented by S(n)=n^s+4n, >=1, n(belong in)N. Determine t(n).

Could really use the answers to these.

1) $\frac{t_{10}}{t_{7}}=2^{10-7}=2^{3}=8$ $\longrightarrow$ $x=-1$ and $t_{10}=-24$.
2) what do you mean by "The sum of an artihmatic sequence..."  the sum of how many terms of it? for example if the sum of first $9$ terms is $-40$ then: $-40=S_{9}=\frac{9}{2}(2t_{1}+8d)$ where $d$ is the common difference. and because $t_{9}=t_{1}+8d=16$ so $t_{1}=\frac{-164}{9}$ and $d=\frac{308}{72}$.
3)$t_{n}=s_{n}-s_{n-1}$ and $t_{n}=n^{s}-(n-1)^{s}+4$

am I right? I mean was this what you wanted? :maybe:


Let $ 0,1,2$ be the states representing north America, the middle east, and eastern Europe respectively.

(a)
$ \left(\begin{array}{ccc}0 & \frac {1}{2} & \frac {1}{2} \\
\frac {1}{3} & \frac {1}{3} & \frac {1}{3} \\
\frac {1}{4} & \frac {1}{4} & \frac {1}{2}\end{array}\right)$

(b) $ p_{00}^{(2)} \equal{} \frac {7}{24}$

For (c) I hope the following is what required
$ \lim_{n\to \infty}P(X_{n} \equal{} i) \equal{} \pi_i, i \equal{} 0,1,2$
where $ \pi \equal{} [\pi_0, \pi_1, \pi_2]$ is obtained by solving $ \pi \equal{} \pi P$ and $ \sum_{i}\pi_i \equal{} 1$, because it is a finite state irreducible chain, it is positive recurrent.
And one gets $ \pi_0 \equal{} \frac {2}{9}, \pi_1 \equal{} \frac {1}{3}, \pi_2 \equal{} \frac {4}{9}$

Please someone  indicate if anything wrong.


Please don't post repeat! http://www.mathlinks.ro/Forum/viewtopic.php?t=151899  :wink:


I would like to know what a good stochastics text would be. I know some measure theory, on the other hand, I would like one maybe with pictures or a lot of motivating material without being too abstract, which these texts tend to be. I have began Doob's Stochastic Processes and it's too abstract for now.

Or is there some papers that are good?

Thanks in advance,

Los

Maybe Oksendal's book could do the trick
According to me he "bypasses" a number of technical point to focus on the core the subject

"Stochastic Differential Equations: An Introduction with Applications" (Universitext)


I am looking for a book has the past AIME problems or a similar book for preparing the AIME. :maybe:

AoPS has a lot of old AIME problems.  AoPS Volume 2 is probably the best book for it.

Consider Contest Problem Book V, by Berzsenyi and Maurer, published by the Mathematical Association of America.  Available through the AoPS Bookstore, the MAA Bookstore, and other usual outlets.

Also consider the 21st Century Contests compact disk, from the AMC.  It has all AIMEs and solutions, 2001-2007.

Steve Dunbar
MAA Director, American Mathematics Competitions

Is Contest Problem Book V too much harder than the AIME test?

No, it's actually just AIME and AHSME tests from 1983-88, if I recall correctly.

Since no one has posted the direct link yet...

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45


Two small carriages are connected by a thin elastic string of unstretched length of 96 cm and of force constant of 18 N/m which at first lies loosely on the ground. One of the carriages is being pulled by a constant velocity of 1.2 m/s.
[color=blue](a)[/color] How long does it take the carriage of mass m=2 kg to catch up with the carriage in front of it after the string has straightened?
[color=blue](b)[/color] Determine the work needed.
[color=blue](c)[/color] What is the distance covered by the second carriage?

In the (inertial) system connected with the 1st carriage, the 2nd carriage has the initial velocity $-v_0 = -1.2\ \text{m/sec}$ and it is subject to the force $F(x) = -kx$ with the potential $V(x) = \frac 1 2 kx^2$, $k = 18\ \text{N/m}$, which results in simple harmonic motion

$x(t) = A \sin (\omega t + \phi_0)$

with the amplitude A, initial phase $\phi_0$, angular velocity $\omega = \sqrt{\frac k m}$ and the period

$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac m k} = 2 \pi \sqrt{\frac{2}{18}} = \frac{2 \pi}{3}\ \text{sec}$

From the initial position $x_0 = 0$ and the initial velocity $-v_0 = -1.2\ \text{m/sec}$, we get $\phi_0 = \pi$, i.e., $x(t) = -A \sin \omega t$ and

$A = \frac{v_0}{\omega} = \frac{v_0 T}{2 \pi} = \frac{1.2}{3} = 0.4 \ \text m$

After half a period, the velocity of the 2nd cartriage has the same magnitude as the initial velocity, but opposite direction, and the force is turned off. The 2nd carriage then travels the length of the string $L = 0.96\ \text m$ to catch up the 1st carriage at rest in the time interval

$\triangle t = \frac{L}{v_0} = \frac{0.96}{1.2} = 0.8\ \text{sec}$

The total time to catch up is

$\frac T 2 + \triangle t = \frac \pi 3 + 0.8 \doteq 1.85\ \text{sec}$

The work needed is equal to the final kinetic energy of the 2nd carriage in the system connected with the ground, i.e., $A = \frac 1 2 m(2v_0)^2 = 2mv_0^2 = 2.88\ \text J$. The distance covered by the 2nd carriage (in the system connected with the ground, I assume) is equal to the distance covered by the 1st carriage in the catch-up time plus the length of the string:

$D = v_0 \left(\frac T 2 + \triangle t\right) + L = \frac{v_0 T}{2} + 2L = 0.4 \pi + 1.92 \doteq 3.18\ \text m$


geometry
circumcircle
incenter
cyclic quadrilateral
geometry solved

G2. Let ABCD be a cyclic quadrilateral. Let O be its circumcenter and M be the intersection of its diagonals. Let AB, CD intersect in E and AD, BC in F. Prove that MO is perpendicular to EF.

[Moderator edit: Replaced "incenter" by "circumcenter" to make the problem make sense.]

We can use 2 well known results:

1 The line EM is the angular polar of point F with respect to angle AED (this is true for any complete quadrilateral), and so is line FM for point E, with respect to angle AFB.

2 for a cyclic quadrilateral the angular polar of the pt of intersection of 2 opposite sides with respect to the angle formed by the other 2 sides is the same as the polar of that point with respect to the circle

We apply 2 for the 2 angular polars at 1 and we get EM the circular polar of F and FM the circular polar of E (circular polar=polar with respect to the circumscribed circle). This means that M, which is the point of intersection of the 2 polars, must be the pole of the line formed by the poles of the lines (FM and EM), which are F and E respectively, so M is the pole of line EF with respect to the circle. Because the circle has center O, this means that OM perpendicular to EF (this is also known about circular polars).

Oh, by the way, cyclic quadrilateral means a quadrilateral inscribed in a circle which is its circumcircle, so the center of this circle is called circumcenter. Incenter is the center of the circle inscribed in the circumscribed quadrilateral (your problem 3).

Thanks for posting these. I love old-fashion geom!  :D

[quote="grobber"]2 for a cyclic quadrilateral the angular polar of the pt of intersection of 2 opposite sides with respect to the angle formed by the other 2 sides is the same as the polar of that point with respect to the circle[/quote]

How do we prove this?


geometry
incenter
circumcircle
geometry unsolved

In an acute-angled triangle $ABC$, we are given that $2\cdot AB = AC + BC$. Show that the incenter of triangle $ABC$, the circumcenter of triangle $ABC$, the midpoint of $AC$ and the midpoint of $BC$ are concyclic.

[hide="Here is a short, complete and metrical solution."]
[color=darkblue][b]Proof.[/b] Denote the middlepoints $M$ and $N$ of the sides $[BC]$ and $[AC]$ respectively. Remember some remarkable relations : $abc=4RS=4Rpr\ ;$ $IO^2=R^2-2Rr\ ;$ $IC^2=\frac{ab(p-c)}{p}\ .$ Therefore, $\mathrm {cyclic\ }\{O,I,M,N\}$ $\Longleftrightarrow$  $\mathrm {cyclic\ }\{C,O,I,M,N\}$ $\Longleftrightarrow$ $IC\perp IO$ $\Longleftrightarrow$ $IO^2+IC^2=OC^2$ $\Longleftrightarrow$ $R^2-2Rr+\frac{ab(p-c)}{p}=R^2$ $\Longleftrightarrow$ $ab(p-c)=2Rpr$ $\Longleftrightarrow$ $ab(a+b-c)=4Rpr$ $\Longleftrightarrow$ $ab(a+b-c)=abc$ $\Longleftrightarrow$ $a+b=2c\ .$[/color][/hide]
[b]Remark.[/b] See the second note of the my reply $(IG\parallel BC\Longleftrightarrow b+c=2a)$
from http://www.mathlinks.ro/Forum/viewtopic.php?t=82813

Here's a synthetic solution:
Let $A',B'$ be the midpoints of the sides $BC,AC$ and let $D$ be a point on $AB$ such that $AB'=AD$. From the hypothesis it follows that $BA'=BD$. We deduce that $IB'=ID$ and $IA'=ID$, hence $IA'=IB'$. Let $E,F$ be the projections of $I$ on the sides $BC,AC$. It follows easily that the triangles $IFB'$ and $IEA'$ are congruent. Suppose WLOG that $AC\le BC$. It follows that $AC\le AB\le BC$. Now, since $AC\le AB$, we have $CE\le EB$ and hence $CE\le CA'$. Similarly, we obtain $CF\ge CB'$. We conclude that $\angle EIF=\angle A'IB'=180^{\circ}-\angle C$, therefore the quadrilateral $CB'IA'$ is cyclic. Since the quadrilateral $CB'OA'$ is obviously cyclic as well, we are done.


vector
calculus
derivative
calculus computations

If directional derivative of $ \phi \equal{} axy^{2}\plus{}byz\plus{}cz^{2}x^{3}$ at $ (1,2,\minus{}1)$ has a maximum magnitude of $ 64$ in the direction parallel to the Z-axis,find a,b,c.
 :coolspeak:

$ grad\  f |_M \equal{} (4a\plus{}3c,4a\minus{}b,2b\minus{}2c) \equal{} V$. Let $ v \equal{} (\alpha,\beta,\gamma).$
Then $ \frac{\partial f}{\partial v} \equal{} \left(\frac{v}{\|v\|}, V\right)$. If $ \frac{\partial f}{\partial v} \to max$, then $ v$ and $ V$ are colinear:  $ v\equal{}kV,k\in \mathbb{R}.$
Then $ 4a\plus{}3c\equal{}0, 4a\minus{}b\equal{}0,|2b\minus{}2c|\equal{}64$. And we can obtain the answer.


Sa se determine o primitiva pentru functia $f(x)=\frac{1-\ln x}{x^2+ln^2 x},\ x\in R^*_+ .$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=46547

Am impresia ca ai gasit o primitiva instantaneu (oral). De fapt, m-a inspirat o primitiva postata de Caesar Lupu. Incearca o compozitie care sa ascunda aceasta idee si sa aiba si un aspect estetic. Iata doua exemple:
$1.\ \int \frac{x^2+x}{(e^x+x+1)^2}dx$.
$2.\ \int \left( \frac{x}{x\sin x+\cos x}\right)^2dx.$

[quote="Ph-An"]
$A=\int \frac{x^2+x}{(e^x+x+1)^2}dx$.
[/quote]

We work on I where $e^x+x+1$ does not vanishes
$A+\int \frac{e^{x}+1}{e^{x}+x+1}=\int \frac{u'v-uv'}{v^2}$

with $u=x^2+1+xe^x+2x$ and $v=e^x+x+1$

This gives $A=\frac{x^2+1+xe^x+2x}{e^x+x+1}-\ln(e^x+x+1)+C$

[quote="Ph-An"]
$B= \int \left( \frac{x}{x\sin x+\cos x}\right)^2dx$[/quote]

We work on a interval where the denominator does not vanishes

Use $\frac{2\tan(x/2)}{1+\tan^2(x/2)}$ for $\sin x$

and $\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ for $\cos x$

B is of the form $\int \frac{u'v-uv'}{v^2}$
identify u and v we get

$B=\frac{-x+x\tan^2(x/2)+2\tan(x/2)}{2x\tan(x/2)+1-2\tan^2(x/2)}+C$


integration
function
calculus
derivative
real analysis
real analysis unsolved

I need to prove that :

$ (L_p [0,1])* = L_q [0,1]$

I proved for every element in $ L_q$ there is an linear functional $ F_g$ in $ (L_p [0,1])*$ with $ F_g (f) = \int\limits_0^1 f(t) g(t) dt$ .

I need to show for every linear functional $ F$ in $ (L_p [0,1])*$ then there is $ g$ in $ (L_q [0,1])$ s.t :

$ F = F_g$ , there is a hint in the question to choose $ g(t) = F(\chi_{[0,t])}$ , I proved that function in $ (L_q [0,1])$ .But , I stuck how could I prove that $ F(f) = \int\limits_0^1 f(t) g(t) dt$

Any help  :)

If $ F\equal{}F_g$ for some $ g\in L^q$, then $ F(\chi_{[0,t]})\equal{}\int_0^t g(s)\,ds$. Therefore, the function $ G(t): \equal{}F(\chi_{[0,t]})$ that you have should be an antiderivative of the function $ g$ that you want.

Note also that $ F_g(f) \equal{} \int_0^1 f(t) \, dG$ where $ G$ is absolutely continuous with derivative $ g$. Also, the Radon-Nikodym derivative of $ dG$ is $ g$. So this idea can be used in any measure space. I.e. let $ \nu(E) \equal{} F(\chi_E)$ and then conclude $ \nu(E) \equal{} \int_E g \, d\mu$

That's help me , Thx both of u .

I need a nice paper for the dual space of $ l^\infty$ and $ L^\infty$

where I can find them ?

 :)


LaTeX
quadratics
algebra
quadratic formula

m/n x^2+n/m=1-2x

my procedure-take a look
m/n x^2-2x+n/m=1
The product of coefficient of x^2 & the constant is 1
the coefficient of x is -2
so, the middle term can be split into -x &-x
m/n x^2-x-x+n/m=1
x(m/n x-1) (x-n/m)=1
x{(mx-n)/n}{(mx-n)/m}=1
I am stuck at this point :?
How can I continue ?

$ \text{\LaTeX}\text{ed}$ so that I can actually see the problem...

$ \frac {m}{n} x^2 + \frac {n}{m} = 1 - 2x$

$ \frac {m}{n} x^2 - 2x + \frac {n}{m} = 1$

$ \frac {m}{n} x^2 - x - x + \frac {n}{m} = 1$

$ x(\frac {m}{n} x - 1) (x - \frac {n}{m}) = 1$

$ x\frac {(mx - n)}{n}\frac {(mx - n)}{m} = 1$

OK, I'll take a look at your problem now  :lol:

Edit: Well I messed with it a little and I ended up with
$ n^2 + 2mnx - mn + m^2 x^2 = 0$
Which gets weirdly factored into
$ n(n + 2mx) + m(mx^2-n) = 0$
And if we want to make it even messier
$ n(n+2mx)+m(mx+\sqrt{n})(mx-\sqrt{n})=0$
Yeah, I'm getting nowhere...  :huh:

i got somewhere, it's pretty basic.
so m/n*x*x+n/m=1-2x
and you get to m^2*x^2+n^2=mn-2xmn
so m^2*x^2+2xmn+n^2=mn
that goes to (mx+n)^2=mn
and mx+n=sqrt(mn)
and x=(sqrt(mn)-n)/m

shouldn't that be +2x on the left side of the equation?

When you have to factorize something like this expression $ ax^{2} \plus{} bx \plus{} c \equal{} 0$  the shortest  way is to use the quadratic formula so we $ ax^{2} \plus{} bx \plus{} c \equal{} (x \minus{} x_{1})(x \minus{} x_{2})\equal{}0$

$ m^2x^2\plus{}2mnx\plus{}n^2\equal{}mn$

$ (mx\plus{}n)^2\equal{}mn$

If $ mn\ge 0$, we can write  $ (mx\plus{}n)^2\minus{}(\sqrt{mn})^2\equal{}0$, then

$ (mx\plus{}n\minus{}\sqrt{mn})(mx\plus{}n\minus{}\sqrt{mn})\equal{}0$

etc.


The solution at the back of the book for this problem is so messy, I doubt it could be done in 10 minutes. Does anyone know of a fast way?

The numbers from 1 through 8 are seperated into two sets, A and B. The numbers in A are multiplied together to get $a$.The numbers in B are multiplied together to get $b$. The larger of the two numbers $a$ and $b$ is written down. What is the smallest possible number that can be written down using this procedure?

There won't be any way that doesn't involve a bit of trial and error / case checking.
[hide="Reasonably quick"]
1*2*3*4*5*6*7*8 is 40320, square root of that is about 200. You want to find the smallest factor thats bigger than 200. Then its just playing around a bit - 210 is a nice place to start and works, so all you have to do is check that numbers between 201 and 209 don't work. Shouldn't take too long.
[/hide]

(edit: I haven't actually seen the official answer, so hopefully its not this :D)

[quote="EFuzzy"]Does anyone know of a fast way?[/quote]
[hide="Other methods"]
TripleM's suggestion is the key.  One factor of $8!=40320$ will be greater than the square root, while the other factor will be less than the square root.

Of course, you don't extract the square root.  Instead, you find the perfect squares that bound 40320:

$200^2=40000$

$???^2=40320$

$201^2=40401$

A National Competitor should be able to square these numbers with ease.  Then, as TripleM suggests, trial-and-error will get you the answer.  In less than 90 seconds if you are quick, and if you know your primes!  This does depend on luck, that the factor you seek will be close to 201.  This is a fine method, which you should consider.

Alternatively, you could try all possible sets.  Excluding the element, 1.  I count 64 distinct pairs of sets.  Brute force will get you the answer, but it may be too time-consuming for a Sprint Round question.

One more...(since you ask for a quick way):

Since this was a National competition, a few competitors might have noticed that $40401-40320=81$.  The solution follows from the difference of squares.  The factors we seek are $201\pm 9$.  If you happen to see this, consider yourself a bit lucky, or a well-qualified National competitor.[/hide]

You have to get the two numbers that are as close as possible to each other...

[hide="This would take me about two minutes but it took four thanks to my sister q_q"]
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8

Well...

multiply all of the numbers together and you get 40320. Not that nice -_-. However, take the square root of it to get an approximate of what we'll get. No calculator needed - just think. 2002 = 40000, which is close enough to 40000. Actually, 2012 is closer at 40401, but 200's a nice number :P .

Now that you have an idea of the smallest possible number, we go about solving it.

You want four numbers closest to 200 as possible. So, we go about looking for those four numbers.

200 is about 8 x 6 x 4 x 1. That's 48 x 4, or 192, which'll be about as close as you can possibly get. Obviously, there's a larger counterpart to 192, as it's lower than the square root of 4320. 2 x 3 x 5 x 7 = 210, which is the answer.
[/hide]

Whoa, what rcv said was nice. Should've thought of that...

[hide]6*5*7=210 I think is the smallest.[/hide]


$ a,b,c\ge 1$ and $ a\plus{}b\plus{}c\equal{}9$
Prove that:$ \sqrt{\sum ab}\ge \sum \sqrt{a}$

[quote="hxy09"]$ a,b,c\ge 1$ and $ a \plus{} b \plus{} c \equal{} 9$
Prove that:$ \sqrt {\sum ab}\ge \sum \sqrt {a}$[/quote]
You are mean the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105823


quadratics
number theory
relatively prime
algebra unsolved
algebra

Prove that there exists an infinite number of relatively prime pairs ($m$,$n$) of positive integers such that the equation:
\[(x+m)^{3}=nx\]
has three distinct integer roots.

Say that $x = 1$ is a root.  Then $n = (1+m)^{3}$ and the cubic is

$(x+m)^{3}= (1+m)^{3}x$

Since $x = 1$ is a root, we can factor out $x-1$ from the cubic

$x^{3}+3mx^{2}+(-m^{3}-3m-1) x+m^{3}= 0$

To get the quadratic

$x^{2}+(3m+1) x-m^{3}= 0$

The discriminant is

$(3m+1)^{2}+4m^{3}= 4m^{3}+9m^{2}+6m+1 = (4m+1)(m+1)^{2}$

Which is a square infinitely often, particularly, when $m = k^{2}+k$ for some integer $k$.  Then the other two roots are given by

$x = \frac{-(3m+1) \pm (2k+1)(m+1)}{2}$

Which are integer iff $m$ is even; but this is always true.  Hence our infinite solutions are given by

$(m, n) = \left( k^{2}+k, (k^{2}+k+1)^{3}\right), k \neq 1,-2$

Why can you assume that $ x=1$ is a root? Sorry if its a stupid question... :)

Actually, I guessed it.  And it worked  :P

Note that I haven't found all possible pairs $ (m, n)$, but the guess that $ x = 1$ was a root was enough to make finding an infinite family easier.


calculus
integration
function
calculus computations

hello, this leads also to an elliptic integral.
Sonnhard.


Let $ S \equal{} \{1,5,9,13,\ldots\} \equal{} \{ 4k\minus{}3 \mid k \in \mathbb{N} \}$. An $ S$[i]-prime[/i] is a number in $ S$ other than $ 1$ that is not the product of two smaller numbers in $ S$. Show that every element of $ S$ that is greater than $ 1$ is a product of $ S$-primes but that such a factorization is not always unique.

[quote="fearailhold"]Show that every element of $ S$ that is greater than $ 1$ is a product of $ S$-primes[/quote]
It follows from the definition. If it is a $ S$-prime it is obvius that it can be a product of itself. And it is only not a $ S$-prime if it's a product of to other elements from $ S$.

[quote="fearailhold"]but that such a factorization is not always unique.[/quote]
$ 9, 21, 49$ are $ S$-primes. $ 9*49 \equal{} 21*21$


abstract algebra
vector
category theory
linear algebra

Let $A$ be a ring, $E$ an $A$-module and $E^{**}$ its bidual. Determine a canonical homomorphism $c_E: E\to E^{**}$ and prove that it is injective if $E$ is free.

Posted in wrong section. Try Superior Algebra.

Nope. Duals are part of linear algebra. This is correct place for this post.

This is a question about modules, so you could classify it as linear algebra (vector spaces) or abstract algebra (rings).
I don't think it really matters.

it may matter whether or not A is commutative (then you have to say left or right module)

but this is a basic result, map every vector e in E onto the bidual vector $e''$ such that, for all dual vectors f in $E*$ :

$e''(f)=f(e)$

now suppose $E$ is free
let us consider a vector v in the kernel
if $v''=0$ then $v''(f)=0$ for all functionals f and thus $f(v)=0$

but for every nonzero vector I can construct a functional not disappearing on this vector (suppose your element is of form$(v_{1},\cdots,v_{n})$ with $v_{i}$ nonzero, just consider the functional that maps each vector onto the i th component)

so therefore $v=0$

does this help you?

by the way, as far as I know, for finite dimensional vector spaces over fields this map is also surjective
it basically then says that taking the dual an even number of times of a space gives you up to a canonical (and that is important!) isomorphism the same space

this could help you (it helped me) to understand contravariant and covariant tensors