Stanford
college
calculus
integration
function
real analysis
linear algebra

So, I look at the sample demos of the Stanford EPGY online math courses. And the real analysis course. And behold - the explanations that you have to udnderstand look the same as the textbook of a standard calculus course. Besides, most college math courses take several quarters to complete, whereas Stanford EPGY only require one course for each (abstract alg, real analysis, etc..). So are the stanford EPGY math courses just brief versions of what you would get in a college math course? And would one of them ever count for credit for a state university one year sequence of say, abstract algebra or real analysis?

http://epgy.stanford.edu/courses/ma...115lecture.html

...Then again, explanations are supposed to simplify what the textbook is saying. So they tend to be more informal. Still, why not let the textbook be more informal itself?

...

And why in the heck do they split up multivariable calculus into integral and differential???[/code]

That is pretty standard for a basic course in real analysis.  A lot of universities have a first real analysis course similar to that one, whereas a second real analysis course would introduce metric spaces, equicontinuity, maybe fourier series, etc.  It really varies by university, but I imagine that most institutions would give you a quarter's worth of credit for the material that that course claims to cover.

So here is my university's 3 quarter real ana sequence:

MATH 424 Fundamental Concepts of Analysis (3) NW
The real number system; field, order, and LUB axioms. Metric spaces: Euclidean space. Bolzano-Weierstrass property. Sequences and limits of Sequences. Cauchy sequences and completeness. The Heine-Borel Theorem. Uniform continuity. Connected sets and the intermediate value theorem. Prerequisite: either 2.0 in MATH 328 or 2.0 in MATH 335. Offered: A.

MATH 425 Fundamental Concepts of Analysis (3) NW
One-variable differential calculus: chain rule, inverse function theorem, Rolle's theorem, intermediate value theorem, Taylor's theorem, and intermediate value theorem for derivatives. Multivariable differential calculus: mean value theorem, inverse and implicit function theorems, and Lagrange multipliers. Prerequisite: either 2.0 in MATH 326 or 2.0 in MATH 335; 2.0 in MATH 424. Offered: W.

MATH 426 Fundamental Concepts of Analysis (3) NW
Lebedgue measure on the reals. Construction of the Lebesgue integral and its basic properties. Monotone Convergence Theorem, Fatou's Lemma, and Dominated Convergence Theorem. Integration of series. Continuity and differentiability theorems for functions defined by integrals. Introduction to general measures and integration. Prerequisite: 2.0 in MATH 425. Offered: Sp.

So... It would actually seem to cover some of all three. But as for rigor - that depends on professor if we're comparing it to uni. And checking the syllabus on http://epgy.stanford.edu/courses/math/M115/M115summary.pdf, they definitely don't cover what a lot of the UWashington course covers - they mostly cover what we already heard about in freshman calculus, only more rigorously. But nothing that we don't hear about in freshman calculus, such as "Heine-Borel Theorem", which is in the Uwashington course.

It's REALLY strange - they give MORE credit for linear algebra and JUST multivariable differential calculus than REAL ANALYSIS. Even though real analysis is supposed to have more.

i took a multivariable calculus course on epgy. it was very thorough. they don't cut any corners.

Many places have less real analysis than that three-quarter sequence standard. I went to UC Berkeley for my undergraduate degree; the requirement there was a one-semester course, which looks comparable to UW 424 plus the Riemann integral.
The Lebesgue material in 426 is usually reserved for graduate courses, and most of 425 was covered in an optional second analysis course.

Also, be wary of the difference between semesters and quarters. Most of the EPGY courses look like semester courses, which are normally offered as one course for the subject. The real analysis class you linked to has about 40 lectures- that's standard for a semester class meeting 3 days a week. This model changes somewhat at the graduate level- courses like algebra and real analysis are usually 1-year sequences anywhere.
At a quarter school, expect 1 quarter of credit and some extra material. The EPGY course definitely contains material equivalent to UW 424, and very little from the other two quarters. At a semester school, you'll get a semester of credit if the topics line up- but you probably won't get useful credit if the course you took is missing a major topic from the course your college requires.

Oh- I'm a grad student and TA in math here at UW. Nice to meet a fellow student.


$ a_1\equal{}1$, $ a_2\equal{}12$, $ a_3\equal{}20$
$ a_{n\plus{}3}\equal{}2a_{n\plus{}2}\plus{}2a_{n\plus{}1}\minus{}a_n$

prove that for every natural $ n$, $ 1\plus{}4a_{n\plus{}1}a_{n}$ is perfect square.


[quote="Humpty_Dumpty"]What do you think about this book?(vol.1 in particular) I'm going to buy it. Customer reviews on amazon are all positive but not very detailed. Can you help me?[/quote]

If you are going to buy it, why not form your own opinion after purchase?
Anyway, it is an all-time classic.

[quote="Humpty_Dumpty"]What do you think about this book?(vol.1 in particular) I'm going to buy it. Customer reviews on amazon are all positive but not very detailed. Can you help me?[/quote]

I don`t know about this book, but I know that there it is an inequality of Polya Szego:D

[quote="Zamfirmihai"]I don`t know about this book, but I know that there it is an inequality of Polya Szego:D[/quote]

They also wrote a book on inequalities in mathematical physics.

[quote="Humpty_Dumpty"]What do you think about this book?(vol.1 in particular) I'm going to buy it. Customer reviews on amazon are all positive but not very detailed. Can you help me?[/quote]

These books are wonderfull !

[color=blue][b]
 Georg P\'olya and  Gabor Szeg"o , [/b][/color]
[b]  Aufgaben  und Lehrs"atze aus der Analysis (I)-(II) ,[/b] Springer-Verlag , 1970,1971.
ISBN 3-540-054656-1
ISBN 0-387-05456-1
Heidelberger Taschenbuecher Band 73,Band 74 ,Vierte Auflage.
Editions 1,2,3 are published in Springer Verlag [vol.20 in series "Grundlehren der Mathematischen Wissenschaften"]
First edition=1925

These nice books were also translated  in English, see:
[b]
G. Polya and G. Szego, "Problems and Theorems in Analysis", volume I,Springer Verlag 1972.
Vol I,  ISBN 0-387-90224-4 or 3-540-90224-4

G.Poya and G. Szego,"Problems and Theorems in Analysis," Volume II,Springer-Verlag, New York, 1976.
Vol II, ISBN 0-387-90291-0 or 3-540-90291-0
[2 Volumes in Springer Study Edition, (c) 1972, 1976]
[/b]


algebra
polynomial
geometry
3D geometry
algebra unsolved

Let $a,b,c\in\mathbb Q$ and suppose $f(x) =x^{3}+ax^{2}+bx+c$ has three real but irrational roots.
Prove or disprove : the roots of $f$ cannot be written in [i]closed form[/i] (i.e. using only rational numbers and – maybe nested – $\sqrt$ & $\sqrt[3]{}$ )

There [b]is[/b] a closed form: Cardano's formula.

:D I think Cardano gives you only a closed form if two roots are complex. As far as I know, if all roots are real, you need to divide a certain angle by 3... it gives a [i]method [/i]but for that case no closed formula.  :maybe:

The roots ARE in closed form with Cardano's formula, but the things being cube-rooted are not necessarily real.  In the end, however, if all 3 roots are real, two things are added together that are conjugates to give the second and third roots, so they are real.

In short the roots are in closed form, but the closed form involves doing arithmetic with (possibly) non-real numbers.

Yes, I should have been more precise. Of course I mean [b]a closed form involving only real numbers[/b]. It is not about giving a formula (because there is nothing other than Cardano's formula), but the following:

[i]is there a polynomial $x^3+ax^2+bx+c$ with its three roots in $\mathbb{R}\backslash\mathbb{Q}$  where the roots can be written with radicals and everything happens inside $\mathbb{R}$?[/i]
Note that $ a,b,c\in\mathbb{Q} $ is essential, otherwise it is trivial to find one, e.g. $(x^2-x-1)(x-\sqrt{5})$.


I need help with 2 of last year's national team problems (#9 & 10 to be specific)! All help is much appreciated, but what I really need is an explanation... :)

1. Five different rockets, A, B, C, D and E, are to be launched from two separate launch pads labeled 1 and 2. Each pad can accomodate only one rocket at a time. The rockets can launch from either pad, in any order and at any time (sequentially or simultaneously). One example of a launch pattern is (C1, A1d2, E1B2) where two commas separate three different launching times. Including the example given, what is the total number of different possible launch patterns?

2. In the figure, what is the area of triangle ABD? Express your answer as a common fraction.
[geogebra]ef20c87a46a183af117bb81f326761c59ca65b2f[/geogebra]

Thanks! :)

see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=606166542&t=268847]this[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=606166542&t=268977]this[/url].

number 9 can be confusing, if you still don't get it, i'll do it using the method that i used when solving this problem.

:idea: Those are very helpful links! (But I still don't under stand the recursion method on number nine)

Hint: If you don't wish to use recursion for #9, use simple casework. Divide up the cases where there are certain numbers of pairs of rockets launched simultaneously, and work from there.

Does anyone know where you can get national mathcounts problems?

okay, here's my solution:

[hide="no recursion."]

[b]CASE ONE: all on different launch pads:[/b]
$ 5!\cdot2^5\equal{}32\cdot120\equal{}3840$

[b]CASE TWO: one set of two, rest on their own[/b]
four sets of two consecutive, so $ \binom52\cdot3\cdot2\cdot4\cdot2^4\equal{}3840$

[b]CASE THREE: two sets of two[/b]
three ways to do that, so $ 5\cdot\binom42\cdot2^3\cdot3\equal{}720$

$ 3840\cdot2\plus{}720\equal{}\boxed{8400}$

if you don't understand where any of those numbers are coming from, feel free to ask.
[/hide]


In my Algebra 2 textbook, it wasn't clear just WHY you have to multiply two matrices the way ou do, like, say: a 3x2 matrix by a 2x2 one. Is there a simple way to remember this?

I don't know how to Latex them.  :blush:

For typing matrices in Latex go here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151274

well, if i understand what u mean, if u have a 3x[b]2[/b] matrix and u wanna multiply it by a [b]2[/b]x2 , the reason u can do this is because the two numbers in bold are the same otherwise u can not multiply the matrices. for example u can not multiply 2x2 by 3x2. in other words the number of columns of the left matrix has to be equal to the number of rows of the right matrix. remember matrix multiplication is not commutative! and ur resulting matrix of the afore-mentioned example will have dimensions 3x2 because u take the rows of the first matrix and the columns of the second.
i hope this helps!


If $ a,b,c$ are the sidelenghts of a triangle prove that :

$ 9a^2\plus{}9b^2\plus{}17c^2\minus{}14ac\minus{}14bc\minus{}6ab>0$

[quote="alex2008"]If $ a,b,c$ are the sidelenghts of a triangle prove that :

$ 9a^2 \plus{} 9b^2 \plus{} 17c^2 \minus{} 14ac \minus{} 14bc \minus{} 6ab > 0$[/quote]
It very weak,not use $ a,b,c$ be the sidelengths of a triangle :wink: ,we can easyli prove that:
$ 9a^2\plus{}9b^2\plus{}17c^2 \ge 6ab\plus{}\sqrt{204}c(a\plus{}b)$


Let $ F: W \to V$ be a linear map. Show that $ F$ is surjective iff transforms a spanning set of $ W$ in a spanning set of $ V$.

Suppose $ F$ transforms spanning set of $ W,: k_{1}, k_{2} ... k_{t}$ into a spanning set of $ V, : l_{1}, l_{2} ... l_{t}$ - i.e. $ F(k_{i}) \equal{} l_{i}$. For any element $ x$ of $ V$, we can choose reals $ a_{i}$ so that $ x \equal{} \sum_{1 \leq i \leq t} a_{i} * l_{i}$, since $ l_{1} ... l_{t}$ spans $ W$. Then we have $ x \equal{} F ( \sum_{1 \leq i \leq t} a_{i} * k_{i})$, so for any $ x$ in $ V$ we can find $ y$ in $ W$ with $ F(y) \equal{} x$, so $ F$ is surjective as required.
For the other direction, suppose $ F$ transforms spanning set of $ W: k_{1} ... k_{t}$ into some list in $ V : l_{1} ... l_{t}$. Suffices to prove $ l_{i}$ is a spanning set. For any $ x$ in $ W$, we can find $ a_{1}, a_{2} ... a_{t}$ so that $ x \equal{} \sum_{1 \leq i \leq t} a_{i} * k_{i}$, so $ F(x) \equal{} \sum_{1 \leq i \leq t} a_{i} * l_{i}$. Thus the range of $ F$ is equal to vectors of the form $ \sum_{1 \leq i \leq t} a_{i} * l_{i}$, which is just $ span (l_{1} ... l_{t})$. But since $ F$ is surjective, the range of $ F$ is $ V$, so thus $ span (l_{1} ... l_{t}) \equal{} V$, and $ (l_{1} ... l_{t})$, is a spanning set, as required.


Okay, I'm going to try to actually run a signups this time. Everyone's starting HP may depend on the number of people who sign up.

Hurt and Heal. Each post you hurt someone and you heal someone (this decreases and increases their health by 1 point, respectively). When someone reaches 0 HP they die.

Also, I will be here as the ever-loving god to keep rule-breakers in control.

You cannot heal yourself.
You cannot hurt yourself.
You cannot double-post in this thread.
Once you hurt and heal, you must recharge (wait) for [b]1 hour[/b] (at least) before hurting & healing again.

Nice, I'll get it started when ~10 people join.

Amendments to rules:

Alliances: This rule is going to be impossible to enforce, so I'm not going to bother with it.

Recharge time: 30 minutes is too short, I have made it 1 hour instead.

Also, no more random events. They will actually be very constructed to keep the game from lasting forever (~200 posts, everyone loses 1)

Oi, that's 10. Game shall commence.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=770965#770965

If people continue signing up another one might commence as well.

sign me in, set me free, insert metalcore lyrics.

I'll play if everyone doesn't gang up on me.

That is 7 ... should I wait for more people or randomly add people in?

Hint: Threats of endless poking from the afterlife should stop people from hurting you at least in the beginning.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=786695

game #2 has started with the next batch of signups.

good luck everyone


What's with the $ .$? WHat do those mean? And it looks slightly nicer if you use \tan{x} as opposed to tanx.

$ \tan{x}$
$ tanx$

[quote="thanhnam2902"]$ \tan^2 2x \cdot \tan^23x\cdot \tan 5x \equal{} \tan^22x \minus{} \tan^23x \minus{} \tan 5x$

[/quote]

This is what OP meant (dots were there for multiplying).

Ok thanks!  :)  Now to solve the problem.

[hide="Hint"]Let $ a \equal{} \tan ^2 2x$, $ b \equal{} \minus{} \tan ^2 3x$, and $ c \equal{} \minus{} \tan 5x$, so $ abc \equal{} a \plus{} b \plus{} c$.  Solve for $ c$.[/hide]
[hide="More"]$ c \equal{} \minus{} \frac{a\plus{}b}{1\minus{}ab} \equal{} \minus{} \frac{\tan ^2 2x \minus{} \tan ^2 3x}{1 \minus{} \tan ^2 2x \tan ^2 3x} \equal{} \minus{}\frac{\tan 2x \plus{} \tan 3x}{1 \minus{} \tan 2x \tan 3x} \cdot \frac{\tan 2x \minus{} \tan 3x}{1 \plus{} \tan 2x \tan 3x}$.[/hide]


I heard somewhere that AoPS is planning on releasing an Into to Geometry book? If so, does anyone know when it will be available for purchase?

If you join the email list [url=http://www.artofproblemsolving.com/StayInformed/AoPS_Stay_Informed.php]here[/url], you'll get an email shortly after it is available.

[quote="xxazurewrathxx"]Is there any chance of an Intermediate Geometry book? I don't know if I would want an intro book because I've never had trouble with amc and mathcounts geometry, but it would be great to learn some of the AIME/olympiad level geometry.[/quote]
Get the "Geometry Revisited" by Coxeter and Greitzer.

[quote="xxazurewrathxx"]Is there any chance of an Intermediate Geometry book? I don't know if I would want an intro book because I've never had trouble with amc and mathcounts geometry, but it would be great to learn some of the AIME/olympiad level geometry.[/quote]

That's a year or two away, unfortunately.  The problems in Challenging Problems in Geometry are good training at that level.


inequalities
floor function
function
number theory proposed
number theory

Let $ n$ be a positive integer, $ q_n$ the product of the primes less than or equal to $ n$.

[b]a[/b] Prove that

\[ \sum_{d|q_n}{2\mu(n)\left\lfloor\frac{n}{d}\right\rfloor}\le\sum_{d|q_n}{\mu(n)\left\lfloor\frac{2n}{d}\right\rfloor}\]

[b]b[/b] For fixed $ n$ determine the greatest integer $ k$ that can substitue number $ 2$ in both hands of the above expression.

 :)

[quote="ElChapin"]Let $ n$ be a positive integer, $ q_n$ the product of the primes less than or equal to $ n$.

[b]a[/b] Prove that
\[ \sum_{d|q_n}{2\mu(n)\left\lfloor\frac {n}{d}\right\rfloor}\le\sum_{d|q_n}{\mu(n)\left\lfloor\frac {2n}{d}\right\rfloor}\]
[b]b[/b] For fixed $ n$ determine the greatest integer $ k$ that can substitue number $ 2$ in both hands of the above expression.

 :)[/quote]
What is $ \mu(n)$?

$ \mu$ is the Môbius function http://en.wikipedia.org/wiki/M%C3%B6bius_function.

Also, the sums consider positive divisors only, as it is usual whit that notation.

:)


How many permutations of $ \{1,2,3,4,5,6,7,8\}$ have multiple cycles?

Note: Such a permutation acts as a function, taking $ k$ to $ \sigma(k)$ ($ k\equal{}1,2,\ldots,8$), where \[ \{\sigma(1),\sigma(2),\ldots,\sigma(8)\}\equal{}\{1,2,\ldots,8\}.\] A cycle is a list of elements $ a_1,a_2,\ldots,a_k$ where $ \sigma(a_1)\equal{}a_2$, $ \sigma(a_2)\equal{}a_3$, $ \ldots$, $ \sigma(a_k)\equal{}a_1$.

Seems advanced, but the answer might be $ 8!$ divided by $ 8$ for the cycles, which is $ 5040$.

I think it is $ 8!\minus{}7!\equal{}\boxed{35280}$.

I sort of guessed, and I never really understood what cycles were...

I don't really understand this problem (so I just gave up :( ), but when is it possible to have only one cycle, and when is it possible to have multiple cycles? Also what exactly is a cycle (I think it is like for example when we have $ (1,2,3,4)$, a cycle of this would be $ (2,3,4,1)$, when is the cycle unique?)?

I seriously think that the people at AoPS should take this problem down because it is tooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo difficult for those who haven't had any experience in group theory (I know this because I researched cycles a billion times).  And cycles are made of permutations like this: (1, 2, 3, 6, 5, 4, 8, 7) and then this permutation can be broken into many different cycles I believe.  i was an inch away from getting the problem correct.  I wasn't sure how to rid of permutations that had only 1 cycle.

A cycle is a property of a permutation. When you permute something, you are rearranging the objects. Cycles show how the objects are rearranged.
For example, consider permuting the set $\{1,2,3,4\}$ into $\{3,1,4,2\}$. What you are doing is putting 1 into the second spot, putting 2 into the fourth spot, putting 4 into the third spot, and putting 3 back into the first spot. If you were to draw this out, making and arrow  from 1 to 2  and from 2 to 4 and from 4 to 3 and from 3 to 1, you would get one "loop" with one beginning and one end. Every "loop" is a cycle so this example had one cycle.
Some permutations are made with more than one "loop". If you permute $\{1,2,3,4\}$ into $\{3,4,1,2\}$, drawing it out, you have an arrow from 1 to 3 and from 3 to 1, and another arrow from 2 to 4 and from 4 to 2. That has two beginnings and ends or two "loops" so it has 2 cycles. Also, if a number stays in the same place, it is a separate cycle. So changing $\{1,2,3,4\}$ into $\{1,2,4,3\}$ takes 3 cycles.
Applying it to the problem, there are $8!$ total ways to permute the set. We need to find the number of ways to make a single cycle. Define "1" as an object we are permuting and define 1 as the first spot. "1" has $7$ places it can go to. It can't stay at the same spot since staying in the same spot creates another cycle. Without loss of generality, let "1" land on "3". "3" now has $6$ options, it can to to every place but 1 and where "3" is currently at. By the same logic, whatever "3" lands on has $5$ options and so on. That makes $7!$ permutations with only one cycle. $8!-7!=\boxed{35280}$

By the way, I think this should be for Holy Grail even thought it's not.

I am surprised how simple the alcumus solution was!

The hardest part is figuring out what a cycle is...


[quote="Fierytycoon"]Why did you post in the Solved Problems section.

Anyways, the answer is 1, but I think it is a bit simple even for beginner's section.  :D[/quote]
i agree..
it's a bit simple
but u hv made a mistake
it should be undefined....
sin pi =0 !

[quote]but u hv made a mistake
it should be undefined....
sin pi =0 ![/quote]

haha, yes, you are right. I should double-check my answer next time before saying its too simple.  :D

[quote]No mistake.  The answer is 1.
[/quote]

You can't divide by 0...

[quote]I thought the solved problems section was for "already solved problems" . . . but now I guess it is for problems solved in the other sections?  The other moderators can move this if they want.[/quote]

The solved problems section is for problems that are posted in the unsolved section which have then been solved by other forum posters. Usually moderators move threads to the solved problems sections. This is my understanding, at least.

[quote]First of all, multiply both sides by sin(pi) to get rid of the fraction in the denominator. [/quote]

You can't do that, since your original expression is blah/sin(pi)=blah/0, which is clearly undefined right from the start.

Integral0:

Find the mistake:

Step 1: a and b > 0
Step 2: a = b
Step 3: a^2 = a^b
Step 4: a^2 - b^2 = ab - b^2
Step 5: (a + b)(a - b) = b(a - b)
Step 6: (a + b) = b
Step 7: b + b = b
Step 8: 2b = b
Step 9: 2 = 1

You're doing the same mistake  :D


I don't see how to derive $\sin\angle{B}=\frac{b}{2R}$ and $\sin\angle{C}=\frac{c}{2R}$ from this diagram (page 3 of Geometry Revisited).  :huh:

Well,
$\frac{a}{2R}= \sin{J}= \sin{\frac{\widehat{BAC}}{2}}= \sin{\frac{360^\circ-\widehat{BJC}}{2}}= \sin{(180^\circ-A)}= \sin{A}$.

We can similarly derive $\sin{B}= \frac{b}{2R}$ and $\sin{C}= \frac{c}{2R}$ in almost exactly the same way, but it's not necessary to show this step in a proof because it follows by symmetry.  :)

EDIT: By the way, if you want to see how someone typed something up in LaTeX, you can just hover your mouse cursor over it and you'll see it, or you can quote the post and take a look at the LaTeX.  :)

by the way, how did you write an unitalicized "sin" in math mode?  :maybe:

[code]sin x[/code] produces $sin x$

[code]\sin x[/code] produces $\sin x$


Okay, the SA of this figure is $ 10(2) \plus{} 5(2) \plus{} 3(2)$ or $ 36$.  Since it takes $ 1$ quart to paint $ 6$ square units, it takes $ \boxed{6}$ quarts to paint $ 36$ square units.


let $ A \equal{} \{a_1,a_2,...,a_n\}$ and $ B \equal{} \{b_1,b_2,...,b_n\}$ be two sets of $ n$ integers, where $ A \neq B$,
if the following two multisets (where the same element can occur twice or more, and in this problem, $ |X|\equal{}|Y|\equal{}{n \choose 2}$) $ X \equal{} \{a_i \plus{} a_j | a_i, a_j \in A, a_i \neq a_j\}$, $ Y \equal{} \{b_i \plus{} b_j | b_i, b_j \in B, b_i \neq b_j\}$ are the same ($ X \equal{} Y$),
prove that $ n \equal{} 2^k$ for some positive integer $ k$

Famous and hard:
http://www.mathlinks.ro/Forum/viewtopic.php?t=150844
http://www.mathlinks.ro/Forum/viewtopic.php?t=13258

  darij


AMC
USA(J)MO
USAMO
AIME
AMC 10
AMC 12
AIME I

I think this year's cutoff will be lower than last year's (assuming this year's AIME will be about the same in difficulty level as last year's AIME), because of the expansion this year and harder AMC12's this year.

Cutoff: 210 to 215?

I'm guessing that it's going to be a 213. Just a guess. It's pretty hard for it to be above a 215, though, since the AMC was hard, as $\beta$ said.

I was thinking 205-210, but what do I know?

I would like a number ending in a "2", since that's the same number my AMC 12 scores ends in, and I'm less confident in my AIME than I would like to be.

I would like a number ending in 2^2, lol, since that is mine.

I'm going to be bold and say the AIME is going to be hard this year in addition to the AMCs, and the cutoff will be 207.

I am thinking it will be in the 210-215 range.

What about the AIME floor value for AMC10 people?
Since this is my first year taking AMC stuff, i think that it's will be $1\le \text{value}\le 15$ :D

The AIME floor will be 8.

[hide=" :ninja:"]Disclaimer: I do not back up any of my statements in this thread![/hide]

I'll take a shot with:

Index: 211
Floor: 7

Oh great, this means I need at least a 10 on AIME, if not an 11...yup, I've got no chance.  :dry:

Ah, the joys of being a freshman and not even knowing how to calculate index.

On a serious note, I think the AMCs were not actually much harder than last year, but the blank-value difference made it seem that way.

[quote="worthawholebean"]]
On a serious note, I think the AMCs were not actually much harder than last year, but the blank-value difference made it seem that way.[/quote]

I'm not sure if they made it seem harder or harder to qualify with.  Instead of having to answer 11 correctly, you have to answer 14.  For those teetering on the edge of qualifying, 3 problems makes a lot of difference because they start getting "difficult" around 12/13.

I predict a 211, because I suspect there will be many people with 210s.

I fee like a contestant on Deal or No Deal...

[size=200]Keep it loooow!!![/size]

[quote="nat mc"][quote="[b]Andrew Kim[/b]"]I am not sure how the whole thing would go, but I hope it is lower than 217.5, so I can get 9 in AIME and pass.[/quote]
Hey I knew an Andrew Kim. He was in this GT thing called PACE in third grade. I don't think its you though.[/quote]

I think that it is one of the most common asian-american names, at least from my experience... (I know many)

I just checked my 7-year record of students taking our local math contest in Southern California, about 700 names. There were 16 people named Kim and 5 named Andrew. There wasn't an Andrew Kim, but there were both an Andrew Lee and an Andrew Park.

Due to the USAMO expansion, I feel the index cutoff will be no greater than 210 and the floor no greater than 8

[quote="Kent Merryfield"]All of these predictions are premature. The single most important factor will be the difficulty of the AIME, and none of you has seen that.[/quote]

Good point.

I based my guess upon the assumption that the AIME will be of the same higher difficulty increase than the AMC increase was.

Well these are pretty approximate but here's my idea:

Floor: 8, maybe 7, doubt it'll be anything other than that, but 6 or 9 is possible I guess.

Index: Hmm I highly doubt it will be any higher than last year seeing as (a) there's [i]more[/i] of an expansion and (b) the AMC scoring system lowers scores.

I think there might be a curve. Maybe 200 for the amc12 b and 210 for the A (most people including I thought the B was harder). But lets say someone gets a 150 on the B and 5 on the AIME. Would the floor only be 5 then?

If he/she got in then yeah. But an index of 200 would be most probably not enough to make USAMO.

They could just make it 200.5 and thus force a floor of 6...

Realistically, I am making a random prediction that the cutoffs will be 218 and 8 for the A and 212 and 8 for the B, for index and floor, respectively.

Would there be four different index floors, since there is the AMC A, AMC B, and AIME and alternate AIME? (assuming that the two AIMEs have enough of a difficulty difference). How about tie breakers for people qualifying by the floor value, since the AMC A and B were different difficulty, will they separate the tie breakers as well?

There might be two different qualifying scores for the USAMO from the AIME I and II depending on difficulty, but I'm pretty sure that they won't make a difference between the AMC 12A and the B, since the qualifying scores were only 2.5 points apart.

[quote="neelnal"]I hope it is 208!

Last year's one was in the 210's I thought?[/quote]
Last year's was 217.5 I think.

If it's <= 217, then I only need 10 on AIME.

My goal is still 14 though.
(wanna break 250 index)

[quote="ra5249"]
Is "index" sum of AMC12 scores or sum of AMC10 and AMC12?[/quote]

Index=AMC 12 +10*AIME score.


[color=darkblue]Let $ f(x) \equal{} a_6x^6\plus{}a_5x^5\plus{}a_4x^4\plus{}a_3x^3\plus{}a_2x^2\plus{}a_1x\plus{}a_0$; $ a_6 \not\equal{} 0$. Know that $ f(1)\equal{}f(\minus{}1)$,$ f(2)\equal{}f(\minus{}2)$,$ f(3)\equal{}f(\minus{}3)$.

Prove that $ f(x)\equal{}f(\minus{}x)$, $ \forall x\in R$.[/color]

consider $ g(x)\equal{}(f(x)\minus{}f(\minus{}x))/2\equal{}a_5x^5\plus{}a_3x^3\plus{}a_1x$

It has the 6 roots x=-3,-2,-1,1,2,3. But it has degree 5. So $ g(x)$ is constant. So f(x)=f(-x) for all x in R.

Q.E.D.


[quote="toanhocmuonmau"]Yes, you are welcome. :) This inequality is not hard at all, but it is related to the hard one
\[\sum\frac{a^{4}}{a^{3}+abc+b^{3}}\ge \frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}\]
:)[/quote]

How are they related, Toanhocmuonmau?  :oops:

Thank you very much.


inequalities
calculus
induction
inequalities solved

well, I think you ommited something I think that a1,a2,..,an are not just real numbers but
positive real numbers <>0. well I am just going to post a solution for n=3 just to show on what my solution is based in case I am right about a1,.. an!

a1+2/(1/a1+1/a2)+3/(1/a1+1/a2+1/a3)<2(a1+a2+a3)

we just apply the fact that 2/(1/a+1/b)<=(a+b)/2
ypu apply it twice and then you get that

a1+2/(1/a1+1/a2)+3/(1/a1+1/a2+1/a3)<a1+(a1+a2)/2+(a1+a2+a3)/3 and now we just prove that
a1+(a1+a2)/2+(a1+a2+a3)/3<2(a1+a2+a3) which is simple, after some calculus I think you get that 0<7a2+10a3 which is true! :D you do the same for the generalization!

I haven't tried induction after n! Do you think it would work?

another thing is that you can't have equality in the inequality, cause we can't have a1=a2=..=an! cheers! hope I am right! :D
bye!

Sorry, a1,a2, ... an be positive real numbers.

By applying your technique , Lagrangia , the coefficient of a1 you end up with is :

1 + 1/2 + .. + 1/n , which is unbounded and will be greater than 2 from some value of n on .

Actually , 1 + 1/2 + .. + 1/n - ln(n) is convergent ! So you can see it grows quite fast ..

That inequality were presented at All-Russia olympiad in 1986 with const 4 instead of 2, but solution was more elegant.

do you know that solution myth and can you post it?

cheers! :D :D

[quote="sqing"]
Let $ a_1,a_2, \cdots,a_n $ be [b]positive[/b] real numbers. Prove that\[a_1+ \frac{2}{\frac{1}{a_1}+\frac{1}{a_2}}+ \frac{3}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}}+ \cdots+ \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}\le 2( a_1+a_2+\cdots+a_n )\]
(All-Russia olympiad in 1986 )[/quote]

We have the following:

\begin{align*}
LHS & \leq \sum_{1 \leq j \leq n} \sum_{1 \leq i \leq j} \frac{a_i}{j} \\
& = \sum_{1 \leq i \leq n}H_{n-i+1} a_i \\
\end{align*}

Here, $H_n$ is the $n^{\text{th}}$ harmonic number,

\[ H_n = \sum_{1 \leq j \leq n} \frac{1}{j} \]

but we have a problem since this series diverges.

Also, another site has the problem statement differently @sqing from the '86 All-Soviet Olympiad:

\[ \sum_{1 \leq j \leq n} \frac{j}{\sum_{1 \leq i \leq n} a_i} \]

See problem 427 from [url=http://olympiads.win.tue.nl/imo/soviet/RusMath.html]here.[/url]


geometry
geometric transformation
homothety
geometry proposed

ABCD has an inscribed circle center O. The lines AB and CD meet at X. The incircle of XAD touches AD at L. The excircle of XBC opposite X touches BC at K. X, K, L are collinear. Show that O lies on the line joining the midpoints of AD and BC.

It looks strange. Aren't the two circles you mention one and the same with the incircle of $ABCD$? Second of all, I must be misunderstanding it at a more profound level, because I drew a sketch on my computer which suggests it's not true. Could you please restate it or something? I picture would help..

[quote="grobber"]It looks strange. Aren't the two circles you mention one and the same with the incircle of $ABCD$?[/quote]

You probably drew a sketch where the point X lies on the rays AB and DC. Try it with another sketch, where the point X lies on the rays BA and CD. With this sketch, the problem makes much more sense. Anyway, I have not tried it since I have no time, but it doesn't look wrong...

  Darij

[quote="grobber"]It looks strange. ...
I picture would help..[/quote]

Here is a picture, on how I interpreted the problem. About the text the only think  I made was cut and paste from the page

[url]http://www.kalva.demon.co.uk/russian/rus00.html[/url].

It looks like the line connecting the midpoints of $AD,BC$ is parallel to $KL$, but I have to give it some more thought..

Edit: I do have a solution now, but I'd like to make it simpler. The idea is proving that, if $M,N$ are the midpoints of $AD,BC$, then $MO\|KL,NO\|KL$.

Let $K'$ the point closer to $L$ where $LK$ cuts the circle and let $U$ be the poine where $CB$ touches the circle. It's easy to see that $K'$ is the antipode of $U$ (because the tangent in $K'$ to the circle is obtained from $CB$ through the homothety taking $K$ to $K'$ and centered at $X$, so this tangent is parallel to $CB$). From here the conclusion $NO\|KK'=KL$ follows easily. $MO\|KL$ is proven analogously.

[quote="grobber"]Let $K'$ the point closer to $L$ where $LK$ cuts the circle and let $U$ be the poine where $CD$ touches the circle. It's easy to see that $K'$ is the antipode of $U$ (because the tangent in $K'$ to the circle is obtained from $CD$ through the homothety taking $K$ to $K'$ and centered at $X$, so this tangent is parallel to $CD$). From here the conclusion $NO\|KK'=KL$ follows easily. $MO\|KL$ is proven analogously.[/quote]

mmh ... ok! I can understand the idea. But, could be interesting to make some comment to that. First of all remark a typo your CD should be, I believe,  CB.

But most important aspect, IMHO, to make clearer the argumentation, yuo should explain because if N is the midpoint of CB then NO//KL. For instance I can see more immediately that if N is the mid point of KU then NO//KL and from this, using well known proprierties, justify the claim.

Do you agree, or am I too pernickety?

Indeed, every occurence of $CD$ in my message should be replaced by $CB$. Terribly sorry about that, I'll edit it.

[quote="jhaussmann5"]sprmnt21,
may I ask, with what software did you produce this nice diagram?[/quote]

WINGEOM which is a FANTASTIC little (in size) tool for drawing dinamic sketchs.

You can dowload, FOR FREE, the program from [url=http://math.exeter.edu/rparris/wingeom.html]here[/url]


Does there exist an infinite sequence of positive integers, containing every positive integer exactly once , such that the sum of the first n terms is divisible by n for every n?

Yes, it does.
Construction is just simple.
Let we choose $a_1,a_2,\ldots$ consequently.
Suppose that we chose $a_1,\ldots,a_n$.
Now suppose that the least number that does not appear in $a_1,\ldots,a_n$ is $k$.
Now we choose $a_{n+1}$ such that
\[ n|a_{n+1}+a_1+\ldots+a_n \]
\[ n+1|a_{n+1}+k+a_1+\ldots+a_n \]
See that by chinese remainder theorem and the fact that we already only chose finite numbers we can choose such $a_{n+1}$.
Then we just make put $a_{n+2}=k$.
It's easy to check that this sequence has the requested property and also the fact that every number appears exactly once in the sequence.

There exists another such sequence, namely the one with $a_1=1$ and such that $a_i$ is the smallest positive integer that has not yet appeared in the sequence satisfying the condition $i|\sum_{k=1}^{i}a_i$. It can be shown that $a_{a_n}=n, \forall n \in \mathbb{N}$ so every positive integer is a term of such a sequence.

To complete Xixas' post:
http://www.mathlinks.ro/Forum/viewtopic.php?t=30381&highlight=


You wrote that one dies if he does not guess. So, if he guess, although it is wrong, he is alive, right?

If you mean that one dies if he does not guess correctly, you didn't mention that they cannot communicate each other. So it is very easy.

If they cannot communicate each other, I think it is very difficult.

[quote="Johan Gunardi"]You wrote that one dies if he does not guess. So, if he guess, although it is wrong, he is alive, right?
[/quote]

No, the wizard stays alive only in case the color he said indeed coincides with the color of his hat.
The purpose is to have at least one wizard, who tells the right color.

[quote="Johan Gunardi"]
If you mean that one dies if he does not guess correctly, you didn't mention that they cannot communicate each other. So it is very easy.

If they cannot communicate each other, I think it is very difficult.[/quote]

No, they cannot communicate each other, i mentioned that, when i said that they don't know the answers of other wizards, they only see the colors of the rest 99 hats.

The point is that there must be a strategy for all of them, doesn't matter which colors they would have so that at least one guesses correct color.

[quote="eugene"]This is perhaps quite famous problem, but nonetheless i consider it very nice:

There are 100 wizards in a dungeon and there are 100 possible colored hats. Every wizard gets a hat, which he doesn't see, but he sees the colors of all the rest 99 hats.
At the same moment every wizard tries to guess the color of his own hat and says the answer, in case he doesn't guess he dies.

It is possible that there are several hats of the same color and all the wizards say the answer at the same time, so that they cannot make decision based on the analysis of the other answers, but they all can invent some strategy the day before.

Prove that there is a strategy so that at least one of the wizard will be alive.[/quote]

Is this formulation really complete and correct?
There seem to be so many details missing:

If the colors of the 100 hats are NOT known a priori, the problem is clearly unsolvable. (Give every wizard
a hat with a different, unique phantasy-color; there is  no way of deducing one's own phantasy color
from looking at the other ones.)

If the colors of the 100 hats are known a priori, the problem is trivial. (Every wizard looks at the
99 other hats, and deduces his own hat color as the missing hat.)

If there are only two colors, but  the number of hats with either color is unknown:
Consider three wizards A,B,C among the 100 wizards.
A looks at B, and guesses B's hat color.
B looks at C, and guesses C's hat color.
C looks at A, and guesses A's hat color.

[quote="test20"][quote="eugene"]This is perhaps quite famous problem, but nonetheless i consider it very nice:

There are 100 wizards in a dungeon and there are 100 possible colored hats. Every wizard gets a hat, which he doesn't see, but he sees the colors of all the rest 99 hats.
At the same moment every wizard tries to guess the color of his own hat and says the answer, in case he doesn't guess he dies.

It is possible that there are several hats of the same color and all the wizards say the answer at the same time, so that they cannot make decision based on the analysis of the other answers, but they all can invent some strategy the day before.

Prove that there is a strategy so that at least one of the wizard will be alive.[/quote]

Is this formulation really complete and correct?
There seem to be so many details missing:

If the colors of the 100 hats are NOT known a priori, the problem is clearly unsolvable. (Give every wizard
a hat with a different, unique phantasy-color; there is  no way of deducing one's own phantasy color
from looking at the other ones.)

If the colors of the 100 hats are known a priori, the problem is trivial. (Every wizard looks at the
99 other hats, and deduces his own hat color as the missing hat.)

[/quote]

It is known that every hat is colored in one of the a priori known 100 fixed colors, say enumerated by 1,2,...,100.

For example here is the solution for the sample problem with two wizards and two colors: 1'st wizard says the color of second wizard and the second wizard says the color,  different to the color of the 1'st. wizard. In this way at least one of them will say his own color.

[quote="eugene"]It is known that every hat is colored in one of the a priori known 100 fixed colors, say enumerated by 1,2,...,100.

For example here is the solution for the sample problem with two wizards and two colors: 1'st wizard says the color of second wizard and the second wizard says the color,  different to the color of the 1'st. wizard. In this way at least one of them will say his own color.[/quote]

Aha!
This changes everything.

Wizard number $ k$ ($ k\equal{}1,2,3,\ldots,100$) guesses that the sum $ S$ of the numbers of all 100 hats equals $ k\bmod100$. Then he deduces his own hat color from $ S$.
Clearly, exactly one of the 100 wizards guesses the correct congruence class for $ S$, and hence will guess his correct hat color.


I stand exactly 10 meters away from a revolving ball of radius 1. I throw three darts at the ball, and my shot is equally likely to hit any part of the ball. What is the probability that all three darts hit the same hemisphere of the ball given that all darts hit the ball?

It's not hard...... :)
[hide="Hint"]What are the possible ways for the points to form a hemisphere?[/hide]

OK..... no one seems to want to do this, so:

[hide="Solution"]$p=\fbox{1}$. Any 3 points on a sphere determine a hemisphere.  Wasn't that bad, was it?[/hide]


Two circles $O_{1},O_{2}$ intersect at $A,B$. The two tangents to $O_{1}$ at $A,B$ meet at $K$. $M$ is a point on $O_{1}$, and $KM$ meets $O_{1}$ again at $C$. $CA$ meets $O_{2}$ again at $Q$, and $AM$ meets $O_{2}$ again at $P$.

Prove that $\frac{CA}{AM}=\frac{QB}{BP}$.

[color=indigo]We have: $CA\cdot BM=CB\cdot AM\;\textrm{(well-known)}\Longrightarrow\frac{CA}{AM}=\frac{CB}{BM}\quad (1)$

On the otherhand, ${\begin{cases}\angle QBP=\angle QAP=\angle CBM\\ \angle QPB=\angle QAB=\angle CMB}\end{cases}\Longrightarrow\triangle BQP\sim\triangle BCM\Longrightarrow\frac{QB}{BP}=\frac{CB}{BM}\quad (2)$

From $(1)$ and $(2)$ we have: $\frac{CA}{AM}=\frac{QB}{BP}\quad\mathbb{QED}$[/color]


If the radius of a circle is a rational number, its area is given by a number which is:

$ \textbf{(A)}\ \text{rational} \qquad\textbf{(B)}\ \text{irrational} \qquad\textbf{(C)}\ \text{integral} \qquad\textbf{(D)}\ \text{a perfect square}$
$ \textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
Since $ A\equal{}\pi r^2$ and $ \pi$ is irrational and $ r$ is rational, $ A$ is irrational, or $ \boxed{\textbf{(B)}}$.
[/hide]


The set $ A\subset l^p(p\ge 1)$ has $ \overline{A}$ is compact iff $ A$ is bounded and $ \forall \epsilon>0$, we find $ N(\epsilon)>0$ such that:
$ \sum_{k\equal{}n}^\infty |x_k|^p<\epsilon$,  $ \forall x\equal{}(x_1,x_2,...)\in A$,  $ \forall n>N(\epsilon)$

How's about the proof?

Which implication are you having trouble with?  One is easier than the other, as I recall.


function
induction
calculus
derivative
algebra
binomial theorem
real analysis

Prove the following:

If $f(t)$ and $g(t)$ are two functions of t, then

$\frac{d^n}{dt^n} fg = \sum_{k=0}^{n} {n \choose k} \left [ \frac{d^{n-k}f}{dt^{n-k}} \right ] \left [ \frac{d^{k}g}{dt^{k}} \right ]$

That is Leibniz formula(rule), i think you can find it in every textbook on Calculus. Alternatively, try induction on $n$.

I have never done induction using derivatives before, so can anyone tell me how should I go about proving that $p_n \Rightarrow p_{n+1}$ after doing the base case?

You differentiate $\frac{d^n}{dt^n} fg = \sum_{k=0}^{n} {n \choose k} \left [ \frac{d^{n-k}f}{dt^{n-k}} \right ] \left [ \frac{d^{k}g}{dt^{k}} \right ]$   :)

Hmmm...

Taking the derivative of both sides be get:

$\frac{d}{dt} \frac{d^n}{dt^n} fg = \frac{d}{dt}\sum_{k=0}^{n} {n \choose k} \left [ \frac{d^{n-k}f}{dt^{n-k}} \right ] \left [ \frac{d^{k}g}{dt^{k}} \right ]$

$\Rightarrow \frac{d^{n+1}}{dt^{n+1}} fg = \sum_{k=0}^{n} {n \choose k} \frac{d}{dt} \left [ \frac{d^{n-k}f}{dt^{n-k}} \frac{d^{k}g}{dt^{k}} \right ]$

Now using the product rule on the R.H.S, we get:

$\Rightarrow \frac{d^{n+1}}{dt^{n+1}} fg = \sum_{k=0}^{n} {n \choose k} \left [\frac{d^{n-k+1}f}{dt^{n-k+1}} \frac{d^{k}g}{dt^{k}} + \frac{d^{n-k}f}{dt^{n-k}} \frac{d^{k+1}g}{dt^{k+1}} \right ]$

Now what?

What now? Split it into two sums and re-index. At some point, you'll need to use Pascal's recursion:

${n \choose k-1}+{n \choose k}={{n+1} \choose k}.$

One thing to note: the algebraic technicalities of the inductive step are nearly identical to the algebraic technicalities involved in proving the Binomial Theorem.

[quote="Kent Merryfield"]One thing to note: the algebraic technicalities of the inductive step are nearly identical to the algebraic technicalities involved in proving the Binomial Theorem.[/quote]I seem to recall that there is a proof of the Leibniz formula (due to Peter Lax, via the Fourier transfrom,) that reduces it to the binomial theorem.

[quote]What now? Split it into two sums and re-index. [/quote]
I am not quite sure how would you do that. I tried but I don't think its getting me anywhere. :(

swapnillium: try backing away from this problem for a moment and see if you can prove the Binomial Theorem - that is, for all $n\in\mathbb{N},$

$(x+y)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k.$

Oh WoW! I think I know how to do this! OK, here it goes:

$(fg)^{(n)} = \sum_{k=0}^{n} {n \choose k} f^{(n-k)}g^{(k)}$

The base case of $n = 0$ is obvious: $fg = fg$

Taking the derivative of both sides and using the product rule on the R.H.S we get:

$(fg)^{(n+1)} = \sum_{k=0}^{n} {n \choose k} \left ( f^{(n-k+1)}g^{(k)} + f^{(n-k)}g^{(k+1)} \right )$

Now we distribute the sum and create a new dummy variable $j$:

$(fg)^{(n+1)} = \sum_{k=0}^{n} {n \choose k} f^{(n-k+1)}g^{(k)} + \sum_{j=0}^{n} {n \choose j} f^{(n-j)}g^{(j+1)}$

Now we set $j = k - 1$ and we get:

$(fg)^{(n+1)} = \sum_{k=0}^{n} {n \choose k} f^{(n-k+1)}g^{(k)} + \sum_{k=1}^{n+1} {n \choose {k-1}} f^{(n+1-k)}g^{(k)}$

Now we take the $k = 0$ term out from the left sum and we take the $k = n + 1$ term out from the right sum. So,

$(fg)^{(n+1)} = f^{n+1}g + \sum_{k=1}^{n} {n \choose k} f^{(n-k+1)}g^{(k)} + \sum_{k=1}^{n} {n \choose {k-1}} f^{(n+1-k)}g^{(k)} + fg^{n+1}$

Combining the sums together we have:

$(fg)^{(n+1)} = f^{n+1}g + \sum_{k=1}^{n} \left [ {n \choose k} + {n \choose {k-1}} \right ] f^{(n+1-k)}g^{(k)} + fg^{n+1}$

From Pascal's identity, we get:

$(fg)^{(n+1)} = f^{n+1}g + \sum_{k=1}^{n} {{n+1} \choose k} f^{(n+1-k)}g^{(k)} + fg^{n+1}$

Now we notice that the right and left terms are exactly the $k = 0$ and $k= n + 1$ terms of the sum, respectively. Thus,

$(fg)^{(n+1)} = \sum_{k=0}^{n+1} {{n+1} \choose k} f^{(n+1-k)}g^{(k)}$

Our Induction is Complete! :D  :lol:  ;)


Let $ ABCD$ be a trapezoid for which $ AD\parallel BC$ and $ AB\perp AD$ . Denote $ I\in AC\cap BD$ and the midpoint

$ M$ of $ [AB]$ . Prove the the point $ I$ is the orthocenter of the triangle $ MCD$ if and only if $ AD^2\plus{}BC^2\equal{}CD^2$ .

[hide="Solution"]$ CA \perp MD \iff \triangle ABC \sim \triangle AMD \iff BC \cdot AD = 2 AM^2$
$ BD \perp MD \iff \triangle ABD \sim \triangle BMC \iff BC \cdot AD = 2 AM^2$
So I is the orthocenter of $ \triangle CMD \iff BC \cdot AD = 2AM^2$.
By the Pythagorean theorem,
$ CD^2 = 4AM^2 + (AD - BC)^2 = AD^2 + BC^2 - 2 \cdot AD \cdot BC + 4AM^2 = AD^2 + BC^2$[/hide]
:)


Prove or give a counterexample: if $ T \in L(V)$, then

$ V \equal{} \text{null }T \oplus \text{range }T$.

Furthermore, prove that if $ T \in L(V)$, then

$ V \equal{} \text{null } T^{n} \oplus \text{range }T^{n}$

where $ n \equal{} \text{dim} V$.

For the first one, consider $ T\equal{}\begin{pmatrix}0&1\\0&0\end{pmatrix}.$

Then $ \text{null }T\equal{}\text{range }T\equal{}\text{span }\begin{pmatrix}1\\0\end{pmatrix}.$


function
linear algebra
linear algebra theorems

Hello all.

I'm thinking about automorphisms - I don't really know much about them though. However, if I have some mathematical object is it true that it is always isomorphic to itself? If so, then if I take a mapping that leaves everything unchanged then is this an automorphism? For instance, if I take a graph G and want to preserve edges and non-edges, i claim the graph is isomorphic to itself I will define functions of vertex set V(G) and edge set E(G) as a:V(G)->V(G) and b: E(G) ->E(G) as a(i)=i, b(i)=i, for all i. So this is an automorphism of my graph, right?

A mathematical object is always isomorphic to itself. The mapping that leaves everything unchanged is the identity function and it's always an automorphism.

The identity function:
$ \mathsf{id}_A: A \to A$
$ a \mapsto a$

It's an isomorphism because it's a biiective function and if $ \varphi(a)$ is true in A, where $ \varphi$ is  a quantifier-free A-formula, then $ \varphi(\mathsf{id}_A(a)) \equal{} \varphi(a)$ is true in $ \mathsf{id}_A(A) \equal{} A$. I used the model theory definition.

In other words, this is some kind of permutation of the elements of a group.


inequalities
superior algebra
superior algebra solved

Let $A,B \in M_{n}(R)$ with the property $AB=BA$. Prove that: $ det[(8n^{2}+4n+1)(A^{2}+B^{2})+4n(A+B)+2B+I_{n}] \geq 0$ $\forall n \geq 3$.

I hope this is original and even more: correct! :)

enjoy

Hi Lagrangia, it is...
[quote="Lagrangia"]... corect!  :)
[/quote]
correct. :P


An escalator in pune central is moving downwards.
sam starts at the bottom of the escalator and takes 40
steps to reach the top of the escalator taking him to
food bazaar. however, while coming down from the top
of the escalator he takes just 10 steps to reach the
bottom. what is the total number of steps on the
escalator when it is not moving? given that sam walks
at a constant speed while going up and coming down.

[quote="236factorial"][hide]uhhh... 25?

this can be solved using a 2-variable equation system. [/hide][/quote]

How exactly????

Ignore my answer, I undercomplicated the problem.

Ah, but some steps are always below the others and rotating.

[hide="my solution"]The effect of the escalator on his speed when he is going up in the opposite of when he is going down, so the number of steps would have to be multplied by one number to get his number of steps on the way up, and its recipricol to get the number of steps he took on the way down.  Thus, $20*2=40$ and $20*\frac{1}{2}=10$, and the answer is $\boxed{20}$.  This is the geometric mean of the two numbers, which is implied by what I said before, but I can't remember the formula for that.  I just found it with logic once I saw that $r=2$.

[hide="geometric mean"]Is this what you do for geometric mean?  Something like $Ar=\frac{B}{r}$ for two numbers $A$ and $B$, so $Ar^2=B\Rightarrow r^2=\frac{B}{A}\Rightarrow r=\sqrt{\frac{B}{A}}$?[/hide][/hide]

[hide]Geometric Mean=$(a_1,a_2...a_n)^{1/n}\Rightarrow(40\cdot20)^{1/2}=20$

Oh, will it always be two and 0.5? I was trying to apply opposite rates.[/hide]

[quote="surge"]
Oh, will it always be two and 0.5? I was trying to apply opposite rates.[/quote]

Yes.

I also came up with 20, by applying pretty simple thinking: let's just say that when he's going down, for every step he actually takes the escalator brings him down the equivalent of two steps. Since he and the escalator are each going at constant rates, when he is going up one of his steps only brings him half a step up. Therefore, 40 equals twice the number of actual steps, and 10 equals half the number of actual steps.


If you have n squares with total area 4 show that they can cover a square with side lenght 1.

Thanks.

[b]cover[/b] means overlap allowed or not ?


hi guys! solve it if u can. here goes:

2n friends namely A0, A1, A2, …, A2n-1 sat around a circular table in this order to have dinner. The waiter served them the dishes and they started eating. Suddenly Aj fainted. On the medical report it was found that his food was poisoned. So the police tried to arrest the waiter but he escaped. Instead they found a letter. It said, "Serve the dishes in the order you see them with the first one to me and the next one after some persons on my right. The next one the same number of persons on his right etc. This number must be such that each of my friends get one of the 2n dishes."

Finding this the police came to the conclusion that one of the 2n dishes was already poisoned and arranged in such an order that if
1. the serving was started with the killer,
2. the interval number was such that each of them got a dish.
Then whatever the number was only Aj would get the poisoned one. With the above information, can you help the police finding the killer?

Unless we consider the possibilty that murderer and murdered are the same person, they could find the killer.

If $j\geq n$ the murderer is $A_{(}j-n)$

If $j<n$ the murderer is $A_{(}j+n)$

Let's suppose that the killer is $A_{0}$

We have to find a number $0<k<2n$ such that for each possible numbers $a,b$ such that $(a,2n)=1$ and $(b,2n)=1$ we have that $k*a\equiv k*b \mod 2n$

We can easily see that $k=n$ works because $n*a\equiv n*b \equiv n \mod 2n$ because we know that $a\equiv b\equiv 1 \mod 2$

There aren't any other possible numbers $k$ such that for each possible numbers $a,b$ such that $(a,2n)=1$ and $(b,2n)=1$ we have that $k*a\equiv k*b \mod 2n$

In fact we can suppose $a=1$ and $b=2n-1$ and $k*1\equiv-k*1 \mod 2n$ is true if and only if $2k\equiv 0 \mod 2n$

So if and only if the killer is $A_{0}$, the murdered is $A_{n}$

So we have that $A_{x}$ killed $A_{j}$ if and only if $x-0 \equiv j-n \mod 2n$ and the police can solve the case!


[quote="sharkman"]Tell me, does precalculus cover algebra 1 or algebra 2 material?

precalculus covers high school trigonometry, right?

Also, is college algebra covered in a precalculus course?[/quote]

Precalculus kind of overlaps with Algebra 2, but most books usually only review it. Precalclus courses expect you to know Algebra 1, though. Trigonometry is covered in Precalculus and college algebra, which is pretty much high Algebra 2, is covered also.

I assume he is referring to intermediate algebra, often called "Algebra 3" in some schools, rather than Abstract Algebra.

College algebra is algebra 2. It's what my algebra 2 class was called. Trig really can be covered anywhere from geometry-precalc.

It appears that algebra 1, algebra 2, and college algebra have different meanings to different people.  At the university level, pre-calculus covers both college algebra and trigonometry.  At my high school, however, trigonometry is barely brushed and algebra 2 is gone over quite a bit.

sharkman, do u know the exact book which will be covered?

evidently we do have varying cases. i took a college algebra class at my community class and it was some review from alg 2, but a lot of stuff from pre-calc. no trig!

Precalculus basically covers important stuff you need to know for calculus, such as algebra, combinatorics, and geometry. Trigonometry is also included there as well. Algebra includes both I and II.

Usually logs and matrices are taught because you use them in calculus as well.

Yup. That's also true, and it's covered in things like linear algebra.


logarithms
calculus
integration
irrational number

For positive numbers x, y, and z,

[tex]
\displaystyle
x = \sqrt{x+\sqrt{x+\sqrt{x \ldots}}}

y^{y^{y^{y^{\ddots}}}} = x

z = y^2 + \frac{1}{y^2+\frac{1}{y^2 \ldots}}
[/tex]

Find x, y, and z.  (note that the diagonal dots are going the wrong way.  How do I fix this?)

\iddots should work, but the package isn't included.

[edit] my mistake, i misread the answer to x as sqrt(2) instead of 2... this makes things nicer:

#2

[tex]\displaystyle{y^x=x} \\
\displaystyle{y=e^{\frac{\ln x}{x}}} \\
\displaystyle{y=x^{\frac{1}{x}}} \\
\displaystyle{y = 2^{\frac{1}{2}}} \\
\displaystyle{y = \sqrt{2}}[/tex]

#3

[tex]\displaystyle{z=y^2+\frac{1}{z}} \\
\displaystyle{z^2-y^2z-1=0} \\
\displaystyle{z=\frac{2+\sqrt{8}}{2}} \\
\displaystyle{z=\sqrt{2}+1}[/tex]

nr1337 wrote:Scrambled wrote:[hide]1 is rather simple.  substitute the second rootx for x so its (2x)^.5 =x so x is equal to 2.

for y, the number should tend toward one simply because the y^y^y^y...=x and y^y^y..=x^1/y.  so x and x^1/y are equal.

for z,  substitute y for 1.  The total should come to (:rt5: +1)/2[/hide]

[hide]For y, the number does not tend toward one, since x is not equal to one.  I think you made a mistake by saying that if y^y^y^y = x, then y^y^y = x^1/y.  Raising both sides of the original equation to the 1/y power, we get y^(y^y^y^y..)/y, not y^y^y^y...[/hide]

Since I assume what is meant is y^y^y^y... and not y^(y^(y^(y^... then what scrambled said is true, and x can also be infinitely large or something undefined (I think).  It all depends on what y is.

This brings me to ask, is a negative number raised to a non-integral (particularly irrational) power defined?  I mean, it would eventually get hard to distinguish between + and - and C and R.

First, y^y^y^... is unclear -- it could mean either (((((y^y)^y)^y)^..., which is not what was meant in this question, or y^(y^(y^(y^...)))), which was what was meant in this question.  In either case, y cannot just take on any values, because 1 fixes the value of x.

Negative numbers raised to powers are only defined for complex numbers.  In the real numbers, it is only possible to raise a negative number to rational powers in which (in reduced form) the denominator of the exponent is odd.  This is because of the way one describes irrational numbers.  In particular, they are represented by the set of all sequences of rational numbers having the irrational number as its limit.  So if you consider a power like [tex](-2)^{\sqrt2}[/tex], you have to look at all sequences of the form [tex](-2)^{a_n}[/tex] where a_n approaches :rt2: while n approaches infinity.  However, many of these sequences do not converge or consistently take non-real values.  Anyhow, the result will be infinitely many complex numbers.

This will make it more clear:  y^y^y^y^y... = y^(y^(y^...)))

Ahh... I had failed to realize that all the equations were for the same variables!


The numbers 2, 4, 6, 8 are a set of four consecutive even numbers.  Suppose the sum of five consecutive even numbers is 320.  What is the smallest of the five numbers?


Let $F=\{f|f: \mathbb{R}\rightarrow\mathbb{R}\}$ and $C(f)=\{g|g\in F;\;g\circ f=f\circ g\}$. Prove that:
a) if $C(f)=F$ then $f=1_{\mathbb{R}}$;
b) there exist $f,g\in F$, $f\neq g$, so that $C(f)=C(g);$
c) for any $f,g\in F$:  $C(f)\cap C(g)\subset C(f\circ g)$.

a) for any $y\in\mathbb{R}$, choose $g(x)=y\forall x\in\mathbb{R}$ , then $f(y)=y$.
c)Trivial from defined of $C(f)$.

b) Take $f\in F$ with $f\not =id_\mathbb{R}=f^{3}$.
Then $C(f)\subseteq C(f^{2})\subseteq C(f^{4})=C(f)$, but $f\not =f^{2}$.

Example $f(x)=\{\begin{array}{ll}x+{1\over 3}& \text{for }\{x\}\in [0,{2\over 3})\\ &\\ x-{2\over 3}& \text{for }\{x\}\in [{2\over 3},1) \end{array}$


The centres of the exscribed circles of triangle ABC are O1, O2, O3. Prove that the area of triangle (O1 O2 O3 )is at least four times the area of triangle ABC.

[quote="ashrafmod"][color=darkred]The exincentres of the triangle ABC are $I_{a}$ , $I_{b}$ , $I_{c}\ .$ Prove that the area $[I_{a}I_{b}I_{c}]$ is at least four times the area $[ABC]\ .$[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote the $A$- exincircle $w_{a}=C(I_{a},r_{a})$ a.s.o. From the well-known relations $S\equiv [ABC]=pr=(p-a)r_{a}$ obtain $ar_{a}=p(r_{a}-r)$ a.s.o.
Also remark the relations  $r_{a}+r_{b}+r_{c}=4R+r$ and $2\cdot [BI_{a}C]=ar_{a}$ a.s.o. Therefore, $[I_{a}I_{b}I_{c}]\ge 4S$ $\Longleftrightarrow$ $\sum [BI_{a}C]\ge 3S$ $\Longleftrightarrow$
$\sum ar_{a}\ge 6pr$ $\Longleftrightarrow$ $p\sum (r_{a}-r)\ge 6pr$ $\Longleftrightarrow$ $\sum r_{a}\ge 9r$ $\Longleftrightarrow$ $4R+r\ge 9r$ $\Longleftrightarrow$ $R\ge 2r\ ,$ what is truly.[/color]


No,[hide="Hint"] Its square is  $5+2\sqrt 6$ and it is easy to show that $\sqrt 6$ is not rational.[/hide]

Using Gyan's hint, wouldn't we also have to prove that the square root of an irrational number is also irrational?

Yeah, but I don't think it's hard to prove that the square root of an irrational number is also irrational... Just let the square root of an irrational number be a rational number $\dfrac{m}{n}$ and look for a contradiction.

Yes, to add to above, easiest to see that $m^2 = 6n^2$ where $m$ and $n$ are natiural numbers is not possible because
if you factor both sides into its prime factors, on LHS number of factors of $2$ is even while on right had side its  odd which is a contradiction... (For any square,   all prime number in its factors have to appear even times)


Long asymptote code, such as that posted in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149405]Now incenter[/url] in the Test forum, is cut off at under 750 characters in the popup window in IE7. Firefox is fine though.

There's nothing that I can do about it. It is created by the limitation in IE's url length. Unfortunately IE is still miles back compared to Firefox, Opera, or other browsers. Right now though almost 50% of the users use Firefox :)


function
real analysis
real analysis unsolved

Construct a set $A \subset \mathbb{R}$ and a function $f: A \to \mathbb{R}$ such that for every $a_{1}, a_{2}\in A$ holds $|{f(a_{1})-f(a_{2})}|\leq|{a_{1}-a_{2}}|^{3}$ and the range of $f$ is uncountable.
(V. Brayman)

[hide="Hint"]$A$ = standard Cantor set.[/hide]
[hide="Solution"]Every element of $A$ has base 3 expansion with only $0$ and $2$ in it. Therefore, for every $a_{1},a_{2}\in A$ the difference $|a_{1}-a_{2}|$ is at least $1/3^{n}$, where $n$ is the first digit (after the dot) where the expansions of $a_{1}$ and $a_{2}$ disagree. Define $g(0.xyzw\dots)=0.00x00y00z00w\dots$. Since $|g(a_{1})-g(a_{2})|$ is at most the $3/3^{3n}$ with $n$ as above, it follows that $|g(a_{1})-g(a_{2})|\le 3|a_{1}-a_{2}|^{3}$. So, $f(a)=g(a)/3$ works.[/hide]


pigeonhole principle
number theory
relatively prime

Prove or disprove the following statement: Every positive integer has a multiple whose decimal representation consists only of the digits 0, 1, and 2 repeated as many times as necessary.

If it isn't clear already, make 11...100....0, where the 0's make the number divisible by any power of 2 or 5 necessary, and the right amount of 1's to make the number divisible by any number relatively prime to 10.

Let our number be $ N$. Consider the sequence $ 1,11,111,...,111...111$ where the last term consists of $ N\plus{}1$ ones. Then by pigeonhole, two have the same remainder on division by $ N$. Subtracting the smaller from the larger, we have a number of the form $ 111...000...$ Which is clearly divisible by $ N$.

[b]Remark[/b]: If $ N$ is not divisible by $ 5$, we can simply divide out the trailing $ 0$s to give a number divisible by $ N$ consisting entirely of ones. Also, we can get numbers with only $ 2,0$ or $ 3,0$ etc by simply multiplying this number by $ 2,3$ etc.

$ 1$ is not a multiple in general, but $ 1$ is always a divisor. e.g. multiples of $ 12$ are $ 0,12,24,36...$ and divisors are $ 1,2,3,4,6,12$ but $ 1$ is certainly not always a multiple of any number.


function
calculus
inequalities
inequalities proposed

Let $a,b,c>0$ such that $a+b+c=\frac{3}{2}$ And
$F=27abc+\frac{8}{ab+bc+ca}$
Here is my question:$minF=?$ :wink:  :wink:
When the equality occur? :wink:

[quote="delegat"]It is easy to find minimum of this function if you replace 27 by 256/27 but for 27???[/quote]
Yes, :wink:  but with 27,  I had a nice pro.

[quote="Minh Can"]q: a,b,c > 0 and a+b+c = 1, find minimum value of f(a,b,c) = 27abc + 8/(ab+bc+ca).
a: not yet a full solution but some thought about it leads me to say that the minimum value is m = f(1/2,1/2,1/2). as you can see if you let a --> 0 then f(0,b,c) attains a minimum at b = c = 3/4. but m <= f(0,3/4,3/4) as you can verify it yourself. the strategy is perhaps the AC method.
-------------------------------------------------------------------------------------[/quote]
The equality don't occur when $a=b=c$ or $a=0$ :wink:

Mixing variables works, assume c is the smallest, then we can check that $f(a,b,c) \ge f(t,t,c)$ where $t = \frac{a+b}{2}$ After that we are to prove $\frac{81}{2}t^{2}-54t^{3}+\frac{8}{3t-3t^{2}}\ge 14$ for $t \in [\frac{1}{2}, \frac{3}{4}]$. Expanding we get
$324t^{5}-567t^{4}+243t^{3}+84t^{2}-84t+16 \ge 0$ which is equivalent to
$(3t-2)^{2}(36t^{3}-15t^{2}-9t+4)\ge 0$ which is true.

[quote="Soarer"]After that we are to prove $\frac{81}{2}t^{2}-54t^{3}+\frac{8}{3t-3t^{2}}\ge 14$ for $t \in [\frac{1}{2}, \frac{3}{4}]$. Expanding we get
$324t^{5}-567t^{4}+243t^{3}+84t^{2}-84t+16 \ge 0$ which is equivalent to
$(3t-2)^{2}(36t^{3}-15t^{2}-9t+4)\ge 0$ which is true.[/quote]

I see, and it is nice, but you only verify that 14 is right after he tells us 14. How would we find the value?

After mixing variables, you are left with an expression of one variable. At least calculus works.

By EV-Theorem, for $a+b+c=3/2$ and $abc=const$, the sum $a^{2}+b^{2}+c^{2}$ (hence E) is minimal when $0<a\leq b=c$.
Therefore, we can consider only this case.
Similarly, we can prove the inequality
$3abc+\frac{30}{ab+bc+ca}\geq 13$
for any positive numbers $a,b,c$ such that $a+b+c=3$.

[quote="Vasc"]$abc=const$[/quote]

Why??
 :?:

Because for $abc=const$ we may apply EV-Theorem. :wink:

EV theorem...What's this ,Vasc?
Was it proved?
Could you tell us about the history of that problem? :)

[quote="Vasc"]Take a look at
http://www.mathlinks.ro/Forum/viewtopic.php?t=7942&start=160

I published a complete proof for Equal Variable Theorem in the last issue (no 2, 2006) of Gazeta Matematica A, in English. In Algebraic Inequalities there is a chapter which treats this method, with a partial proof, 7 corolarries and 34 applications.
[/quote]
I'm sorry but can you post it here?

[quote="Vasc"]Take a look at
http://www.mathlinks.ro/Forum/viewtopic.php?t=7942&start=160

I published a complete proof for Equal Variable Theorem in the last issue (no 2, 2006) of Gazeta Matematica A, in English. In Algebraic Inequalities there is a chapter which treats this method, with a partial proof, 7 corolarries and 34 applications.[/quote]
Thank you,but where can I find this book?
[b]a little problem[/b]:Do we can use EV theorem with $ab+bc+ca=const$?

See
http://www.mathlinks.ro/Forum/viewtopic.php?t=100399.

And,
$2(ab+bc+ca)=(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})$.


geometry
analytic geometry
perimeter
search
trigonometry
incenter

Let $ D$ be the tangency point of the incircle $ (I)$ of a $ \triangle ABC$ with $ BC.$ Let $ E, F$ be tangency points of the C-, B-excircles $ (I_c), (I_b)$ with $ CA, AB.$ Let $ \triangle XYZ$ be the anticomplementary triangle of the $ \triangle ABC$ with $ A \in YZ, B\in ZX, \C \in XY$ and $ YZ \parallel BC, ZX \parallel CA, XY \parallel AB$  The lines $ AD, BE, CF$ concur at the X-excenter of the $ \triangle XYZ.$

Remark: $ AD, BE, CF$ are concurrent by Ceva's theorem at external analog of the Nagel point. It is very easy to show with barycentric coordinates that this is the X-excenter of the $ \triangle XYZ.$ What would be a synthetic proof ?

Dear yetti, thank you for a nice problem.
[hide="Solution"]See at yetti's figure.
Denote Jx is the X-excenter of triangle XYZ and M,N,P be the tangency points of the circle (Jx) with YZ,ZX,XY respectively. Let's prove that CF,AD,BE all pass through Jx.
p(ABC) is the semi-perimeter of triangle ABC.
Since CD=p(ABC)-AB=p(XBC)-CX, so XD is the Nagel line of triangle XBC, XD passes through M, in addtion, D is the midpoint of XM. Let X' such that Jx is the midpoint of MX'. The tangent at X' of (Jx) meets XY,XZ at H,K. Then, (Jx) is the incircle of triangle XHK, XX' is the Gergonne line of this triangle, this line cuts YZ at point M' which is symmetric to M wrt A. So, A,D,Jx are collinear.
With F is the midpoint of PZ, E is the midpoint of NY, we can solved these cases similarly.
In short, we have the result of the problem.[/hide]

Dear Yetti and Mathlinkers,
the synthetic proof of your concurrency can be seen in my article "Les cinq théorèmes de Nagel, p. 26" vol. 3 in my site http://perso.orange.fr/jl.ayme.
Sincerely
Jean-Louis

[size=134][color=darkblue][b]Nice and interesting problem ! It isn't even hardly. Thank you,[/b][/color] [color=darkred][b]Yetti[/b][/color] [color=darkblue][b]![/b][/color][/size]

[quote="yetti"] [color=darkred]Let $ D$ be the tangency point of the incircle $ C(I,r)$ of $ \triangle ABC$ with $ BC$ . Let $ E, F$ be tangency points of the $ C$ - excircle $ C(I_c,r_c)$ and

$ B$  - excircle $ C(I_b,r_b)$ with $ CA$ , $ AB$ respectively. Let $ \triangle XYZ$ be the anticomplementary triangle of the $ \triangle ABC$ , i.e.  $ A \in YZ$ , $ B\in ZX$ ,

 $ C \in XY$ and $ YZ \parallel BC$ , $ ZX \parallel CA$ , $ XY \parallel AB$ . Prove that the lines $ AD$ , $ BE$ , $ CF$ concur at the $ X$ - excenter of the $ \triangle XYZ$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Observe that $ \left\|\begin{array}{c} BF \equal{} CE \equal{} p \\
\ DB \equal{} AE \equal{} p \minus{} b \\
\ DC \equal{} AF \equal{} p \minus{} c\end{array}\right\|$ . Apply the [b]Ceva's theorem[/b] to the points $ \left\|\begin{array}{c} D\in BC \\
\ E\in CA \\
\ F\in AB\end{array}\right\|$ for the triangle $ ABC$ :

$ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB} \equal{} \frac {p \minus{} b}{p \minus{} c}\cdot\frac {p}{p \minus{} b}\cdot\frac {p \minus{} c}{p} \equal{} 1$ , i.e. $ AD\cap BE\cap CF\ne\emptyset$ . Denote $ L\in AD\cap BE\cap CF$ . Apply the [b]Menelaus' theorem[/b]

to the transversals: $ \left\|\begin{array}{c} \overline {LAD}/BCE\ : \ 1 \equal{} \frac {LE}{LB}\cdot \frac {DB}{DC}\cdot\frac {AC}{AE} \equal{} \frac {LE}{LB}\cdot \frac {p \minus{} b}{p \minus{} c}\cdot \frac {b}{p \minus{} b}\implies \frac {LE}{p \minus{} c} \equal{} \frac {LB}{b} \equal{} \frac {BE}{p \minus{} a}\implies \boxed {\ \frac {EL}{EB} \equal{} \frac {p \minus{} c}{p \minus{} a}\ } \\
 \\
\overline {LAD}/BCF\ : \ 1 \equal{} \frac {LF}{LC}\cdot \frac {DC}{DB}\cdot\frac {AB}{AF} \equal{} \frac {LF}{LC}\cdot \frac {p \minus{} c}{p \minus{} b}\cdot \frac {c}{p \minus{} c}\implies \frac {LF}{p \minus{} b} \equal{} \frac {LC}{c} \equal{} \frac {CF}{p \minus{} a}\implies \boxed {\ \frac {FL}{FC} \equal{} \frac {p \minus{} b}{p \minus{} a}\ }\end{array}\right\|$ .

Apply the [b]Aubel's relation[/b] : $ \frac {AL}{AD} \equal{} \frac {EL}{EB} \plus{} \frac {FL}{FC} \equal{} \frac {p \minus{} c}{p \minus{} a} \plus{} \frac {p \minus{} b}{p \minus{} a}$ $ \implies$ $ \boxed {\ \frac {AL}{AD} \equal{} \frac {a}{p \minus{} a}\ }$ . Denote the [b]distance $ \delta_d(P)$ of the point $ P$ to the line $ d$[/b] .

Observe that $ \delta_{YZ}(D) \equal{} h_a$ . Thus, $ \frac {\delta_{YZ}(L)}{\delta_{YZ}(D)} \equal{} \frac {AL}{AD} \equal{} \frac {a}{p \minus{} a}\implies$ $ \delta_{YZ}(L) \equal{} \frac {ah_a}{p \minus{} a} \equal{} \frac {2S}{p \minus{} a}\implies$ $ \boxed {\ \delta_{YZ}(L) \equal{} 2r_a\ }$ .

$ \frac {\delta_{AB}(L)}{\delta_{AB}(C)} \equal{} \frac {FL}{FC} \equal{} \frac {p \minus{} b}{p \minus{} a}$ $ \implies$ $ \delta_{AB}(L) \equal{} \frac {h_c(p \minus{} b)}{p \minus{} a}$ $ \implies$ $ \delta_{XY}(L) \equal{} \delta_{AB}(L) \plus{} h_c \equal{} \frac {h_c(p \minus{} b)}{p \minus{} a} \plus{} h_c \equal{} \frac {ch_c}{p \minus{} a} \equal{} \frac {2S}{p \minus{} a}$ $ \implies$ $ \boxed {\ \delta_{XY}(L) \equal{} 2r_a\ }$ .

$ \frac {\delta_{AC}(L)}{\delta_{AC}(B)} \equal{} \frac {EL}{EB} \equal{} \frac {p \minus{} c}{p \minus{} a}$ $ \implies$ $ \delta_{AC}(L) \equal{} \frac {h_b(p \minus{} c)}{p \minus{} a}$ $ \implies$ $ \delta_{XZ}(L) \equal{} \delta_{AC}(L) \plus{} h_b \equal{} \frac {h_b(p \minus{} c)}{p \minus{} a} \plus{} h_b \equal{} \frac {bh_b}{p \minus{} a} \equal{} \frac {2S}{p \minus{} a}$ $ \implies$ $ \boxed {\ \delta_{XZ}(L) \equal{} 2r_a\ }$ .

In conclusion, $ \delta_{YZ}(L) \equal{} \delta_{XY}(L) \equal{} \delta_{XZ}(L) \equal{} 2r_a$ , i.e. the point $ L$ is the $ X$ -excenter of $ \triangle\ XYZ$ and the its $ X$ - exradius is equally to $ 2r_a$ . [/color]

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ D,\ E,\ F$ be, the tangent points of its $ A$-excircle, to the sidelines $ BC,\ AC,\ AB,$ respectively. Prove that the line connecting the midpoints of the segments $ AB,\ CF,$ passes through the $ A$-excenter of $ \bigtriangleup ABC.$[/color][/size][/b]

I think that this Lemma is well known, but I don’t remember the proof. So, I must search my archive and if I am not mistaken, I will post here later the proof of the Lemma.

$ \bullet$ In our problem now ( see the [b][size=100]yetti’s schema[/size][/b] ), let be the point $ P\equiv XY\cap ZF$ and it is easy to show, by the argument of the semi perimeter result ( mentioned also by [b][size=100]mr.danh[/size][/b] ), that this point is the tangent point of the sideline $ XY,$ to the $ X$-excircle of $ \bigtriangleup XYZ.$

So, we have $ FZ \equal{} FP,$ because of $ AZ \equal{} AY$ and hence, based on the above Lemma, we conclude that the line segment $ CF,$ passes through the point $ J_{x},$ as the $ X$-excenter of $ \bigtriangleup XYZ.$

Similarly the line segment $ BE,$ also passes through the point $ J_{x}.$

Hence, because of the line segments $ AD,\ BE,\ CF,$ are concurrent at one point, we conclude that the line segment $ AD,$ passes through the point $ J_{x}\equiv BE\cap CF$ and the proof is completed.

Kostas Vittas.

Thanks to [b]Virgil Nicula[/b] for his detailed proof. Thanks to [b]jayme[/b] for his proof that $ Na_a, G, I_a$ are collinear and $ \overline{GNa_a} \equal{} \minus{} 2\ \overline{GI_a},$ where $ G$ is the common centroid of $ \triangle ABC, \triangle XYZ$ and $ Na_a$ the concurrency point of $ AD, BE, CF.$ I needed this result in the proof of the problem [url]http://www.mathlinks.ro/viewtopic.php?t=213970[/url], the (hidden) excircle case, and all I could come up with was the barycentric proof. I got as far as [b]vittasko's[/b] lemma but could not prove it synthetically.

To [b]mr.dang[/b]: While it is easy to see that the A-Gergonne line $ AD$ of the $ \triangle ABC$ passes through the X-excenter $ J_x$ of the $ \triangle XYZ,$ it is not necessary to prove this, because $ AD, BE, CF$ (in my notation) are concurrent by Ceva and if $ BE$ goes through $ J_x,$ so does $ CF.$ However, the lines $ BE, CF$ are of different type than $ AD,$ therefore you cannot say $ BE$ goes through $ J_x$ similarly as $ AD.$

It is even easy to see that $ E$ is the midpoint of $ YN,$ where $ N$ is the tangency point of $ (J_x)$ with $ ZX.$ To finish your proof depends on [b]vittasko's[/b] lemma. Please, could you or [b]vittasko[/b] or anyone post a proof of this lemma ? Tx.

[size=134][color=darkblue][b]Very nice and interesting lemma ! Thank you,[/b][/color] [color=darkred][b]Vittasko.[/b][/color] [color=darkblue][b]I like it.[/b][/color] [/size]

[size=134][color=darkblue][b]In the my opinion, it is even more difficult than the[/b][/color] [color=darkred][b]Yetti[/b][/color][color=darkblue][b]'s proposed problem.[/b][/color][/size]

[quote="vittasko"][color=Darkred][b][u]Lemma.[/u][/b] A triangle $ \bigtriangleup ABC$ is given and let $ D$ , $ E$ , $ F$ be the tangent points of its $ A$ - excircle $ C(I_a,r_a)$ to the sidelines

$ BC$ , $ AC$ , $ AB$ respectively. Denote the midpoints $ M$ , $ N$ of the segments $ [AB]$ , $ [CF]$ . Prove that $ I_a\in MN$ .[/color] [/quote]

[color=darkblue][b][u]Proof (metric).[/u][/b] Denote $ \|\begin{array}{c} L\in IA\cap BC \\
\ S\in MN\cap BC \\
\ V\in MI_a\cap BC\end{array}\|$ . Observe that $ BL = \frac {ac}{b + c}$ . Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {MSN}/BCF$ :

$ \frac {MB}{MF}\cdot \frac {NF}{NC}\cdot\frac {SC}{SB} = 1\ \implies\ \frac {SB}{SC} = \frac {MB}{MF} =$ $ \frac {\frac c2}{p - c + \frac c2}\ \implies\ {\boxed {\ \frac {SB}{SC} = \frac {c}{a + b}\ }\ (3)}$ . Apply the [b]Menelaus' theorem[/b] to the transversal

$ \overline {VMI_a}/ABL\ \ : \ \ \frac {I_aL}{I_aA}$ $ \cdot\frac {MA}{MB}\cdot\frac {VB}{VL} = 1\ \implies\ \frac {VB}{VL} =$ $ \frac {I_aA}{I_aL} = \frac {I_aA}{I_aI}\cdot\frac {I_aI}{I_aL} = \frac pa\cdot\frac {r + r_a}{r_a} =$ $ \frac pa\cdot\frac {\frac 1p + \frac {1}{p - a}}{\frac {1}{p - a}} = \frac {b + c}{a}$ $ \implies$

$ \frac {VB}{b + c} = \frac {VL}{a} = \frac {BL}{2p} = \frac {ac}{2p(b + c)}$ $ \implies$ $ VB = \frac {ac}{2p}$ $ \implies$ $ VC = a - \frac {ac}{2p} = \frac {a(a + b)}{2p}$ $ \implies$ $ \boxed {\ \frac {VB}{VC} = \frac {c}{a + b}\ }\ (4)$ .

From the relations $ (3)$ , $ (4)$ obtain $ \frac {SB}{SC} = \frac {VB}{VC}$ , i.e. $ V\equiv S$ , what means $ \boxed {\ I_a\in MN\ }$ .

[b][u]Remark.[/u][/b] Denote the point  $ U\in AC$ so that $ C\in (AU)$ , $ CU = CB$ . If $ W\in BU\cap AI$ , then $ SW\parallel AC$ because $ \frac {SB}{SC} = \frac {AB}{AU} = \frac {WB}{WU}$ .[/color]

[hide="Another longer metrical proof."]
[color=darkblue] Show easily the identity in $ \triangle\ ABC\ \ : \ \ \boxed {\ \frac {bc}{bc + 2p(p - c)} = \frac {c}{\cos A(a + b + 2r_a\tan A)}\ }\ (*)$ . Indeed, it is equivalently with the relation

$ b\cos A(a + b + 2r_a\tan A) = bc + 2p(p - c)$ $ \Longleftrightarrow$ $ b(a + b)\cos A + 2br_a\sin A = bc + 2p(p - c)\ \|\ \ \bigodot\ 2c$ $ \Longleftrightarrow$

$ (a + b)(b^2 - a^2 + c^2) + 8r_aS = 2bc^2 + 4cp(p - c)$ $ \Longleftrightarrow$ $ c^2(a + b) + (b - a)(b + a)^2 + 8p(p - b)(p - c) = 2bc^2 + 4cp(p - c)$ $ \Longleftrightarrow$

$ (b - a)(a + b)^2 = c^2(b - a) + 4p(p - c)(b - a)$ $ \Longleftrightarrow$ $ (a + b)^2 = c^2 + 4p(p - c)$ $ \Longleftrightarrow$ $ (a + b)^2 = c^2 + (a + b)^2 - c^2$ , what is truly.

====================================================================================================

Denote the intersections $ \|\begin{array}{c} P\in MN\cap AC \\
\ R\in MI_a\cap AC \\
\ T\in AB\cap EI_a\end{array}\|$ . Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {PMN}$ in $ \triangle\ ACF$ : $ \frac {PA}{PC}\cdot\frac {NC}{NF}\cdot\frac {MF}{MA} = 1$ .

Thus, $ \frac {PA}{PC} = \frac {MA}{MF} = \frac {\frac c2}{p - \frac c2} = \frac {c}{a + b}$ $ \implies$ $ \frac {PA}{c} = \frac {PC}{a + b} = \frac {b}{2(p - c)}$ $ \implies$ $ PE = PA + AE = \frac {bc}{2(p - c)} + p = \frac {bc + 2p(p - c)}{2(p - c)}$ $ \Longrightarrow$

$ \boxed {\ \frac {PA}{PE} = \frac {bc}{bc + 2p(p - c)}\ }\ (1)$ . Observe that  $ FT = r_a\tan A$ and $ MT = MF + FT = AF - AM + FT = p - \frac c2 + r_a\tan A$ $ \implies$

$ MT = \frac 12\cdot (a + b + 2r_a\tan A)$ . Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {RMI_a}$ in $ \triangle\ AET$ : $ \frac {RA}{RE}\cdot\frac {I_aE}{I_aT}\cdot\frac {MT}{MA} = 1$ $ \implies$

$ \frac {RA}{RE} = \frac {I_aT}{I_aF}\cdot \frac {MA}{MT}$ $ \implies$ $ \boxed {\ \frac {RA}{RE} = \frac {1}{\cos A}\cdot\frac {c}{a + b + 2r_a\tan A}\ }\ (2)$ .  From the relations $ (*)$ , $ (1)$ , $ (2)$ obtain $ \frac {PA}{PE} = \frac {RA}{RE}$ , i.e. $ \boxed {\ I_a\in MN\ }$ .

====================================================================================================[/color][/hide]

[color=darkblue][b][u]Proof (with barycentrical coordinates).[/u][/b] Since $ \|\begin{array}{c} 2\cdot M = A + B \\
 \\
2\cdot N = C + F \\
 \\
c\cdot F = (c - p)\cdot A + p\cdot B\end{array}\|$ , where $ \|\begin{array}{c} A\ (\ 1\ ,\ 0\ ,\ 0\ ) \\
\ B\ (\ 0\ ,\ 1\ ,\ 0\ ) \\
\ C\ (\ 0\ ,\ 0\ ,\ 1\ )\end{array}\|$ obtain $ \|\begin{array}{c} M\ (1\ ,\ 1\ ,\ 0\ ) \\
 \\
N\ (\ c - p\ ,\ p\ ,\ c\ ) \\
 \\
I_a\ (\ - a\ ,\ b\ ,\ c\ )\end{array}\|$ .

Therefore, $ I_a\in MN\ \Longleftrightarrow\ |\begin{array}{ccc} 1 & 1 & 0 \\
 \\
c - p & p & c \\
 \\
- a & b & c\end{array}| = 0$, what is truly because $ L_2 = (p - b)\cdot L_1 + L_3$ , where $ L_1$ , $ L_2$ , $ L_3$ are the rows of the determinant.
[/color]

[quote="yetti"]
To [b]mr.danh[/b]: While it is easy to see that the A-Gergonne line $ AD$ of the $ \triangle ABC$ passes through the X-excenter $ J_x$ of the $ \triangle XYZ,$ it is not necessary to prove this, because $ AD, BE, CF$ (in my notation) are concurrent by Ceva and if $ BE$ goes through $ J_x,$ so does $ CF.$ However, the lines $ BE, CF$ are of different type than $ AD,$ therefore you cannot say $ BE$ goes through $ J_x$ similarly as $ AD.$
[/quote]
Dear yetti, why not ?
For example, I will prove that BE goes through Jx similatly as AD.
Since AE=p(ABC)-AC=p(YAC)-AC, so E is the tangency point of C-excircle of triangle YCA with CA, so YE passes through N, in addtion, E is the midpoint of YN.
Let N' such that Jx is the midpoint of NN', YN' passes through the point which is symmetric to N wrt B (this point is tangency point of the Z-excircle of triangle XYZ with XZ). So, B,E,Jx are collinear.

maybe, my solution is not nice as yours, but I beleive it is correct.

Dear Virgil Nicula, my proof seems solved your problem. Is that true?

[quote="vittasko"][b][size=100]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ D,\ E,\ F$ be, the tangent points of its $ A$-excircle, to the sidelines $ BC,\ AC,\ AB,$ respectively. Prove that the line connecting the midpoints of the segments $ AB,\ CF,$ passes through the $ A$-excenter of $ \bigtriangleup ABC.$[/size][/b][/quote]

[b][size=100][color=DarkBlue]EQUIVALENT PROBLEM. – An isosceles triangle $ \bigtriangleup ABC$ with $ AB = AC$ is given and let $ X,\ Z$ be, the orthogonal projections of the midpoint $ M,$ of the side-segment $ BC,$ on $ AB,\ AC,$ respectively. We draw the circle $ (M),$ centered at $ M$ with radius $ MX = MZ$ and let $ D,\ E,\ F$ be, the intersection points of the sidelines $ BC,\ AC,\ AB$ respectively from an arbitrary line, tangents to $ (M).$ Prove that the line connecting the midpoints $ K,\ L$ of the segments $ AF,\ EX$ respectively, passes through the point $ M.$[/color][/size][/b]

[b][size=100]PROOF[/size][/b]. – We consider the degenerated quadrilateral $ AFBC$ and because of $ K,\ L,\ M,$ are the midpoints of the segments $ AF,\ EX,\ BC$ respectively, it is enough to prove that $ \frac {EA}{EC} = \frac {XF}{XB}$ $ ,(1)$ based on the power theorem which has already been posted in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97493]Collinearity of orthocenters[/url] (post #10), of [b][size=100]shobber[/size][/b].

We can prove the result of $ (1),$ as follows ( see the schema [b][size=100]t=214545(a)[/size][/b] ):

$ \bullet$  From $ \frac {EA}{EC} = \frac {XF}{XB}$ $ \Longrightarrow$ $ \frac {EA + EC}{EC} = \frac {BX + XF}{BX}$ $ \Longrightarrow$ $ \frac {AC}{EC} = \frac {BF}{BX}$ $ ,(2)$

From $ (2)$ $ \Longrightarrow$ $ (BF)\cdot (CE) = (BX)\cdot (AC) = (BM)^{2}$ $ ,(3)$ because of $ BX = CZ$ and $ (CZ)\cdot (CA) = (MC)^{2} = (MB)^{2},$ $ ($ from the right triangle $ \bigtriangleup MAC$ $ ),$

From $ (3),$ it is enough to prove that $ (BF)\cdot (BE') = (BM)^{2}$ $ ,(4)$ where $ E'$ is the point on $ AB,$ such that $ E'E\parallel BC.$

But, the relation $ (4)$ is true, because of the quadrilateral $ MFE'E$ is cyclic, from $ \angle EE'M = \angle E'EM = \angle EMC = \angle EFM$ as well ( and easy to prove by angle changing ) and clearly, the line segment $ BM$ tangents to its circumcircle, at point $ M.$

So, we conclude that the relation $ (1)$ is true and the proof of the equivalent problem is completed.

Kostas Vittas.

[quote="jayme"]Dear Mathlinkers,
I think that the Vittas's lemma is an extraversion of this well know result:
the midpoint of the side of a triangle, the incenter and the midpoint of the gergonnian correspondind to this side, are collinear.
Sincerely
Jean-Louis[/quote]
We can also prove this result by the same way as before :

Let $ P,\ Q$ be, the intersection points of $ BC,\ AB$ respectively, from the line through the incenter $ I$ of $ %Error. "bigtrinalgeup" is a bad command.
ABC$ and perpendicular to the angle bisector of $ \angle B.$

So, we consider the degenerated quadrilateral $ ABCD,$ where $ D$ is the tangency point of $ (I)$ to $ BC$ and because of $ M,\ I,\ N,$ are the midpoints of the segments $ BC,\ PQ,\ AD$ respectively, it is enough to prove that it is true the relation $ \frac {QA}{QB} = \frac {PD}{PC}$ $ ,(1)$

But, I think $ (1)$ is possible to be proved , by the similar arguments as in the Lemma has already been proved in the previous post.

$ \bullet$ Finally, these simple but difficult results are new to me ( I didn’t find something like that in my archive).

Many thanks to [b][size=100] Vladimir[/size][/b] who proposed this nice problem and to all the other friends who shared out here, their interesting ideas and solutions.

Kostas Vittas.

[quote][color=darkred][b][u]Lemma.[/u][/b] Let $ ABC$ be a triangle. Denote the midpoint $ M$ of the side $ [BC]$ . The incircle $ w \equal{} C(I,r)$ and $ A$ - excircle $ w_a \equal{} C(I_a,r_a)$ touch the

sideline $ BC$ in the points $ E$ , $ F$ respectively. Denote the midpoints $ P$ , $ R$ of the segments $ [AE$ , $ [AF]$ respectively. Then $ P\in IM$ and $ R\in I_aM$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote the diameters $ [ES]$ , $ [FG]$ of the circles $ w$ , $ w_a$ respectively. Prove easily that $ S\in (AF)$ and  $ G\in AE$ . Since $ P$ , $ I$ , $ M$ are the middle

of $ [EA]$ , $ [ES]$ , $ [EF]$ respectively, obtain $ P\in IM$ . Analogously, since $ R$ , $ I_a$ , $ M$ are the middle of  $ [FA]$ , $ [FE]$ , $ [FG]$ respectively, obtain $ R\in I_aM$ .[/color]

My dear friends,

   Newton

  Thank you.
  M.T.

[quote][color=Darkred][b][u]Lemma (Vittasko).[/u][/b] A triangle $ ABC$ is given and let $ F$ be the tangent point of its $ A$ - excircle $ C(I_a,r_a)$  to the

 sideline $ AB$ . Denote the midpoints $ M$ , $ N$ of $ [AB]$ and $ [CF]$ respectively . Prove that $ I_a\in MN$ .[/color][/quote]

[color=darkblue][b][u]Proof (shorter).[/u][/b] Denote the incenter $ I$ of $ \triangle ABC$ and ${ \left\|\begin{array}{c} D\in AI\cap BC \\{
\ E\in CI\cap AB}\end{array}\right\|$ . Observe that $ \frac {MF}{MB} = \frac {a + b}{c} = \frac {IC}{IE}$ . Let  $ X\in BC$ be a

point for which $ IX\parallel AB$ . Thus,  $ \left\{\begin{array}{ccccc} \frac {XB}{XD} = \frac {IA}{ID} = \frac {I_aA}{I_aD} & \implies & \frac {I_aA}{I_aD}\cdot \frac {XD}{XB}\cdot\frac {MB}{MA} = 1 & \implies & \boxed {\ I_a\in MX\ } \\
 \\
\frac {XC}{XB} = \frac {IC}{IE} = \frac {MF}{MB} & \implies & \frac {MF}{MB}\cdot\frac {XB}{XC}\cdot\frac {NC}{NF} = 1 & \implies & \boxed {\ N\in MX\ }\end{array}\right\|$ $ \implies$ $ \boxed {\ I_a\in MN\ }$ .[/color]

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u][/b] Let$ ABC$ be a triangle. Choose a point $ D\in (BC)$ and the points $ U\in (AD)$ , $ V\in AD$ so that $ D\in (AV)$

and the division $ (A,U,D,V)$ is harmonically, i.e. $ \frac {UA}{UD} = \frac {VA}{VD}$ . Denote the intersection $ E\in CU\cap AB$ and define the point $ F\in AB$

so that $ B\in (AF)$ and $ \frac {MB}{MF} = \frac {UE}{UC}$ . Denote the midpoints $ M$ , $ N$ of $ [AB]$ and $ [CF]$ respectively . Prove that $ V\in MN$ .[/color][/quote]

[color=darkblue][b][u]Remark.[/u][/b] For $ U: = I$ and $ V: = I_a$ obtain the upper nice [b]Vittasko's lemma.[/b][/color]

After yetti's problem I have some idea...

We had $ N_a$ be A-Nagel of $ ABC$ and others Nagel points are $ N_b,N_c$. We got $ N_a,J_a,G$ are colinear we call it by A-Nagel line, follow a theorem, $ S_p$ - Spieker center (incenter of median triangle) lies on Nagel line (the line connecting $ N,I,G$), if we call $ S_{pa}$ be the A'-excenter of median triangle $ A'B'C'$ then $ S_{pa}$ lie on A-Nagel line. Actually they are divided by some constants ratio.

For Gergon problem, if we let A-excircle touch the line $ BC$ and ray $ AC,AB$ at $ D$ and $ E,F$, resp. Then $ AD,BE,CF$ are concurrent by $ G_a$ we call as A-Gergon point, similar we got $ G_b,G_c$. The lines $ J_aG_a,J_bG_b,J_cG_c$ are concurrent at a point $ Ge'$ moreover $ Ge',I,Ge$ - Gergon point are colinear.

The above results I think we can prove by barycentric coordinate maybe not easy, but is there any synthetic proof?

To achieve the proof I shall use the following two lemmas:

[b][color=darkblue]Lemma 1[/color][/b]: [i][size=100]Within the conditions of the problem, the following equalities hold:[/size][/i]
$ \begin {eqnarray*} BC \cdot \overline {OD} = BD \cdot \overline {OC} + CD \cdot \overline {OB} \& (\ 1 \ ) \\
AB \cdot \overline {OF} = AF \cdot \overline {OB} - BF \cdot \overline {OA} \& (\ 2 \ ) \end {eqnarray*}$

Proof: The proof comes very easy from the well known fact that $ \overline {BD} = k \cdot \overline BC$ ( 3 ), but $ \overline {BD} = \overline {OD} - \overline {OB} \& , \ \overline {BC} = \overline {OC} - \overline {OB} \ \& and \ k = \frac {BD}{BC}$; substituting all these in (3) we get (1).
For (2) (when $ F$ is beyond the limits of $ \overline {AB}$) we work similarly.

[b][color=darkblue]Lemma 2[/color][/b]: [i][size=100]If the incenter of the triangle $ \triangle ABC$ touches its sides $ BC, \ CA \and \ AB$ at $ X, \ Y \ and\ Z$ respectively, there exists the relation:[/size][/i]
$ \sum AB \cdot \overline {IZ} = \overline 0$ ( 4 )

Proof: As $ AB \cdot \overline {IZ} \perp IZ \cdot \overline {AB}$ and they have the same module, a $ 90^\circ$ transforms the first vector into the second one. If the left part of (4) is noted by $ \overline v$, the vector obtained after the rotation around $ I$ is:
$ \overline v' = \sum IX \cdot \overline {BC} = r \cdot \sum \overline {AB} = \overline 0$ and relation (4) is thus proven.
The relation (4) shows that, multiplying the lengths of sides of a triangle by vectors of equal module and perpendicular to the respective side, the aggregate vector is null.
When $ I$ becomes instead $ O$, our $ A$-excircle, $ \overline {IX}$ and $ \overline {OD}$ have opposite orientations, hence for O the relation (4) is re-written as follows:
$ AB \cdot \overline {OF} + AC \cdot \overline {OE} - BC \cdot \overline {OD} = \overline 0$ ( 5 ) and we are now ready to prove the Lemma of Vittasko:
Let’s name w the length of tangent from W to O, hence the relations (1 )and (2) become:
$ BC \cdot \overline OD = b\cdot \overline {OC} + c \cdot \overline {OB}$( 6 ) ,  $ AC \cdot \overline {OE} = a\cdot \overline{OC} - c\cdot \overline {OA}$ ( 7 ) and, for easier calculation, $ AB \cdot \overline {OF} = (a - b) \cdot \overline {OF}$ ( 8 ); replace the relations (6) to (8) in (5) and get $ (a - b) \cdot (\overline {OF} + \overline {OC}) - c \cdot (\overline {OA} + \overline {OB}) = \overline 0$ ( 9 )
But, if $ M$ and $ N$ are midpoints of $ CF$ and $ AB$ respectively, (9) becomes $ 2(a - b) \cdot \overline {OM} = 2c \cdot \overline {ON}$ and, since not both $ a - b$ and $ c$ can be $ Zero$, it follows that $ O, M$ and $ N$ are collinear.

Note: the idea of this proof does not belong to me, but to professor Dinu Serbanescu, proving by vectors this remarkable theorem of Newton.

Best regards,
sunken rock


How many polynomials of the from x^3-8x^2+cx+d such that c and d are real numbers and the three roots are distinct positive integers?

Let $ m,n,p$ be the polynomial's roots.

Then, $ x^3\minus{}8x^2\plus{}cx\plus{}d \equal{} (x\minus{}m)(x\minus{}n)(x\minus{}p)$.

If we expand the RHS we get $ m\plus{}n\plus{}p\equal{}8$, $ mn\plus{}np\plus{}pm\equal{}c$, $ \minus{}mnp\equal{}d$.

Notice that since $ m,n,p$ are positive integers, $ c$ and $ d$ are necessarily real numbers.

So the only condition here is that $ m\plus{}n\plus{}p\equal{}8$. There are just two possible values for $ \{m,n,p\}$: $ \{1,2,5\}$ and $ \{1,3,4\}$. These give $ c\equal{}17, d\equal{}\minus{}10$ and $ c\equal{}19, d\equal{}\minus{}12$ respectively.

Hence, the answer is $ 2$.

Aren't there also sets of m, n, p like (2,3,3), (2,4,2), etc.  I don't understand how you got to the conclusion there were only two sets of m,n,p..

The problem statement says "and the three roots are [b]distinct [/b]positive integers".

So without loss of generality $ m<n<p$, $ m\equal{}1$ because if not, $ m\plus{}n\plus{}p \geq 2\plus{}3\plus{}4 \equal{} 9$, also $ n\equal{}2,3$, because if not, $ m\plus{}n\plus{}n \geq 1\plus{}4\plus{}5 \equal{} 10$. So those are the only two cases.

Whoops I didn't see the "distinct" part, sorry about that.


function
floor function
combinatorics open
combinatorics

let $n\geq m\geq 1$ is  integers and $A={1,2,...,n}$ ,$M={m,n-m}$.find $k$ is smallest such that in every set $B$ has $k$ elements is subset of $A$ ,always there exist $i<j$ such that $j-i$ to belong to $M$

It easy to see that A can be replaced with $\mathbb{Z}_n$.
We will construct a subset $C\subset\mathbb{Z}_n$ with the [b]maximum[/b] number of elements such that there are no two elements $x,y\in C$ with $x-y=m$. Then $k=|C|+1$.

Consider a function $f(x)=x+m$ defined for $x\in\mathbb{Z}_n$.
Let $O(x)=\{x,f(x),f(f(x)),\dots\}=\{x,x+m,x+2m,\dots\}$ be an orbit of element $x$ under the action of function $f$.
Note that the length of any orbit is $|O(x)|=\frac{n}{\gcd(n,m)}$ and there are $\gcd(n,m)$ [b]distinct[/b] orbits which we denote $O(x_1), \dots, O(x_{\gcd(n,m)})$.

Note that $O(x)\cap C$ can have at most $\left\lfloor \frac{|O(x)|}{2}\right\rfloor$ elements. Moreover, from any orbit $O(x)$  we can select $\left\lfloor \frac{|O(x)|}{2}\right\rfloor$ elements such that there are no two with the difference $m$. And elements of different orbits cannot have difference $m$ either.
Therefore,
\[ |C|=\sum_{i=1}^{\gcd(n,m)} \left\lfloor \frac{|O(x_i)|}{2}\right\rfloor = \gcd(n,m)\cdot \left\lfloor \frac{n}{2\gcd(n,m)}\right\rfloor.  \]

There are two cases depending on whether $\frac{n}{2\gcd(n,m)}$ is integer or not.
If it is, then $|C|=\frac{n}{2}$ and $k=\frac{n}{2}+1$.
If not, then $|C|=\frac{n-\gcd(n,m)}{2}$ and $k=\frac{n-\gcd(n,m)}{2}+1$.

[quote="dieuhuynh"]you were wrong.
 because              {x,x+m,x+2m,...,x+(n-1)m}(mod n)  isn't similar to
        {1,2,...,n}[/quote]
I did not say that.
$O(x)=\{x,x+m,x+2m,...,x+(n-1)m\}$ may or may not be equal to the whole set $\mathbb{Z}_n$.
The size of $O(x)$ is $\frac{n}{\gcd(n,m)}$. We have $|O(x)|=n$ only when $\gcd(n,m)=1$.

i am vietnam.email:trinhcongdien_511
  can you  send to me?

as i know  you replace {1,2,...n}  :arrow: {x,x+m,...,x+(n-1)m}mod n
  yes?

you can see
http://www.mathlinks.ro/Forum/viewtopic.php?p=267973#p267973