calculus
induction
algebra
polynomial
geometry
rectangle
strong induction

[img]http://img251.imageshack.us/img251/7664/ps3q3questionk.jpg[/img]

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Hello there,

This question is for a first-year calculus course, which does not involve and assumes no previous background in statistics and combinatorics. But we are currently reviewing mathematical induction so this problem has been assigned.

I can do part a).

For part b), I have found the following (I’m not sure about #3 though).

I) If I place one 2 x 2 square tile on a 2 x n board, then I have $ T_{n\minus{}2}$ possibilities to fill up the rest of the board.

2) If I place one 2 x 1 tile (horizontal one), then I have $ T_{n\minus{}1}$  possibilities to fill up the rest of the board.

3) If I place one 1 x 2 tile (vertical one), then I have  $ T_{n\minus{}2} \plus{} 1$ possibilities to fill up the rest of the board.

Also, it seems that $ T_n$ = (case 1 + case 2 + case 3) many possibilities.

However, I’m not sure how to rewrite $ T_n$ using these facts and thus how to complete b).

I think I need these statements to help me with part c).

Thank you very much.

[quote="catalan"]I) If I place one 2 x 2 square tile on a 2 x n board, then I have $ T_{n \minus{} 2}$ possibilities to fill up the rest of the board.
2) If I place one 2 x 1 tile (horizontal one), then I have $ T_{n \minus{} 1}$  possibilities to fill up the rest of the board.
3) If I place one 1 x 2 tile (vertical one), then I have  $ T_{n \minus{} 2} \plus{} 1$ possibilities to fill up the rest of the board.
Also, it seems that $ T_n$ = (case 1 + case 2 + case 3) many possibilities.[/quote]  Yes, yes, no, and yes.  What part of writing down the sum of three known quantities is proving problematic for you?

If you start with a vertical tile in your upper-left corner, you are obliged too use another one next to it, so you are left with a $ 2 \times (n \minus{} 2)$ table (you miscounted there).
So your recurrence relation is $ T_n \equal{} T_{n \minus{} 1} \plus{} 2T_{n \minus{} 2}$, with clearly $ T_1 \equal{} 1$, $ T_2 \equal{} 3$.
But all this leads you nowhere towards c) if you don't know the theory behind linear recurrence relations.
[hide="Hint"]The characteristic polynomial is $ \lambda^2 \minus{} \lambda \minus{} 2 \equal{} 0$, with roots $ 2$ and $ \minus{} 1$; then $ T_n \equal{} \alpha2^n \plus{} \beta( \minus{} 1)^n$. How do you determine the coefficients $ \alpha$ and $ \beta$ ?[/hide]

[quote="mavropnevma"]But all this leads you nowhere towards c) if you don't know the theory behind linear recurrence relations.[/quote]  This isn't true: you can guess the answer (since your professor kindly told you what it is), so you just induct!  (Not that this is more [i]satisfying[/i], but it works fine for any particular instance like this one.  And, of course, it's the method that the exercise is trying to illustrate.)

Hello there,

Thank you for your replies. I have a few more questions based on them:

1. [quote]If you start with a vertical tile in your upper-left corner, you are obliged too use another one next to it, so you are left with a 2 \times (n - 2) table (you miscounted there). [/quote]

Could you please explain why it is still $ 2 \times (n \minus{} 2)$? At the top, there would still be two grids beside the vertical (1 x 2) tile which must be filled, with another vertical (1 x 2) tile? If I express the possibilities as $ 2 \times (n \minus{} 2)$, am I not ignoring the lack of the other vertical (1 x 2) tile?

2. [quote]This isn't true: you can guess the answer (since your professor kindly told you what it is), so you just induct! (Not that this is more satisfying, but it works fine for any particular instance like this one. And, of course, it's the method that the exercise is trying to illustrate.)[/quote]

I think that this is the intended approach, since I don't believe that we will learn linear recurrence relations in this course.

I am familiar with basic mathematical induction, but how would I go about strong induction?

Would I have to use the formula given in part (c) to express $ T_n$ in terms of $ T_{n \minus{} 1}$ and $ T_{n \minus{} 2}$? Also, I know that in strong induction, all values up to k must be assumed to be true. Therefore, would I show the base cases (part a), assume that all values up to k are true, and then use my induction hypothesis to show that (k + 1) holds?

Thanks a lot!

To part 1. Yes, as you describe, after using a vertical tile, you are not ignoring the lack of the other vertical $ 1 \times 2$ tile, since, as I said, you are FORCED to occupy this space with yet a vertical tile!
 What is left now is a $ 2 \times (n\minus{}2)$ table, etc.

To part 2., jbl, whom you quoted, noted that you teacher supplied the formula for $ T_n$. Indeed, use strong induction: check it for the small values ($ T_1$ and $ T_2$); assume it works for all $ k \leq n\minus{}1$, and after writing the formula $ T_n \equal{} T_{n\minus{}1} \plus{} 2T_{n\minus{}2}$ apply the induction hypothesis for the lower values, the do the computations to check it provides the answer required.

Thanks for your reply, mavropnevma.

I think that I have the idea behind the proof but I am stuck on a step involving algebra. Could you please correct any mistakes or explain that step?

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$ T_n \equal{} \frac {2^{n \plus{} 1} \plus{} ( \minus{} 1)^{n}}{3}$

I have found my base cases to be true and my induction hypothesis is that the above will hold true for all values $ < n$.

By induction hypothesis, $ T_n \equal{} T_{n \plus{} 1} \plus{} 2 \times T_{n \plus{} 2}$

$ RS \equal{} \frac {2^{n} \plus{} ( \minus{} 1)^{n \minus{} 1}}{3} \plus{} 2 \times \frac {2^{n \minus{} 1} \plus{} ( \minus{} 1)^{n \minus{} 2}}{3}$

$ \equal{} \frac {2^{n} \plus{} ( \minus{} 1)^{n \minus{} 1} \plus{} 2^n \plus{} 2( \minus{} 1)^{n \minus{} 2}}{3}$

$ \equal{} \frac {2^{n \plus{} 1} \plus{} ( \minus{} 1)^{n \minus{} 2}}{3} \neq T_n$

Have I erred in algebra? Thanks again.

Ah at which part are you confused? Suppose you have $ T_{n \equal{} 1}$. Does this make sense for the formula you gave? The $ ( \minus{} 1)^{n \minus{} 2}$ factor acts a bit odd. (Haha :P)
You want that to have an exponent of $ n$ which is confusing you I suppose. Well why can't you just make it $ n$? Instead of $ n \minus{} 2$. Think about whether or not the sign of $ ( \minus{} 1)$
changes if you change from $ n$ to $ n \minus{} 2$.
Hint : Suppose you substract $ 2$ from an [i]odd[/i] number, what is the parity of the remaining number?
 Suppose you substract $ 2$ from an [i]even[/i] number, what is the parity of the new number?

Thanks for your reply, Fermat's Little Turtle.

I missed that intuitive step.

For the above problem, would $ ( \minus{} 1)^{n \minus{} 2} \Rightarrow ( \minus{} 1)^n$ since the effects of both exponents result in the same parity of any given integer?

If n is odd, then n - 2 is odd. If n is even, then n - 2 is even.

Yep. It's a really good thing to remember, usually in combinatorics. For factors of $ ( \minus{} 1)$, it's exponent is only relevant being congruent to $ 0$ or $ 1 \mod 2$. Thus you can change the exponent to sort of meet the needs of your question, ie. to make sense.
In fact if you want to test your professor's view on such an idea, leave the exponent as $ n\minus{}2$ and see how he marks it. Or be safe and change it (justified) and then ask him for his input. I'm sure his experience can offer plenty of insight into lots of stuff indirectly related to exponents.

Sort of aside think about the question still. Just to know more math =)
You have an even number with certain exponents, and either you're subtracting $ 1$ or adding $ 1$ to this even number.
Also you're dividing this sum by $ 3$, and hopefully it's understood $ T_n$ is always an integer.
Don't you find that neat, how $ 3$ divides either into $ 1$ less than a power of $ 2$ or $ 1$ more? Some fun number theory stuff there.
Consider $ 8,16,32,64$.  Subtracting $ 1$ from each gives $ 7,15,31,63$, and adding $ 1$ to each gives $ 9,17,33,65$. Notice the divisibility issues?
If $ n$ is even then $ 2^n \minus{} 1$ is divisible by $ 3$, but not $ 2^n \plus{} 1$. Similarly if $ n$ is odd, $ 2^n \minus{} 1$ is not divisible by $ 3$ but $ 2^n \plus{} 1$ is.
This can help you realize that parity arguments are very powerful, and lots of mathematics relates to this basic idea; parity.

I don't recall professors in first year or second offering ANY insight into ANY questions. It's disappointing looking back at them. I've learned
to think about such things, even the basic stuff. You should too =) Try to see what certain things are telling you.

Sorry its long and a little off topic =)

Not at all! Your supplement is very useful and I'm always interested in thinking more about what I am doing.

Thank you very much again!

Another way to do this problem would be to write it as a sum involving some binomial coefficients and powers of 2 (to count the number of ways to break the $ 2 \times n$ rectangle up into $ 2 \times 2$ blocks and $ 2 \times 1$ horizontal rectangles, and then to choose whether the $ 2 \times 2$ blocks should be whole blocks or two vertical rectangles).

I wonder, though, can we give a direct heuristic explanation for this answer, e.g., can we find some set of size $ n \plus{} 1$ such that about one third of its subsets correspond to good tilings of this rectangle?  (Or, you know, a set of size $ n$ for which about two thirds of the subsets correspond to good tilings, or whatever.)


function integration probability inequalities

Can some [b]please [/b] explain to me the uncertainity priniciple?

All it says is that it is impossible to measure both momentum and position at the same time with exact certainty.

What I'm about to describe isn't the uncertainty principle, not exactly. It's an analogue in classical mechanics. However, it is related to the quantum mechanical principle in that notions of waves underlie both phenomena.

Play a note on a musical instrument. What pitch was it - what frequency? Say it's an A 440 Hz. But is that really its frequency, 440 Hz? If the signal (the local variation in air pressure, cause by some vibrating object in the instrument) is a perfect sinusoidal curve that has been going on forever and will go on forever - the function $ f(t)\equal{}A\cos(440\cdot2\pi t\minus{}\delta),$ defined on the whole real line - then yes, we can say that the frequency is exactly 440. But real sounds don't go on forever, they start and stop. And in the act of starting and stopping, the function differs from that perfect sinusoid and in differing from that perfect sinusoid, it can't be said to be exactly that frequency.

One way to look at this: think of the signal function $ f(t)$ as being composed of a linear combination of sinusoids: $ f(t)\equal{}\int_{\minus{}\infty}^{\infty}\widehat{f}(\nu)e^{2\pi i \nu t}\,d\nu.$

(This is called the Fourier transform.)

Now, scale things so that $ \int_{\minus{}\infty}^{\infty}|f(t)|^2\,dt\equal{}1.$ That means we can think of $ |f(t)|^2$ as being a probability distribution for the time $ t.$ By a theorem called the Plancherel theorem, we can now also regard $ |\widehat{f}(\nu)|^2$ as a probability distribution for the frequency $ \nu.$

Now suppose these probability distributions have means and standard deviations. The mean of $ t$ is "the time" at which the sound occurred. The standard deviation $ \sigma_t$ of this distribution is a measure of the "width" of the sound - the uncertainty in the time it occurred. Similarly, the mean of $ \nu$ is "the frequency" of the sound, and $ \sigma_{\nu}$ is the uncertainty of that frequency.

I can then prove the following, as a theorem in Fourier analysis: for all such signal profiles $ f(t),$ we have $ \sigma_t\cdot \sigma_{\nu}\ge \frac1{4\pi}.$

(Note: the exact form of the inequality comes from me choosing to express frequency in Hz and write sinusoids in the form $ e^{2\pi i\nu t}.$ If I used radians/sec and wrote sinusoids as $ e^{i\omega t},$ the details would be slightly different; the quantity on the other side of the inequality would be $ 1/2$ instead of $ 1/(4\pi).$)

The essence of an uncertainty relation is in that inequality. If we want to be very sure of the pitch - to have a very small frequency uncertainty, then the uncertainty in time must be large. It must be a long steady note. On the other hand, if we play a very short note, then the frequency uncertainty must be large.

The bottom two notes on a piano, A and A#, nominally have frequencies of 27.5 and 29.1 Hz. If we try to trill these notes back and forth faster than about 4 times a second, the resulting frequency uncertainty is so large that that it overwhelms the 1.6 Hz nominal difference in pitch. We don't perceive two different notes. But the difference between A and A# four octaves higher is between 440 and 466 Hz, and that 26Hz difference is large enough that any humanly possible trill still sounds like two notes.

A mathematical formulation of the Uncertainty Principle is
$ \Delta x\Delta p\geq \frac{h}{4\pi}$
where $ h$ is Planck's constant.  For more, see
http://en.wikipedia.org/wiki/Uncertainty_principle

[quote="JRav"]A mathematical formulation of the Uncertainty Principle is
$ \Delta x\Delta p\geq \frac {h}{4\pi}$
where $ h$ is Planck's constant.  For more, see
http://en.wikipedia.org/wiki/Uncertainty_principle[/quote]

actually the mathematical formulation is exactly what kent said it was -- if $ f\in L^2(\mathbb{R})$ with $ \|f\|_{L^2(\mathbb{R})}\equal{}1$, then

\[ \int_{\mathbb{R}} x^2|f(x)|^2dx\int_{\mathbb{R}} \xi^2|\widehat{f}(\xi)|^2d\xi \geq \frac{1}{16\pi^2}
\]

you don't need to assume that $ \|f\|_{L^2(\mathbb{R})}\equal{}1$, but the cost is a scaling factor of $ \|f\|^4_{L^2(\mathbb{R})}$ on the RHS.  there are slightly more general versions of this this theorem which come from replacing $ f(x)$ with $ e^{\minus{}2\pi ic_1 x}f(x\plus{}c_2)$, but once you have the above you basically have everything.

I include the proof of the theorem blahblahblah stated (and its generaliztion, which he hinted at) in my class notes when I teach Fourier Analysis.

One note: if $ f$ is Gaussian, then the inequality becomes an equality. But "rougher" functions than that tend to exceed the inequality. If you put in the "box" function (the characteristic function of an interval balanced at the origin), then its Fourier transform, the "sinc" function, gives us

$ \int_{\mathbb{R}}\xi^2|\widehat{f}(\xi)|^2\,d\xi\equal{}\infty$

which satisfies the inequality, all right, but in a bad way.

The form I included is one that is easy to understand if you don't know Fourier analysis.  It says that the product of the uncertainty in position and momentum must exceed some constant.


calculus
inequalities
derivative
function
Gauss
algebra
system of equations

I do not know much about multivariable calculus, but can it be used to solve multivariable inequalities (using brute force) like how single variable calculus can be used to solve single variable inequalities?

Yes, but the conditions for having a maximum or minimum in many variables are much more complicated than they are for one variable.

Oh, so you can't just go and find when the derivative of a multivariable function = 0?

Well, the problem with multivariable functions is that the derivative is dependant on the direction you're heading.  In a single-variable function, you can go left or right.  But in a bivariable function, there are infinitely many directions you can head from each point.  I don't remember what the conditions under which you could have maximums and minimums are, but they aren't nearly so nice looking as f' = 0.  I'm sure someone else can answer your question more specifically (gauss?  ravi?  simon?) but I'm actually just learning this right now, so I won't be able to tell you until Friday :)

Ah, I see. Naw, I don't need to know the specifics, I was just wondering whether many olympiad inequalities could be brute-forced using calculus.

Thanks for the response.

You can always brute-force inequality problems with calculus. In most situations though, I think I'd rather get a 0 than do a calculus solution and get a 7 on an olympiad inequality (unless I thought it would make the difference between getting into MOP and not getting in or something like that). It's still f'=0 for extrema in multivariable calculus, but there are other situations in which f' can be 0 other than extrema. (Well, that's the case in normal calculus too, but in multivariable they're even weirder sometimes.)

It's possible to use multidimensional calculus for some inequalities, but it can be messy.  Some issues are:

(1) It may not be easy to find all local mins.  Yes, you can set all n partial derivatives to 0, but then you have a system of n equations to solve.  Also, as ComplexZeta alluded to, besides local mins and maxes, there are saddlepoints, which don't arise in one dimension.

(2) Even if you find all local mins, how do you prove they are global mins?

(3) Many inequalities have some side conditions.  ("If something :ge: 0, then something else :ge: 0.")  Those conditions be hard to deal with.  There is a technique called Lagrange multipliers that sometimes handles them.

I don't think there should be any de facto aversion to using calculus for an Olympiad problem - but most of the time it's simply impractical. This is because finding the critical points of a multivariable function often turns into solving a system of (not necessarily linear) equations; and this system of equations is rarely solvable by hand (or even if it is, at least not any more easily than the original inequality itself).

Plus once you find the critical points you have to apply a test analogous to the "second derivative test" to determine if they're relative max's, min's, or neither. This test can be a pain in the butt. And if there are boundary conditions it makes life even more comlicated.

So in the end, solve inequalities using calculus isn't really practical for contest purposes. And most Olympiad inequalities are specifically designed to resist such attacks. Although every once in a while, you [i]might[/i] catch lightening in a bottle and be able to solve one nicely this way. I wouldn't recommend it as your first (or even second or third) line of attack though.

unless you really know what you are doing, dont even attempt lagrange multipliers on an olympiad contest


Όπως είδατε είναι πολύ καλύτερα να  έχουμε τις ασκήσεις μαζεμένες , κάτι σαν Γεωμετρικό Μαραθώνιο .Ας κρατήσουμε τη σειρά , ώστε κάποια στιγμή που οι ασκήσεις θα επαναλαμβάνονται  και να μπορούμε να τις βρίσκουμε , αλλά και να δίνουμε εύλολα την υπόδειξη.

Μπάμπης

Δίνεται τετράπλευρο ΑBCD και σημεία Ε , F στο εσωτερικό του τέτοια ώστε
                   AE = BE , CE = DE , γωνAEB=γωνCED
και
                   AF = DF , BF = CF , γωνAFD = γωνBFC
να δείξετε ότι γωνAFD + γωνAEB = π .

[quote="nikos"]Δίνεται τετράπλευρο $ABCD$ και σημεία $E , F$ στο εσωτερικό του τέτοια ώστε
                   $AE = BE , CE = DE , \angle AEB = \angle CED$
και
                   $AF = DF , BF = CF$ , $\angle AFD = \angle BFC$
Να να αποδείξετε ότι  $\angle AFD+\angle AEB = 180$ .[/quote]

[size=100][b]ΑΣΚΗΣΗ 12 - ΕΚΦΩΝΗΣΗ.[/size][/b]

Δίδεται τρίγωνο $%Error. "bigtriangle" is a bad command.
ABC,$ σημείο $P,$ στο εσωτερικό του και έστω $D,$ $E,$ $F,$ τα σημεία τομής των πλευρών του $BC,$  $CA,$ $AB$ αντιστοίχως, από τις ευθείες $AP,$ $BP,$ $CP.$ Δια του $P,$ φέρνουμε τις παράλληλες ευθείες προς τις $DE,$ $DF,$ οι οποίες τέμνουν την $BC,$ στα σημεία $K,$ $L$ αντιστοίχως. Αποδείξτε ότι $DK = DL.$

Κώστας Βήττας.

[size=100][b]ΑΣΚΗΣΗ 13 – ΕΚΦΩΝΗΣΗ.[/size][/b]

Δια των κορυφών $B,$ $C,$ δοσμένου τριγώνου $\bigtriangleup ABC,$ φέρνουμε τις κάθετες ευθείες επί την $BC,$ οι οποίες τέμνουν την εξωτερική διχοτόμο της γωνίας $\angle A,$ στα σημεία $F,$ $E,$ αντιστοίχως. Αποδείξτε ότι οι ευθείες $BE,$ $CF$ και η δια του $A,$ διάμετρος του περιγεγραμμένου κύκλου του $\bigtriangleup ABC,$ τέμνονται στο ίδιο σημείο.

Κώστας Βήττας.

Πολύ ωραίες ασκήσεις . (Ελπίζω να μη γράψουν πάλι ότι αντέγραψα) .
[color=brown]ΑΣΚΗΣΗ 12-ΛΥΣΗ[/color]
Έστω ότι η παράλληλη από το P στην DE τέμνει την ΑC στο Μ και η παράλληλη από το P στην DF τέμνει την ΑB στο N .
Από το θεώρημα Θαλή έχουμε $\frac{EM}{EA}=\frac{DP}{DA}$
kai $\frac{FN}{FA}=\frac{DP}{DA}$ άρα
$\frac{EM}{EA}=\frac{FN}{FA}$ άρα ΝΜ//ΕF και
$\frac{EM}{FN}=\frac{EA}{FA}$ (1).
Από το θεώρημα Ceva έχουμε ότι $\frac{EA}{FA}=\frac{BD\cdot CE}{BF\cdot CD}$ .Άρα η (1) γίνεται
$\frac{EM}{FN}=\frac{BD\cdot CE}{BF\cdot CD}$
ή $\frac{EM\cdot CD}{CE}=\frac{BD\cdot FN}{BF}$ (2)

Όμως από το θεώρημα Θαλή και πάλι
$DK=\frac{EM\cdot CD}{CE}$  και $DL=\frac{BD\cdot FN}{BF}$

Άρα η (2) δίνει $DK=DL$  , το ζητούμενο
Πώς σας φάνηκε ?

[b][u]ΓΕΩΜΕΤΡΙΑ 12- ΛΥΣΗ[/u][/b]

Έστω $G$ η τομή της $PK$ με την $AC$ και $H$ η τομή της $PL$ με την $AB$.

$HP // FD \Leftrightarrow \frac{AF}{HF}=\frac{AD}{PD}$ $(1)$.

$PG // DE \Leftrightarrow \frac{AE}{GE}=\frac{AD}{PD}$ $(2)$.

Από $(1)$ και $(2)$ προκύπτει $\frac{AF}{HF}=\frac{AE}{GE}\Leftrightarrow \frac{AF}{AE}=\frac{HF}{GE}$ $(3)$.

Επειδή $KG // DE$ ισχύει $\frac{CE}{CD}=\frac{EG}{KD}$ $(4)$.
Επειδή $HL // FE$ ισχύει $\frac{BD}{FB}=\frac{DL}{FH}$ $(5)$.

Από το θεώρημα Ceva για το τρίγωνο $ABC$ παίρνουμε ότι ισχύει $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$ και αντικαθιστώντας τις $(3),(4)$ και $(5)$ βγαίνει $\frac{HF}{GE}\cdot\frac{EG}{KD}\cdot\frac{DL}{FH}=1 \Leftrightarrow DK=DL$.

Στέλιος

[Edit]: Σιλουανέ με πρόλαβες και έστειλες καθώς έγραφα τη λύση μου... .   :mad: :P

Μπράβο κι στους δυο σας κι ειδικα στον Σιλουανό για την λύση του [hide](ασ με συγχωρήσαει ο stedes γιατι θελω να  πειράξω τον κομπλέξα)[/hide]Συνεχίστε έτσι

[b][color=red]ΑΣΚΗΣΗ 14[/color][/b]

 [i]Λοιπόν , ο Θαλής πέρασε με χαμόγελα! Για να επαναληφτούν τα χαμόγελα στον Ευκλείδη , ξανά στη δουλειά! Ορίστε ένα πολύ καλό θέμα για γερή μελέτη.Το παίρνω από ένα CRUX του 1993.[/i]

Έστω $P$ ένα σημείο του περιγεγραμμένου κύκλου του τριγώνου $ABC$.Έστω $D,E$ τα ίχνη των καθέτων από το $P$ προς τις ευθείες $BC, CA$  αντίστοιχα.Αν $L,M$ είναι τα μέσα των τμημάτων $AD, BE$ , αντίστοιχα , να αποδείξετε ότι  $DE \perp LM$

 ΜΠΑΜΠΗΣ

[b][color=red]ΑΣΚΗΣΗ 15[/color][/b]

 [color=blue]Για τον Ευκλείδη ( Α΄, Β΄Λυκείου )[/color]

 Δίνεται τρίγωνο $ABC$ , στο οποίο ο εγγεγραμμένος κύκλος έχει κέντρο $I$ και εφάπτεται των πλευρών $BC,CA,AB$  στα σημεία $D,E,F$ αντίστοιχα. Η παράλληλη από το $A$ προς τη $BC$ τέμνει την ευθεία $DE$  στο σημείο $G$.Αν $M$ είναι το μέσο του τμήματος $DG$ , να αποδείξετε ότι :
   [b]a) [/b]$AM // DF$

   [b]b)[/b] $IM \perp AM$

   [b]c) [/b]Τα σημεία  $B,I,M$ είναι συνευθειακά.

ΜΠΑΜΠΗΣ

[b][color=red]ΑΣΚΗΣΗ 15 - ΛΥΣΗ[/color][/b]

[hide]
Το $\triangle CDE$ είναι προφανώς ισοσκελές. Μάλιστα, είναι όμοιο με το $\triangle AGE$ αφού έχουν τις γωνίες τους μία προς μία ίσες. Άρα και το $AGE$ είναι ισοσκελές $(AG=AE)$
Έστω $G'$ το σημείο τομής της $DF$ με την $AG$.
Με τον ίδιο τρόπο βρίσκουμε ότι και το $AG'F$ είναι ισοσκελές $(AF=AG')$

Άρα $A$ μέσον του $GG'$ επομένως η $AM$ είναι μεσοπαράλληλος του τριγώνου $GDG'\Rightarrow AM \parallel DF$

Έστω $w$ ο κύκλος με διάμετρο το $GG'$.
$AG=AG'=AE=AF \Rightarrow F,E \in w \Rightarrow FG \perp DF$

Άρα ο κύκλος με διάμετρο το $DG$ και κέντρο το $M$ διέρχεται από το $F \Rightarrow MF=MD$ δηλαδή το $M$ ανήκει στη μεσοκάθετο του $FD$ που δεν είναι άλλη από την ευθεία $BI$. Άρα $B,I,M$ συνευθειακά

Και αφού $MI \perp DF \Rightarrow MI \perp AM$[/hide]

[size=100][b]ΑΣΚΗΣΗ 16 – ΕΚΦΩΝΗΣΗ.[/size][/b]

Δίδονται δύο παραλληλόγραμμα $ABCD,$ $A'B'C'D',$ τυχαία συσχετισμένα στο επίπεδο ή στο χώρο. Αποδείξτε ότι τα μέσα των ευθυγράμμων τμημάτων $AA',$ $BB',$ $CC',$ $DD',$ είναι επίσης, κορυφές παραλληλογράμμου.

Κώστας Βήττας.

[size=100][b]ΑΣΚΗΣΗ 17 – ΕΚΦΩΝΗΣΗ.[/size][/b]

Δίδεται τετράπλευρο $ABCD$ και έστω $E,$ $F,$ τα σημεία τομής των απέναντι πλευρών του $AB,$ $CD$ και $BC,$ $AD,$ αντιστοίχως. Επί των πλευρών $AB,$ $BC,$ $CD,$ $AD,$ λαμβάνουμε αντιστοίχως τα σημεία $A',$ $B',$ $C',$ $D'.$
Αποδείξτε ότι εάν οι διαγώνιες $B'D',$ $A'C',$ του $A'B'C'D',$ διέρχονται αντιστοίχως δια των σημείων $E,$ $F,$ τότε τα σημεία $E'\equiv A'B'\cap C'D',$ $F'\equiv B'C'\cap A'D',$ κείνται επί των ευθειών των διαγωνίων $AC,$ $BD$ αντιστοίχως, του τετραπλεύρου $ABCD.$

Κώστας Βήττας.

Me arxi anaforas to simeio $O$, ean $M_{1},M_{2},M_{3},M_{4}$ einai ta mesa twn $AA',BB',CC',DD'$ exoume

\begin{eqnarray}\overrightarrow{M_{2}M_{1}}=\overrightarrow{OM_{1}}-\overrightarrow{OM_{2}}&=&\frac{\overrightarrow{OA}+\overrightarrow{OA'}}{2}-\Big(\frac{\overrightarrow{OB}+\overrightarrow{OB'}}{2}\Big) \nonumber \\ &=& \frac{\overrightarrow{OA}-\overrightarrow{OB}}{2}+\frac{\overrightarrow{OA'}-\overrightarrow{OB'}}{2}=\frac{\overrightarrow{BA}}{2}+\frac{\overrightarrow{B'A'}}{2}\nonumber \\ &=&\frac{\overrightarrow{CD}}{2}+\frac{\overrightarrow{C'D'}}{2}=\ldots=\overrightarrow{M_{3}M_{4}}\nonumber \end{eqnarray}

Ara to tetrapleuro me korifes ta $M_{1},M_{2},M_{3},M_{4}$ exei $2$ apenanti korifes ises kai paralliles. Ara einai parallilogrammo.

Alexandros

Αλέξανδρε σ' ευχαριστώ για την πολύ σύντομη και όμορφη διανυσματική λύση και αξίζει να σημειωθεί ο μικρός χρόνος απάντησης (τριάντα λεπτά!!).
Η συνθετική λύση που έχω υπόψη μου, είναι εφαρμογή γνωστού θεωρήματος, δύο αποδείξεις του οποίου δημοσιεύτηκαν πρόσφατα στο Αγγλικό φόρουμ.

http://mathlinks.ro/Forum/viewtopic.php?t=97493

Κώστας Βήττας.

ΥΓ. Αλέξανδρε, πολύ λίγα θυμάμαι από τη θεωρία των διανυσμάτων και έτσι δεν καταλαβαίνω πως προκύπτει η παραλληλία των $\overrightarrow{M_{2}M_{1}},$ $\overrightarrow{M_{3}M_{4}}.$ Ζητώ συγνώμη και σας παρακαλώ αν γίνεται να μου το εξηγήσετε.

[size=100][b]ΑΣΚΗΣΗ 18 – ΕΚΦΩΝΗΣΗ.[/size][/b]

Επί ευθείας $(\ell),$ δίδονται κατά σειρά τα σημεία $A,$ $B,$ $C,$ και έστω $(O_{1}),$ $(O{2}),$ τα ημικύκλια με διαμέτρους τα ευθύγραμμα τμήματα $AB,$ $BC,$ αντιστοίχως, προς το αυτό μέρος της $(\ell),$ $($ $AB\neq BC$ $).$
Με διάμετρο, τη διάκεντρο $O_{1}O_{2},$ των δοσμένων ημικυκλίων, γράφουμε το ημικύκλιο $(O),$ το οποίο τέμνει τα $(O_{1}),$ $(O{2}),$ στα σημεία $D,$ $E$ αντιστοίχως και έστω $D',$ $E',$ τα σημεία στα οποία η ευθεία $DE,$ τα επανατέμνει. Αποδείξτε ότι $DD' = EE'.$

Κώστας Βήττας.

Κalispera Kwsta,

Loipon,
eks'orismou, otan dio dianismata einai isa tote ektos apo to oti exoun isa metra, simainei oti einai kai siggramika (idia dieuthinsi kai fora). Isotita dianismatwn diladi den simainei mia sxesi apla pou afora ta metra alla einai kati isxirotero. Afora kai tin parallilia twn dianismatwn kathws episis kai tin omoia fora tous.
Ara pairnoume kai to zitoumeno se auto pou (ipothetw) rwtas.

An i erwtisi exei na kanei me tis teleies pou leipoun simplirwnw tin apodeiksi :  $\frac{\overrightarrow{CD}}{2}+\frac{\overrightarrow{C'D'}}{2}= \frac{\overrightarrow{OD}-\overrightarrow{OC}}{2}+\frac{\overrightarrow{OD'}-\overrightarrow{OC'}}{2}=\frac{\overrightarrow{OD}+\overrightarrow{OD'}}{2}-\Big(\frac{\overrightarrow{OC}+\overrightarrow{OC'}}{2}\Big)=\overrightarrow{OM_{4}}-\overrightarrow{OM_{3}}=\overrightarrow{M_{3}M_{4}}$

Na eksigisw kai 2-3 vasika pragmatakia epipleon gia plirotita.

Genika dio dianismata legontai parallila otan exoun ton idio forea 'h parallilous foreis.

Gi'auto apodeiknietai (eukola) oti ta dianismata $\vec{a}$ kai $\vec{b}\neq\vec{0}$ einai parallila an-n iparxei arithmos $\lambda\in\mathbb{R}$ tetoios wste $\vec{a}=\lambda\cdot\vec{b}$.

Episis na thimisw oti an $\overrightarrow{AB}$ ena dianisma tote gia opoiodipote simeio $O$ tou epipedou (kai kat'epektasin kai tou xwrou) exoume $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$. Ta $\overrightarrow{OB},\overrightarrow{OA}$ onomazontai dianismata thesews twn $B,A$ antistoixa kai to $O$ onomazetai simeio anaforas. Gi'auto i parapanw sxesi ekfrazei tin gnwsti protasi "Kathe dianisma ston xwro einai iso me tin dianismatiki aktina tou peratos meion tin dianismatiki aktina tis arxis".

Telos an $\overrightarrow{AB}$ ena dianisma tou xwrou kai $O$ ena simeio anaforas, tote gia tin dianismatiki aktina tou mesou $M$ tou tmimatos $AB$ exoume an $\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}$ kai $\overrightarrow{OM}=\overrightarrow{OB}+\overrightarrow{BM}$ opote me prosthesi kata meli prokiptei i gnwsti sxesi gia tin dianismatiki aktina mesou tmimatos (Dioti $\overrightarrow{AM}+\overrightarrow{MB}=\overrightarrow{0}$ ws antitheta dianismata)

$\boxed{\overrightarrow{OM}=\frac{\overrightarrow{OA}+\overrightarrow{OB}}{2}}$

(Vasiki efarmogi tou parapanw: Estw trigwno $ABC$. To simeio $G$ einai to kentro varos tou trigwnou an kai mono an isxiei $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}$ kai gia opoiodipote simeio $O$ isxiei $\overrightarrow{OG}=\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}$).

Asfalws eimai ipermaxos tis Eukleideias Gewmetrias. Apla kamia fora ta dianismata mas dinoun omorfes kai kompses liseis gia kapoia thewrimata pou stin Eukleideia Gewmetria einai ligo pio diskolo na apodeixthoun (FISIKA isxei kai to antistrofo).

Alexandros

[quote="stergiu"][b][color=blue]ΑΣΚΗΣΗ 18, ΛΥΣΗ[/color][/b]

[b]Καλή και διδακτική ! [/b]

[color=red]ΥΠΟΔΕΙΞΗ [/color]

 [hide]Αν $K, L, M$  είναι οι προβολές των $O_{1}, O, O_{2}$ πάνω στην ευθεία  $DE$ , τότε το $O_{1}KMO_{2}$  είναι τραπέζιο , το $OL$ είναι διάμεσος και έτσι :
 $LD = LE , LK = LM$. Άρα $DK = EM$ , οπότε και τα διπλάσιά τους είναι ίσα.[/hide]

[u]Μπάμπης[/u][/quote]

Poli kali lisi!!!

Opws paratirw apo tin apodeiksi, nomizw oti den einai anagaio oi kikloi $(O_{1})$ kai $(O_{2})$ na einai diadoxikoi (na efaptontai diladi ekswterika sto $B$). Nomizw oti arkei kai mono na paroume dio tixaious kiklous $(O_{1})$ kai $(O_{2})$ pros tin idia meria tis $l$ kai meta me diametro tin diakentro na paroume ton kiklo $(O)$ kai na sinexisoume opws leei i ekfwnisi tis askisis. Tote nomizw to apotelesma den allazei... Xanw mipws tipota?

Alexandros

[quote="cretanman"]Κalispera Kwsta,

Loipon,...

Asfalws eimai ipermaxos tis Eukleideias Gewmetrias. Apla kamia fora ta dianismata mas dinoun omorfes kai kompses liseis gia kapoia thewrimata pou stin Eukleideia Gewmetria einai ligo pio diskolo na apodeixthoun (FISIKA isxei kai to antistrofo).

Alexandros[/quote]

Αλέξανδρε σε ευχαριστώ για το χρόνο σου. Τελικά, είναι σφάλμα το να ξεμακραίνει κανείς τόσο, ώστε να ξεχνάει ακόμα και τα βασικά, με άλλοθι το βιοπορισμό του.

Και πάλι σε ευχαριστώ, Κώστας Βήττας

[quote="cretanman"]
Poli kali lisi!!!

Opws paratirw apo tin apodeiksi, nomizw oti den einai anagaio oi kikloi $(O_{1})$ kai $(O_{2})$ na einai diadoxikoi (na efaptontai diladi ekswterika sto $B$). Nomizw oti arkei kai mono na paroume dio tixaious kiklous $(O_{1})$ kai $(O_{2})$ pros tin idia meria tis $l$ kai meta me diametro tin diakentro na paroume ton kiklo $(O)$ kai na sinexisoume opws leei i ekfwnisi tis askisis. Tote nomizw to apotelesma den allazei... Xanw mipws tipota?

Alexandros[/quote]

Έτσι είναι. Όταν ένα χαρακτηριστικό της εκφώνησης δεν μας χρειάζεται για την απόδειξη, τότε το πιθανότερο είναι ότι το πρόβλημα ισχύει και χωρίς τον περιορισμό αυτού του χαρακτηριστικού.
Στό πρόβλημα αυτό, το μέσο του $DE$ $($ δηλαδή το σημείο $L,$ $),$ ανήκει στον ριζικό άξονα των κύκλων $(O_{1},$ $(O{2}).$

Κώστας Βήττας.

[quote="silouan"]ΑΣΚΗΣΗ 17 -ΛΥΣΗ
Οι ευθείες $AB,\ B'D',\ CD$ σχηματίζουν δέσμη .
Επίσης οι ευθείες $DA,\ CB,\ C'A'$ σχηματίζουν δέσμη .
Άρα και οι ευθείες $C'D',\ CA,\ B'A'$ σχηματίζουν δέσμη . :)
Όμοια οι ευθείες $D'A',\ DB,\ C'B'$ σχηματίζουν δέσμη

Σωστή δεν είναι ? :)[/quote]

Σιλουανέ, αν με τη φράση ''αποτελούν δέσμη'', εννοείς ότι τρεις ευθείες είναι συντρέχουσες, αυτό είναι το ζητούμενο. Αν πάλι αναφέρεσαι στη θεωρία περί ''διπλού λόγου'', με προβληματίζει το σκεπτικό σου, μιας και το προσπάθησα χωρίς αποτέλεσμα και δεν μπορώ να βρώ τι είναι αυτό που μου ξεφεύγει.

Παρακάτω δίνω τη μία από τις δύο αποδείξεις που έχω υπόψη μου, βασισμένη στο θεώρημα του Πάππου $($ όπως είναι πλέον διεθνώς καθιερωμένο, ενώ στα χρόνια μας το μαθαίναμε σαν θεώρημα του Pascal $),$ η οποία ταιριάζει με τις δέσμες που αναφέρεις, ως συντρέχουσες ευθείες $($ στο σχήμα μου είναι $AB = 5.5,$ $BC = 3.0,$ $AC = 5.0,$ $CD = 4.0,$ $AD = 4.5$ $).$

[b][size=109][color=red]ΑΣΚΗΣΗ 17 - ΛΥΣΗ 1.[/color][/size][/b]

Επί των ευθειών $BF,$ $DE,$ θεωρούμε τις τριάδες των σημείων $A_{1}\equiv F,$ $B_{1}\equiv B,$ $C_{1}\equiv B'$ και $A_{2}\equiv E,$ $B_{2}\equiv C',$ $C_{2}\equiv D,$ αντιστοίχως. Σύμφωνα με το θεώρημα του Πάππου, προκύπτει ότι τα σημεία $A'\equiv A_{1}B_{2}\cap A_{2}B_{1},$ $D'\equiv A_{1}C_{2}\cap A_{2}C_{1},$ και $F'\equiv B_{1}C_{2}\cap B_{2}C_{1},$ ανήκουν στην ίδια ευθεία.

Δηλαδή οι ευθείες $A'D',$ $BD,$ $B'C',$ τέμνονται στο ίδιο σημείο. Ομοίως αποδεικνύεται ότι και οι ευθείες $A'B',$ $AC,$ $C'D',$ είναι συντρέχουσες και η πρόταση έχει αποδειχθεί.

Κώστας Βήττας.

ΥΓ. Η δεύτερη λύση που έχω υπόψη μου, βασίζεται στο θεώρημα του Μενελάου και θα τη βάλλω αργότερα στο φόρουμ, αν δεν έχει λάθος και τύχει να μη δημοσιευτεί από κάποιον άλλον.

Δέσμη παραπάνω εννούσα συντρέχουσες ευθείες και συγγνώμη που δεν το διευκρίνησα .Απλά χρησιμοποίησα το γνωστό (τουλάχιστον για μένα) θεώρημα που η απόδειξη είναι αυτή με το θεώρημα του Πάππου που βάλατε παραπάνω. :wink:

[quote="silouan"]ΑΣΚΗΣΗ 19 -(Χρήστος Πατήλας)
Δίνεται κύκλος (Ο,R) και τέσσερα σημεία $A,\ B,\ C,\ D$  πάνω στον κύκλο ώστε oι διαγώνιοι του τετραπλεύρου $ABCD$ να τέμνονται κάθετα . Να βρεθεί ο γεωμετρικός τόπος των μέσων των πλευρών του τετραπλεύρου $ABCD$[/quote]

Silouane mipos den einai apolyta orthi i ekfonisi?diladi exeis 4 vathmous eleutherias??

Ναι έχεις δίκαιο πρέπει να τη διατυπώσω αλλιώς .
Τα πράγματα εχουν ως εξής .
Έστω δύο σταθερές ευθείες $m_{1},\ m_{2}$  που τέμνονται κάθετα και σταθερό σημείο $O$ .Με κέντρο το $O$ και μεταβλητές ακτίνα γράφουμε κύκλο που τέμνει την $m_{1}$ στα $A,\ C$ και την $m_{2}$ στα $B,D$ κάθε φορά το $ABCD$ να ειναι κυρτό .Να βρεθεί ο γεωμετρικός τόπος των μέσων των πευρών του εγγραψιμου  τετραπλεύρου $ABCD$  καθώς η ακτίνα του κύκλου μεταβάλλεται .

[size=100][b]ΑΣΚΗΣΗ 20 – ΕΚΦΩΝΗΣΗ.[/size][/b]

Δίδεται ορθογώνιο τρίγωνο $ABC$ $($ $\angle BAC = 90^{o}$ $)$ και έστω $AD,$ το ύψος του. Με διάμετρο $AD,$ γράφουμε τον κύκλο $(O_{1})$ ο οποίος τέμνει τις πλευρές $AC,$ $AB,$ στα σημεία $E,$ $F$ αντιστοίχως και τον περιγεγραμμένο κύκλο $(O),$ του $\bigtriangleup ABC,$ στο σημείο $M.$ Αποδείξτε ότι οι ευθείες $BE,$ $CF$ και $DM,$ τέμνονται στο ίδιο σημείο.

Κώστας Βήττας.

[quote="fotis"]Εχει βρει κανείς λύση για την άσκηση 13;[/quote]

Υπάρχει μία σύντομη λύση, ως άμεση εφαρμογή $($ αρκεί να θεωρήσουμε τη διάμετρο $AA'$ $)$ του παρακάτω όχι πολύ γνωστού (στην Ελληνική βιβλιογραφία) θεωρήματος, των ισογώνιων ευθειών δια των κορυφών ενός τριγώνου.

[b][size=117]ΘΕΩΡΗΜΑ.[/size][/b] - Από κάθε κορυφή ενός τριγώνου, φέρνουμε δύο ισογώνιες ευθείες ως προς την αντίστοιχη γωνία. Αποδείξτε ότι οι ευθείες που συνδέουν τις κορυφές του τριγώνου, με τα σημεία τομής των ισογώνιων ευθειών που πρόσκεινται στι απέναντι πλευρές του, τέμνονται στο ίδιο σημείο.

Μία απλή συνθετική απόδειξη αυτού του θεωρήματος έχει δημοσιευτεί στο περιοδικό ''ΑΠΟΛΛΩΝΙΟΣ'', τεύχος 4, σελίδα 172. Είναι ένα δυνατό θεώρημα, που αληθεύει για κάθε περίπτωση των ισογώνιων ευθειων (στο εσωτερικό ή στο εξωτερικό μέρος του τριγώνου ή άλλες στο εσωτερικό και άλλες στο εξωτερικό) και των σημείων τομής τους (ανάλογα σε ποια πλευρά, κάθε ισογώνια ευθεία θεωρείται προσκείμενη).

Έχω υπόψη μου και άλλη μία συνθετική απόδειξη, νομίζω πολύ όμορφη, για το θεώρημα αυτό και θα τη βάλλω αργότερα στο φόρουμ, μαζί με κάποια σχόλια και όσες βιβλιογραφικές σημειώσεις γνωρίζω.

Μία λύση όμως που είναι άμεση εφαρμογή κάποιου θεωρήματος, δεν είναι ιδιαίτερα διδακτική (έστω και αν το θεώρημα είναι).

Η άσκηση 13 λύνεται συνθετικά, όχι δύσκολα και με απλά μέσα, αρκεί να θεωρήσουμε τις κατάλληλες βοηθητικές ευθείες από την κορυφή $A$ και ανατρέξουμε στη γνωστή ιδιότητα της διχοτόμου γωνίας τριγώνου, η οποία διχοτομεί επίσης και τη γωνία μεταξύ του ύψους και διαμέτρου του περιγεγραμμένου κυκλου, που αντιστοιχούν στην ίδια κορυφή.

Εύχομαι καλή και δημιουργική χρονιά σε όλους, με υγεία.

Κώστας Βήττας.


Is there a simpler solution to this problem? (It is taken from the resources section of this site.)

Prove that for all integers n > 1, the number $\sqrt{\overline{1 \cdots 14 \cdots 4}}$, where 1 appears n times and 4 appears 2n times, is irrational.

[hide]
That number can be written as $q_{n}= (10^{3n}+3 \times 10^{2n}-4)/9$.

We have $9q_{n}= (x-1) (x+2)^{2}$, where $x=10^{n}$. It is therefore  enough to prove that numbers 11, 111, 1111, ... are not perfect squares.

We can do this by looking at the last digit. To get 1 we must square 1 or 9. So we have that $\overline{1 \cdots 1}$ is equal to $(10 m+1)^{2}$ or $(10 m+9)^{2}$ for some m. In the both cases the second to last digit has to be even, because it is the last digit of $2 m (5 m+1)$, or the last digit of $2(9 m+4)$.

[/hide]

[hide="Yes"] Factor out $4$.  The resulting number ends $...1...1$, and for $n > 1$ there will be at least two $1$s.  Then the number is congruent to $11 \bmod 4 \equiv 3 \bmod 4$, not a quadratic residue. [/hide]

Nit-pic .. about the above solution ..
[hide][quote] Factor out $4$.  The resulting number ends $...1...1$, and for $n > 1$ there will ...  [/quote]
One has to be  a little careful here in  the wording ( for example I would write this as: [quote]Factor out $4$. And for   $n > 1$ .. The resulting number ends $...1...1$,  there will ... [/quote]

($144$  divided by 4 does not end in 1 .. n>1 is needed not only for ending in  (2)  1's but even 1  1)[/hide]


Does there exist a function $f: \mathbb{R}\to\mathbb{R}(f\not\equiv\text{const})$ such that \[|f(x)-f(y)|\leq (x-y)^{2}\;\forall x,y\in\mathbb{R},x\not = y?\]

$\left |{f(x+h)-f(x)\over h}\right |\leq |h|$ for all $x,h\in\mathbb{R},h\not =0$.
So $f'(x)=\lim_{h\to 0}{f(x+h)-f(x)\over h}=0$ exists for all $x\in\mathbb{R}$.
So $f$ is differentiable with $f'\equiv 0$.
So $f$ is constant by the mean-value theorem.


$ n,m$ are positive integers.
prove that $ \sum_{k \equal{} 0}^{n}(\minus{}1)^{k}{m\plus{}1\choose k}{m\plus{}n\minus{}k\choose m}\equal{} 0$

this is the solvation you want to see....[/img]

There is a combinatorial proof of this identity. Just follow the lines of the one given here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1445332714&t=20206

Pierre.


inequalities
function
Cauchy Inequality
inequalities unsolved

[quote="Hong Quy"]it's equavilent
 $ 9 \sum ab \plus{} 1 \ leq 108abc$[/quote]
Why  :?:

[i]Hong Quy

  I think you just want to spam

  Please do not particpate mathlinks

   People can think bad about Vietnamese[/i]

[quote="Hong Quy"]i  don't  spam.
 i use Cauchy
$ \frac {a}{abc \plus{} a^2 \plus{} 1} \equal{} \frac {a}{abc \plus{} \frac {3a^2}{3} \plus{} \frac {27}{27}} \leq \frac {1}{31^2}(\frac {1}{bc} \plus{} \frac {9}{a} \plus{} 27^2)$
hence
$ LHS \leq \frac {1}{31^2} ( \sum \frac {1}{bc} \plus{} 9\sum \frac {1}{a} \plus{} 3.27^2 ) \leq RHS$
it's  equavilent with ineq $ 9 \sum ab \plus{} 1 \leq 108abc$[/quote]
But $ 9\sum{ab}\plus{}1 \geq 108abc$

$ f"(x)  \equal{} ( \frac{1}{bc\plus{}x\plus{}\frac{1}{x}})" \equal{}  \frac{ 2(x^3\plus{}x^2\minus{}3x\minus{}1\minus{}bc)}{(xbc\plus{}x^2\plus{}1)^3} \leq 0$  $ \forall x \in [0;1]$
Use Jensen's  ineq
 we  have :  $ f(a) \plus{}f(b) \plus{} f(c) \leq 3 f(\frac{a\plus{}b\plus{}c}{3})  \equal{}3f(\frac{1}{3})  \equal{} \frac{27}{31}$
 seem  it 's  wrong !  but i don't sure

[i]Hong Quy

I think you just want to spam

Please do not particpate mathlinks

People can think bad about Vietnamese

 If you don 't sure please don't post

    It take many time to check a solution[/i]

[quote="Hong Quy"]$ f"(x) \equal{} ( \frac {1}{bc \plus{} x \plus{} \frac {1}{x}})" \equal{} \frac { 2(x^3 \plus{} x^2 \minus{} 3x \minus{} 1 \minus{} bc)}{(xbc \plus{} x^2 \plus{} 1)^3} \leq 0$  $ \forall x \in [0;1]$
Use Jensen's  ineq
 we  have :  $ f(a) \plus{} f(b) \plus{} f(c) \leq 3 f(\frac {a \plus{} b \plus{} c}{3}) \equal{} 3f(\frac {1}{3}) \equal{} \frac {27}{31}$
 seem  it 's  wrong !  but i don't sure[/quote]
Hong Quy, it's obviously wrong because $ x$ is function of $ b$ and $ c.$ :wink:

[quote="alexilic"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove inequality:
\[ \frac {1}{bc \plus{} a \plus{} \frac {1}{a}} \plus{} \frac {1}{ac \plus{} b \plus{} \frac {1}{b}} \plus{} \frac {1}{ab \plus{} c \plus{} \frac {1}{c}} \leqslant \frac {27}{31}.
\]
[/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199831]See my solution here[/url] :)

later,  i  have solution  for this problem.
 use Schur's ineq   with $ a\plus{}b\plus{}c \equal{}1$ we have : $ abc  \geq 4(ab\plus{}bc\plus{}ca) \minus{}1$ (*)
by other  form , this ineq's  equavilent  with :
 $ \sum \frac{a}{abc \plus{}a^2\plus{}1}  \leq \frac{27}{31}$
use (*)  $ \to$   we  need  prove that :
 $ \sum (1\minus{}\frac{31a(a\plus{}b\plus{}c)}{\sum ab \plus{} 9a^2 \plus{}8}  \leq 0$
or $ \sum \frac{(7a\plus{}8c\plus{}10b)(c\minus{}a) \minus{}(7a\plus{}8b\plus{}10c)(a\minus{}b)}{a^2 \plus{} k} \leq 0$
  where  $ k \equal{} \frac{4 \sum ab \plus{}8}{9}$
later, $ \sum (a\minus{}b)^2\frac{17ab\plus{}8(a62\plus{}b^2)\plus{}10c(a\plus{}b)\plus{}k }{(a^2\plus{}k)(b^2\plus{}k)} \geq 0$
 occur  equal  when : $ a\equal{}b\equal{}c\equal{}1/3$

Please , someone confirm my solution! thank you  :)

[quote="anas"]hello evry body , here is my solution ( hope it's right  :D ) , it's very ugly  :(

the inequality is equivalent to :
\[ \sum{\frac {1}{bc \plus{} a \plus{} \frac {1}{a}} \minus{} a} \leq \frac { \minus{} 4}{31}
\]
( because $ a \plus{} b \plus{} c \equal{} 1$)
\[ \sum{\frac {a^2(bc \plus{} a)}{abc \plus{} a^2 \plus{} 1}} \geq \frac {4}{31}
\]

\[ \sum{\frac { (a^2(bc \plus{} a)^2)}{(abc \plus{} a^2 \plus{} 1)(bc \plus{} a)}} \geq \frac {4}{31}
\]
but  by cauchy shwarz :
\[ LHS(\sum{(abc \plus{} a^2 \plus{} 1)(bc \plus{} a)}) \geq (\sum{a(bc \plus{} a)})^2
\]
setting : $ r \equal{} abc,q \equal{} ab \plus{} ac \plus{} bc$ and since $ a \plus{} b \plus{} c \equal{} 1$ it's easy the check that:

$ \sum{(abc \plus{} a^2 \plus{} 1)(bc \plus{} a)} \equal{} 2 \minus{} 2q \plus{} 5r \plus{} qr$
$ \sum{a(bc \plus{} a)} \equal{} 3r \plus{} 1 \minus{} 2q$
so we need to prove that:
$ (3r \plus{} 1 \minus{} 2q)^2 \minus{} \frac {4(2 \minus{} 2q \plus{} 5r \plus{} qr)}{31} \geq 0$
or:
$ 9r^2 \plus{} \frac {166r}{31} \minus{} \frac {376qr}{31} \plus{} \frac {23}{31} \minus{} \frac {116q}{31} \plus{} 4q^2 \geq 0$
now we put:
$ f(r) \equal{} 9r^2 \plus{} \frac {166r}{31} \minus{} \frac {376qr}{31} \plus{} \frac {23}{31} \minus{} \frac {116q}{31} \plus{} 4q^2$
we have:
$ f(r)' \equal{} 18r \plus{} \frac {166}{31} \minus{} \frac {376q}{31}$
it's easy to check that : $ f(r)'\geq0$ since   $ q\leq \frac {1}{3}$
so f is an increasing function , and by shur : $ r\geq \frac {4q \minus{} 1}{9}$
thus :$ f(r)\geq f(\frac {4q \minus{} 1}{9}) \equal{} \frac {12q^2}{31} \minus{} \frac {28q}{31} \plus{} \frac {8}{31}$
hence it sufficies to prove , that : $ \frac {12q^2}{31} \minus{} \frac {28q}{31} \plus{} \frac {8}{31} \geq 0$
wich is equivalent to : $ \frac {4(3q \minus{} 1)(q \minus{} 2)}{31} \geq 0$ , wich is true since $ q\le\frac {1}{3}\leq 2$
@can_hang2007 : I really like your solution , is it a Intuition or a technique that I should learn? , thank you very much  :)[/quote]
ur solution's  right  and  nice.

[quote="anas"]hello evry body , here is my solution ( hope it's right  :D ) , it's very ugly  :(

the inequality is equivalent to :
\[ \sum{\frac {1}{bc \plus{} a \plus{} \frac {1}{a}} \minus{} a} \leq \frac { \minus{} 4}{31}
\]
( because $ a \plus{} b \plus{} c \equal{} 1$)
\[ \sum{\frac {a^2(bc \plus{} a)}{abc \plus{} a^2 \plus{} 1}} \geq \frac {4}{31}
\]

\[ \sum{\frac { (a^2(bc \plus{} a)^2)}{(abc \plus{} a^2 \plus{} 1)(bc \plus{} a)}} \geq \frac {4}{31}
\]
but  by cauchy shwarz :
\[ LHS(\sum{(abc \plus{} a^2 \plus{} 1)(bc \plus{} a)}) \geq (\sum{a(bc \plus{} a)})^2
\]
setting : $ r \equal{} abc,q \equal{} ab \plus{} ac \plus{} bc$ and since $ a \plus{} b \plus{} c \equal{} 1$ it's easy the check that:

$ \sum{(abc \plus{} a^2 \plus{} 1)(bc \plus{} a)} \equal{} 2 \minus{} 2q \plus{} 5r \plus{} qr$
$ \sum{a(bc \plus{} a)} \equal{} 3r \plus{} 1 \minus{} 2q$
so we need to prove that:
$ (3r \plus{} 1 \minus{} 2q)^2 \minus{} \frac {4(2 \minus{} 2q \plus{} 5r \plus{} qr)}{31} \geq 0$
or:
$ 9r^2 \plus{} \frac {166r}{31} \minus{} \frac {376qr}{31} \plus{} \frac {23}{31} \minus{} \frac {116q}{31} \plus{} 4q^2 \geq 0$
now we put:
$ f(r) \equal{} 9r^2 \plus{} \frac {166r}{31} \minus{} \frac {376qr}{31} \plus{} \frac {23}{31} \minus{} \frac {116q}{31} \plus{} 4q^2$
we have:
$ f(r)' \equal{} 18r \plus{} \frac {166}{31} \minus{} \frac {376q}{31}$
it's easy to check that : $ f(r)'\geq0$ since   $ q\leq \frac {1}{3}$
so f is an increasing function , and by shur : $ r\geq \frac {4q \minus{} 1}{9}$
thus :$ f(r)\geq f(\frac {4q \minus{} 1}{9}) \equal{} \frac {12q^2}{31} \minus{} \frac {28q}{31} \plus{} \frac {8}{31}$
hence it sufficies to prove , that : $ \frac {12q^2}{31} \minus{} \frac {28q}{31} \plus{} \frac {8}{31} \geq 0$
wich is equivalent to : $ \frac {4(3q \minus{} 1)(q \minus{} 2)}{31} \geq 0$ , wich is true since $ q\le\frac {1}{3}\leq 2$
@can_hang2007 : I really like your solution , is it a Intuition or a technique that I should learn? , thank you very much  :)[/quote]

I have another solution, everybody can enjoy it!
We have the equivalent inequality :
$ \sum {\frac {9a^2 \plus{} 9abc \plus{} 9 \minus{} 31a}{a^2 \plus{} abc \plus{} 1}} \geq 0$
Now we have the comment:
 Because of $ a \plus{} b \plus{} c \equal{} 1$ so that $ 0\leq a,b,c \leq 1$
Without loss of generality, we assume $ a \geq b \geq c$

But we also have the two monotonous sequences ( you can check it)
* $ 9( a^2 \plus{} abc \plus{} 1) \minus{} 31a \leq 9(b^2 \plus{} abc \plus{} 1) \minus{} 31b \leq 9(c^2 \plus{} abc \plus{} 1) \minus{} 31c$   ( Note that $ 9( a \plus{} b) \leq 31)$

** $ \frac {1}{ a^2 \plus{} abc \plus{} 1} \leq \frac {1}{ b^2 \plus{} abc \plus{} 1} \leq \frac {1}{ c^2 \plus{} abc \plus{} 1}$

Applying Chebyshev inequality for two sequences above, we have

$ \sum{\frac {9(a^2 \plus{} abc \plus{} 1) \minus{} 31a}{ a^2 \plus{} abc \plus{} 1}} \geq \sum{\frac {9(a^2 \plus{} abc \plus{} 1) \minus{} 31a}{ 3}}. \sum{\frac {1}{ a^2 \plus{} abc \plus{} 1}}$

But it is easy to check by homogenious method the following inequality:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc \geq \frac {4}{9}$

An elementary proof! Can you agree with me?

[quote="omerftekin"]This inequality is a special case of İlham's Cauchy. Putting n=31 in Famous (Meshuur) İlham's Cauchy we get the solution.[/quote]
Post please, the İlham's Cauchy inequality here. Thx!

[quote="omerftekin"]cauchy schwarz bunyakovski inequalıty is specıal case of ılham's cauchy.are you serıous that you don't know this inequality.[/quote]

Do you enjoy to post spam like this? All your posts are full of non-sense contents like that. Please stop to do that. :mad:

Let $ abc\minus{}ab\minus{}bc\minus{}ac\plus{}1\equal{}m$ and $ 2abc\plus{}2\equal{}n$. Inequality equilavents to

$ \frac{\minus{}mn\plus{}m\plus{}n}{\frac{nm^2}{2}\plus{}1} \leq \frac{27}{31}$  $ \Longleftrightarrow$ $ 62(m\plus{}n) \leq n(27m^2\plus{}62m)\plus{}54$

$ n\leq \frac{56}{27}$ $ \Longrightarrow 62(m\plus{}n) \leq 62(m\plus{}\frac{56}{27})$. We will prove that $ 62(m\plus{}\frac{56}{27}) \leq n(27m^2\plus{}62m)\plus{}54$

By Schur's inequality $ (a\plus{}b\plus{}c)^3\plus{}9abc \geq 4(ab\plus{}bc\plus{}ac)(a\plus{}b\plus{}c)$   $ \Longrightarrow$

$ n \geq \frac{16\minus{}8m}{5}$

By $ AM$-$ GM$ $ (a\plus{}b)(b\plus{}c)(a\plus{}c) \leq \frac{8}{27}(a\plus{}b\plus{}c)^3\equal{} \frac{8}{27} \Longrightarrow$

$ m\equal{}abc\minus{}ab\minus{}bc\minus{}ac\plus{}1 \geq \frac{19}{27}$

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \geq \frac{9}{(a\plus{}b\plus{}c)^2}>1 \Longrightarrow m<1 \Longrightarrow (2m\plus{}1)^2\minus{}\frac{292}{27}<0 \Longrightarrow$

$ (27m\minus{}19)(4m^2\plus{}4m\minus{}\frac{265}{27})\leq0 \Longrightarrow$

$ 108m^3\plus{}32m^2\minus{}341m\plus{}\frac{5035}{27}\leq0 \Longleftrightarrow$

$ 62(m\plus{}\frac{56}{27}) \leq \frac{16\minus{}8m}{5}.(27m^2\plus{}62m)\plus{}54 \leq n(27m^2\plus{}62m)\plus{}54$  :wink:

How can you prove it? It seems like harder than original one. Could you please post that AM-GM solution?

[quote="nguoivn"][quote="alexilic"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove inequality:
\[ \frac {1}{bc \plus{} a \plus{} \frac {1}{a}} \plus{} \frac {1}{ac \plus{} b \plus{} \frac {1}{b}} \plus{} \frac {1}{ab \plus{} c \plus{} \frac {1}{c}} \leqslant \frac {27}{31}.
\]
[/quote]

I think it's easy. Using Am-Gm, we are easily to prove the stronger:
$ (\frac {1}{bc \plus{} a \plus{} \frac {1}{a}})^{\frac {10}{9}} \plus{} (\frac {1}{ca \plus{} b \plus{} \frac {1}{b}})^{\frac {10}{9}} \plus{} (\frac {1}{ab \plus{} c \plus{} \frac {1}{c}})^{\frac {10}{9}} \leq\ 3 (\frac {9}{31})^{\frac {10}{9}}$
(with the same conditon)
With this stronger, all of the proofs above can't solve it.
 :)[/quote]

I cannot find the solution either. Could you please elaborate a little? :blush:

Very sorry, my friend. I'll post my proof (but not now) because of some reason  :oops:

[quote="nguoivn"][quote="alexilic"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove inequality:
\[ \frac {1}{bc \plus{} a \plus{} \frac {1}{a}} \plus{} \frac {1}{ac \plus{} b \plus{} \frac {1}{b}} \plus{} \frac {1}{ab \plus{} c \plus{} \frac {1}{c}} \leqslant \frac {27}{31}.
\]
[/quote]

I think it's easy. Using Am-Gm, we are easily to prove the stronger:
$ (\frac {1}{bc \plus{} a \plus{} \frac {1}{a}})^{\frac {10}{9}} \plus{} (\frac {1}{ca \plus{} b \plus{} \frac {1}{b}})^{\frac {10}{9}} \plus{} (\frac {1}{ab \plus{} c \plus{} \frac {1}{c}})^{\frac {10}{9}} \leq\ 3 (\frac {9}{31})^{\frac {10}{9}}$
(with the same conditon)
With this stronger, all of the proofs above can't solve it.
 :)[/quote]
Now, here is my proof:  :)

[quote="BaronShadeNight"][quote="alexilic"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove inequality:
\[ \frac{1}{bc \plus{} a \plus{} \frac{1}{a}} \plus{} \frac{1}{ac \plus{} b \plus{} \frac{1}{b}} \plus{} \frac{1}{ab \plus{} c \plus{} \frac{1}{c}} \leqslant \frac{27}{31}.\][/quote]

I only use the AG-GM and Cauchy-Schwarz:

$LS=\sum\frac{a}{a(a+b)(a+c)+1}\\ =\sum\frac{a}{(1-b-c)(a+b)(a+c)+1}\\ =\sum\frac{a}{(a+b)(a+c)+1-(a+b)(b+c)(c+a)}$

By CauChy-Schwarz, conspicuous:

$LS \le \sum\frac{a}{31^2}[\frac{12^2}{(a+b)(a+c)}+\frac{19^2}{1-(a+b)(b+c)(c+a)}]\\ =\sum\frac{12^2}{31^2}[\frac{a}{(a+b)(a+c)}]+\frac{19^2}{31^2}[\frac{a+b+c}{1-(a+b)(b+c)(c+a)}]=L$

On the other hand:
$\sum\frac{a}{(a+b)(a+c)}=\frac{2(ab+bc+ca)}{(a+b)(b+c)(c+a)}\le \frac{9}{4}$

[because $(a+b)(b+c)(c+a)\ge \frac{8}{9}(ab+bc+ca)$]

And: $\frac{1}{1-(a+b)(b+c)(c+a)}\le \frac{1}{1-[\frac{2(a+b+c)}{3}]^3}=\frac{27}{19}$        [b](AG-GM)[/b]

So:  $L\le \frac{12^2}{31^2}.\frac{9}{4}+\frac{19^2}{31^2}.\frac{27}{19}=\frac{27}{31}$

$\Rightarrow Q.E.D$   :lol:[/quote]

very nice


How would you write multiplication process with Latex?

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367787#367787

Thanks.

I would write that problem differently. Let $a,b$ be digits in decimal representation. Knowing that $\overline{2a} \cdot \overline{b3} = 989$ find $a+b$.

This works:

\begin{tabular}{ccc} & 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{tabular}

[code]
\begin{tabular}{ccc}
& 2 & a\\
\times & b & 3\\ \hline
& 6 & 9\\
9 & 2\\ \hline
9 & 8 & 9\end{tabular}[/code]

If you put @{} between the columns, you'll eliminate the spacing:
\[ \text{ \begin{tabular}{r@{}c@{}c} & 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{tabular}} \]
[code]\begin{tabular}{r@{}c@{}c}
& 2 & a\\
\times & b & 3\\ \hline
& 6 & 9\\
9 & 2\\ \hline
9 & 8 & 9\end{tabular}[/code]


for these problems i like to add numbers until i get to a number divisble by both 9 and 7.

is the answer 39? first i tried adding 6, to make the remainder when i divide by 9 0. however, there is still a remainder of 3 for the seven. so now add 9 more, for a total of 15. the remainde now is 5 for the seven part. now add 9 more, for a total of 24. now there is a remainder of 0 for the 7's and 0 for the 9's

lcm(7,9)-24=63-24=39 :wink:

See [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese remainder theorem[/url].

[hide]so
$ x \equal{} 4 (mod 7)$
$ x \equal{} 3 (mod 9)$

u should get 39 after using ur modular arithemetic.  :lol:[/hide]

[hide]$ x\equiv 4 \pmod{7}$
$ x\equiv 3 \pmod{9}$

The smallest $ x$ satisfying the first condition is $ 18$, which is congruent to 0 mod 9.

$ 7\equiv \minus{}2 \pmod{9}$, so for every 7 we add to 18 the remainder becomes -2 more.

$ x\equiv 3 \equiv \minus{}6\pmod{9}$, so we add $ \frac{\minus{}6}{2}\equal{}3$ sevens.

$ 18\plus{}21\equal{}39$[/hide]

[hide]we know our number has to be divisible by 3. so now just list out all numbers that are 4 more than a multiple of 7.

4
11
18
25
32
39<-- divisible by 3. so 39 is our answer.[/hide]

$ x\equiv 4 \mod 7$
$ x\equiv 3 \mod 9$

$ x\equal{}7k\plus{}4$
$ 7k\plus{}4\equiv 3\mod 9$
$ \minus{}2k \equiv 8\mod 9$
$ k\equiv 5\mod 9$

$ k\equal{}9y\plus{}5$

$ x\equal{}63y\plus{}39$

Smallest is then $ x\equal{}39$

[quote="mathcrazed"]$ x\equiv 4 \mod 7$
$ x\equiv 3 \mod 9$

$ x \equal{} 7k \plus{} 4$
$ 7k \plus{} 4\equiv 3\mod 9$
$ \minus{} 2k \equiv 8\mod 9$
$ k\equiv 5\mod 9$

$ k \equal{} 9y \plus{} 5$

$ x \equal{} 63y \plus{} 39$

Smallest is then $ x \equal{} 39$[/quote]

yay thats what i did  :lol:


geometry
rotation
parallelogram
geometric transformation
LaTeX

In triangle ABC, E is the midpoint of side BC and D is on side AC.
If the length of AC is 1, and $\angle BAC = 60^{\circ}$, $\angle ABC = 100^{\circ}$, $\angle ACB = 20^{\circ}$, and $\angle DEC = 80^{\circ}$, then the area of triangle ABC plus twice the area of triangle CDE equals?

There's an interestingly easy solution to this. No trig.

rotate it 180 degrees about E, you have a parallelogram, and part of it is the desired area ($ABD'C$) [hide="answer"] \frac{\sqrt{3}}{8}[/hide]

Can you explain the rotation solution better?

Like I understand that you are trying to find the area of ABD'C, but how exactly do you do that?

[hide]Draw an altitude from B.  Draw DE and a line from E to the foot of the altitude from B.  You'll form 2 iscoceles triangles and 2 30-60-90 triangles.  Let EC = S and you can solve it all with pythagorean theorem.  Remember that the median from the right angle in a right triangle = half the hypotenuse.  Since I don't know latex yet I'll just leave it at this.  The rest of the solution just takes arithmetic.[/hide][/hide]


:arrow: [color=darkred]Prove or Disprove:[/color] The Beatty Sequence of $ 2\plus{}\sqrt{2}$ is a subsequence of the Beatty sequence of $ \sqrt{3}$.

 :arrow: In general, is there any way to know whether a beatty sequence is a subsequence of another or not.
 :arrow: Can anyone please give any source in internet about properties of Beatty Sequences?

[hide]The problem is in connection to http://www.mathlinks.ro/viewtopic.php?t=265288.

[/hide]

see here

en.wikipedia.org/wiki/Beatty_sequence

Someone please help. :(  or atleast give a hint. :maybe:

I don't see a general way of showing that a certain sequence is a non-trivial subsequence in terms of Beatty sequences.

And there's always a good method in showing the opposite: finding a counterexample. That's how I'll show your statement about these two sequences isn't true:

$ \{2\plus{}\sqrt{2}\}$ sequence: [b]3, 6, 10, 13, 17, 20, [color=red]23[/color], 27, [color=red]30[/color], 34, [color=red]37[/color][/b]
$ \{\sqrt{3}\}$ sequence: 1, [b]3[/b], 5, [b]6[/b], 8, [b]10[/b], 12, [b]13[/b], 15, [b]17[/b], 19, [b]20[/b], 22, 24, 25, [b]27[/b], 29, 31, 32, [b]34[/b], 36, 38

So, there are terms in first sequence that aren't in the second one. Hence the first sequence is not a subsequence of the second one.

See this post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136049[/url]


Two cylindrical cans have the same volume. The height of one
can is triple the height of the other. If the radius of the
narrower can is 12 units, how many units are in the length of
the radius of the wider can? Express your answer in simplest
radical form.


calculus
integration
floor function
real analysis
real analysis unsolved

How to show that (both equalities)
\[ \sum_{k \equal{} 1}^{n}\int_1^k\frac {dx}{x} \equal{} \int_1^n\frac {n \minus{} \lfloor x \rfloor}{x}dx \equal{} \int_1^n \frac {n \plus{} \frac12 \plus{} (\{x\} \minus{} \frac12 ) \minus{} x}{x}dx
\]
($ \lfloor x \rfloor \equiv \max{(n\in \mathbb{Z}: n\leq x)}$ and $ \{x\} \equal{} x \minus{} \lfloor x \rfloor$)

the right equaitly is obvius ,for the left equality both side are equal to $ ln(n!)$

Just elaborate what drapt said for left hand side.

Break the integral in the middle into
$ \int_1^2 \frac{n\minus{}[x]}{x} \plus{} \cdots \plus{} \int_{n\minus{}1}^n \frac{n\minus{}[x]}{x}$
Now in each integral, you know that what [x] is. In fact [x] = k if on [k,k+1).
The rest is easy.


let ABC be atriangle and O is its circumcenter.a circle goes through A,O,B.and intersects CA and CB in points P and Q.

prove that:

               CO is prependicular to PQ

As far I know this problem is from Turkey, 1993.

Since $\angle BOC$ is the central angle of $\angle A$, we have $\angle BOC=2\angle A$, so that $\angle OCB = \pi/2 - A$.

In addition the quadrilateral ABPQ is cyclic; hence $\angle QPC=\angle A$. Therefore, if $N=OC \cap PQ$, in triangle $CNP$ we have $\angle NCP+\angle NPC=\pi/2$, i.e., PQ and CN are perpendicular, and we're done.

A similar argument works if $\angle C$ is obtuse.


How many four-digit odd integers greater than 6000 can be formed from the digits 0, 1, 3, 5, 6 and 8, if no digit may be used more than once?

The last digit can be $ 1,3,5$ and the first digit can be $ 6$ or $ 8$.

There are $ 4$ digits remaining after this, so our answer is $ 3\cdot2\cdot4\cdot3\equal{}\boxed{72}$.

<insert incorrect solution here>

<the above was edited by me>


board games: Chess, checkers, monopoly, Life, parcheesi
sports: frisbee
computer: Roller Coaster Tycoon, BZFlag
other: Rock-Paper-Scissors

board: chess, go, checkers
sport: mountain biking, soccer
comp: SimCity, AOE II
other: random online games

board: chess, bughouse,
sport: tennis, ping pong
comp: Halo, Halo 2, soon to be Halo 3 (yeah its on xbox 360 but that counts)
other: umm...

board: i dunno, chess, kangaroo chess, go (i dont know how to play, im just interested in it kinda :D )
sport: i suck at sports, and therefore dont really like them
comp: starcraft (lost my CD key though, so only 2 computers can play it), continuum/subspace, random MMORPGs.
other: twilight princess (actually, it was pretty fun, but terribly easy. i mean, BABIES can beat it), super paper mario, super smash brothers melee, etc.

[quote="abacadaea"][quote="Prime factorization"]I once played a German rock-paper scissors called akin-blakin-shplakin (at least that was how it was pronounced).[/quote]
i take german i think its rocke papier, schere. i dont know what rock is though. but thats the litteral translation. YOU ALL LOSE THE GAME![/quote]

[u][b][color=red][size=200]YOU ARE CORRECT.[/size][/color][/u][/b]


superior algebra
superior algebra unsolved

I'm not sure the problem really belongs here, but the statement is the following:

Prove that there are uncountably many linear orders on $\mathbb{Z}$.

In fact there are uncountably many wellorders on countable sets:

Consider the smallest uncountable ordinal $\Omega_{1}$. It is the set of all finite and all countable ordinals (because an ordinal consists exactly of all smaller ordinals) but is itself of course not countable.
If you take the (countably many) finite ordinals aways, it remaining set of all countable ordinals is still not countable and therefore you have uncountably many countable, pairwise non-isomorphic wellorders.

By an bijection to $\mathbb{Z}$ each of this gives rise to one well ordering of $\mathbb{Z}$ and conservly every wellordering of $\mathbb{Z}$ is isomorphic to exactly one of this ordinals.


Given a fixed circle $ (O,R)$ and a fixed point M outside $ (O)$, a fixed point $ N$ inside $ (O)$. $ A,B \in (O)$ such that $ AB$ goes through $ N$. $ MA$ and $ MB$ meet $ (O)$ at the second point $ C,D$ respectively.
[hide="part 1"]Prove that: $ CD$ goes through a fixed point[/hide]
[hide="part 2"]Prove that: the circumsribed circle of $ \triangle{MAB}$ goes trough two fixed point (included M)
just tell me where is the fixed point in this part[/hide]

Let ray $MN$ cut $\odot(MAB)$ at $ M'.$  From power of $N$  to $\odot(MAB)$ we get $\overline{NM} \cdot \overline{NM'} \equal{} \overline{NA} \cdot \overline{NB} \equal{} p(N,(O)).$ Since $ M,N$ are both fixed, then $M'$ is fixed $\Longrightarrow$  $\odot(MAB)$ goes trough $ M$ and its inverse image $M'$ through the inversion with center $N$ and power $\overline{NA} \cdot \overline{NB}.$ Inversion $ \mathcal{I}$ with center $M$ and power equal to the power of $M$ with respect to $(O)$ takes $(O)$ into itself and the line $CD$ into the circle $\odot(MAB),$ passing through the fixed point $ M'.$  Thus $CD$ goes through the fixed inverse image of $ M'$ under $\mathcal{I}.$


After a lot of work I finally managed to solve the last problem. Now I have another one I find difficult:

Find all circles that are orthogonal to $|z|=1$ and $|z-1| = 4$.

I realize that I am somehow supposed to use that Möbius transforms are conformal maps $C^{*}\to C^{*}$, but I don't really know how. Hints?

[hide="Wrong initial response"]Pick a point of intersection of those two circles, and use a transformation that takes it to infinity.  Then your two circles become intersecting lines (unless the two circles were tangent, in which case they'll become parallel lines), and finding the circles (in the second case, lines) orthogonal to both is easy.  Then invert the transformation to find the preimage of each of those circles (lines), and you're done.[/hide]

Sorry, what's hidden there is what I initially wrote, and it doesn't help in this case because the circles don't intersect.  Instead, find some transformation which makes the circles concentric (f.l.t.s preserve circles, but not centers of circles) -- then the problem is easy to solve, and then you invert the transformation to get the preimages which are what you want.

Ok, thanks. This problem was easier than I thought.

Unless I made a mistake all circles are given as the images of $R\cup\{\infty\}$ under
$f_{a}(t) = (-7+4\sqrt{3}) \frac{at-96-56\sqrt{3}}{at-1}\quad a \neq 0$.

Is there some clever way of determining which circles these are in the form $|z-b| = R$?


Politicians are obsessed with guilt by association don't you think?

Obama's been linked to unrepentant terrorists, Palin to a secessionist group, and tons more.

why do they do this?

it is obvious that Obama will not take over Washington and turn it into a Muslim terrorist camp, and Palin will not allow secession

Any negative connotation associated with someone can potentially change a voter's decision, perhaps subconsciously.

so any putative linking that has come up so far between obama and terrorists has been proven definitively false, but if the charge were true, that would have a huge impact and rightly so. perhaps he wouldn't turn washington into a muslim training camp, but if he were open to the terrorist view of america, that would obviously have immense negative consequences up to the possibility of giving away american positions, information, and so forth to our enemy. man, it would be even worse than that. I can't even fathom what a catastrophe that would be.

similarly, if palin were receptive to secessionist philosophy (I'll admit I don't know to what extent this may be true since I already have a plethora of other valid reasons to not want her in office), that would be similarly sucky.

hence if there is any possible cause for concern about possible associations, then yes, we as voters do have a full right to know the extent. of course, this doesn't mean we need to spend all our time worrying about fabricated or utterly immaterial assertions, as it always falls to us which issues possess greater relative importance.


limit
logarithms
algebra proposed
algebra

Just in case you have no problems to solve ...  :)

[b]Problem. [/b] Let $\{a_n\}_{n \geq 1}$ be a strictly increasing sequence of positive reals such that $\lim_{n \to \infty} a_n = +\infty$ and
\[
(a_1 +a_2 + \cdots + a_n)/a_n \to_{n \to \infty} +\infty.
\]
If $N(x)$ is the number of members of the sequence not exceeding $x$, prove that $N(x)/\log{x} \to +\infty$ as $x \to +\infty$.

--Vesselin

This is just beautiful! Let $x_n=\frac{a_1+...+a_n}{a_n}$ Then it is easy to see that $ a_n=\frac{a_1 x_1}{x_n-1}\cdot \frac{x_2}{x_2-1}\cdot...\cdot\frac { x_{n-1}}{x_{n-1}-1}$. Now take $M>1$ and $n_0$ such that for $ n\geq n_0$ we have $ x_n>M+1$. Then for $n>n_0$ we have $a_n<A(1+\frac{1}{M})^n$ where A is a constant which depends only on the first $n_0$ initial terms. Then it is easy to see that $ N(x)> \frac{lnx-lnA}{ln(1+\frac{1}{M})}-n_0-3$ for x greater than $n_0$. Use the fact that $ln(1+\frac{1}{M})<\frac{1}{M}$ and I thiink it's done.


JK!
Okay, there are 18 balls in a box. 3 are green, 10 are white and 5 are red. What is the chance of getting a red and a green? express as a decimal.

ok so this is gonna get a lil messy cause I KNOW i did it wrong lol

$ \frac{5}{18}*\frac{3}{17}*2$

this is $ 0.0980392...$

check ne1?

I agree with MCNational! i got the same thing.

I hate to break your bubble (Copied from AIME15), but you posted the same thing 2 times...  :D

It says he just draws two (randomly probably), so it should be $ \binom{18}{2}$, right?  It doesn't say he draws one and then without replacing it he draws another.  MAKE YOUR PROBLEM CLEAR!

tcjy... you've got to stop doing this. spamming the forums with problems that dont even make sense!  :(

[quote="jonathanchou711"]tcjy... you've got to stop doing this. spamming the forums with problems that dont even make sense!  :mad:[/quote] Well usually I assume you don't put it back if the problem doesn't say put it back. That's how most problems in my books make you do :noo: :noo:

[quote="ThinkFlow"]Well, here, just to make sure that your problem is not misinterpreted, please make it as specific as possible. This includes specifying whether order matters, items are drawn with replacement, etc. Though many problems are common and have implied restrictions, its best to be as clear and as specific as possible. :)[/quote]

GAH! I give up. I'll just go back and stay in MATHCOUNTS. No more MOEMS for me. :)

When I said "here", I was referring to the entire forum. It's best to be clear and concise no matter where you're posting. :)

007Math i think mines right lol

its $ \frac {(5 \plus{} 3)*2}{18*17}$

y would it be $ \binom{18}{2}$

His problem did not clarify that he choose one, and drew another one without replacement.  So mine could be right.  $ \binom{18}{2}$ would be the denominator.

[size=150]OMG I AM RIGHT FOR ONCE[/size]

uh, no, AIME and MCNational!, 007math is actually right given the conditions stated by tcjy.


pigeonhole principle
number theory
prime numbers
relatively prime

We select 11 positive integers that are less than 29 at random.  Prove that there will always be two integers selected that have a common divisor larger than 1.

[hide]The prime numbers below 29 are total 9 in number. so, the numbers we choose will be these primes or their multiples. we should remeber that 1 also. hence if 1 number extra than these 10 comes it will have common factor>1 with one of the 9 other than 1.
so, we are done[/hide]

@analytic: That's one case. You have to prove that is an extreme case.

[hide="solution"]Let us try to construct a set of positive integers less than 29 such that no two have common factors greater than 1. We would want 1 to be in, since it is relatively prime to every integer. Now we would want the primes, since no distinct primes have common factors greater than 1. So we have 1, 2, 3, 5, 7, 11, 13, 17, 19, and 23. That's 1, and all the primes less than 29. Now we [i]must[/i] choose a composite number less than 29, and is thus a multiple of at least one prime less than 29. Thus we cannot construct a set of positive integers less than 29 where no two have common factors greater than 1, and thus we have proven our original statement.[/hide]


topology
real analysis
real analysis unsolved

Let $A,B \subset \mathbb{R}$ countable and dense (in $\mathbb{R}$) sets. Prove that $A$ and $B$ are homeomorphic.

Who helps me? :blush:
$Tipe$

[quote="mlok"]With the topology induced from $\mathbb R$, the sets $\mathbb Z$ and $\mathbb Q$ are not homeomorphic. Singletons are open in $\mathbb Z$, but not in $\mathbb Q$.[/quote]
He said dense sets....Is $\mathbb Z$ dense in $\mathbb R$?

$X$ is dense in $\mathbb{R}$ if $\overline{X}=\mathbb{R}$.

$\mathbb{Z}$ is not dense since $\overline{\mathbb{Z}}=\mathbb{Z}$.

$Tipe$


AMC USA(J)MO USAMO articles induction Putnam probability

Hi, this is my first year qualifying for the USAMO.
I was wondering for all those people that have either taken it before or know a lot about it, what are some good ways to study?  Do you guys think doing old USAMO's will help; because I was looking over the ones for the past few years and I dont even think I could get started on about 95% of the questions.  Or do you think someone like me should start with easier more basic proofs and try to build up?  What are some good books I could use/other competitions that are good with proofs?

And how exactly does the grading system work?  For example, on last years number 4, I manage to find that the answer was 8025, but I couldnt really prove it without using some "stretched" information.  Since I got the answer and some basic parts of the proof, how many points would I get?

Also, do the graders care how you write the proof?  If your organization is bad and your wording is bad, do they count off alot?  Because I have never had any formal training with proofs.

You probably would have gotten a 2 on that solution for #4, but I may be wrong. I don't know exactly how they graded that problem, but evidence with several people's scores suggests that finding the answer with an intuitive proof merited a 2.

I would assume that with a completely correct, but badly written, solution, you would get a 6. Again, I'm not sure. I recall somewhere in the USAMO handbook it says that they only consider how well-written the solution is when breaking ties among winners. Then again, I'm not expert. But to be safe, do your best to write up a solution that reads cohesively and makes sense. You may want to read the AoPS article "How to Write a Solution," found under "Resources" on the bar at the left of your screen.

As for how/what to study, you said that for the majority of the problems, you didn't know how to get started. It may just be that you need a little bit of time to put down your pen and think. That's the case for almost all people.

Congratulations on qualifying.

Here is a rough guide to scoring.

7 :  Correct solution, no matter how badly written
6 :  Essentially correct solution with minor but significant flaws
5 :  Essentially correct solution, but with big flaws or gaps.  Usually badly written, as well.
3 and 4 :  Rarely given.  Sometimes for multi-part problems where you only solved one or two of the parts.  Once I got a 3 for an induction solution that did the induction in the wrong direction, but the right direction wasn't too different.  Note that the Putnam uses a 10-point scale in which the only grades given are 0, 1, 2, 8, 9, 10.
2 :  Lots of progress, many parts of a correct solution.
1 :  Significant progress; nontrivial observations which could lead to a correct solution
0 :  Most frequent grade

That said, the better written your solution is, the greater the probability that it will get all the points it deserves, so you should practice taking a lot of time to write nice proofs.  In the contest, you probably won't have that much time, but it's best to be in the habit of producing clean writeups.  A poorly written proof will make the grader unhappy.  This is bad.  Unhappy graders are mean graders.

How much a bald numerical answer will get varies, depending on whether or not it's considered significant progress.  On this particular problem, if you were able to find the right answer, you probably had the right general idea, but couldn't seize on the right elements of a rigorous proof.

I recommend practicing old USAMOs.  The farther into the past you go with those, the more likely you will be able to solve them.  However, the transition to olympiads involves an increase in patience.  In middle school competitions, you're uncomfortable if you can't solve a problem in a few minutes.  On the AMCs and at contests like ARML, this becomes more like five or seven minutes.  On the AIME, this tolerance is extended to around half an hour.  But on an olympiad, if an hour goes by and you've solved the problem, it was an easy problem.  So try doing some without a time limit.  Chances are you won't be able to see where to go from the beginning.  But you should still experiment with different approaches.  You can also use other national and international olympiads.  In general, the farther back in the past they go, the easier they are.

Also when you have written a solution, show it to people (who have math experience...) and see how rigorous they think it is.

Thanks for the advice!

The grading seems pretty harsh though...it seems like even if i make some  decent progress within a problem, i might still get 0

I also have AoPS vol. 2 so Ill try to study that.
I used the 'search function' and it seems to me that many people also recommend Art and Craft of Problem solving and the Engel book so maybe ill try and buy those books.

Also, when i read some of the posts, it seems to me that the test makers now do not test on inequalities anymore (the algebric inequalities).  Is this true?  Thanks (i was probalby going to study inequalities first)

[quote="matrixeater"]Hi, this is my first year qualifying for the USAMO.
I was wondering for all those people that have either taken it before or know a lot about it, what are some good ways to study?  Do you guys think doing old USAMO's will help; because I was looking over the ones for the past few years and I dont even think I could get started on about 95% of the questions.  Or do you think someone like me should start with easier more basic proofs and try to build up?  What are some good books I could use/other competitions that are good with proofs?[/quote]

If your goal is to solve one or two problems completely (7-14 points), you should be practicing with mostly #1's and #4's, and maybe a few #2's and #5's if they look approachable.  Or you can go back to when the USAMO only had 1 day and 5 problems, since those problems are significantly easier than the problems from more recent years.

Get used to spending a lot of time on a single problem.  Don't worry if you look at a problem and don't know where to start - most of the time that's the whole challenge of an olympiad-level problem.  The first thing you try on the problem usually won't work, so just keep trying stuff and you'll eventually solve the problem.  When I was preparing for the USAMO for the first time, a lot of times I would work on a problem for 30 minutes, not really get anywhere, and then give up and look at a solution.  This is the worst thing you can do, since you don't get anything out of reading a solution to a problem you haven't spent a VERY long time working on, and you've basically wasted a USAMO problem by doing this.  If you want to give up on a problem, you don't have to look at a solution, just put the problem aside and work on it another time.

When you think you've solved a problem, actually write up a solution.  This will help you practice proof-writing, and also sometimes you'll notice a hole in your proof while writing it up that you wouldn't have noticed otherwise.

One more thing - the USAMO, especially problems #1 and #4, is more focused on solving problems than using advanced techniques.  The only techniques you really need to know are induction and contradiction.  Take a look at (by this I mean "try to solve"  :) ) last year's problems #1 and #4 - there's not much math at all needed on either of these problems.  This means that you should be focusing on solving problems, not learning more math.

Good luck!

[quote=zfsmathematica]I do not have any advice for how to study,as I am a 7th grader who has never even taken the AMC 10 or 12,but I know how the USAMO/USAJMO work. The scoring is the same as in the IMO. There are 6 problems. Depending on if you come up with the answer,and then the quality of your proof or solution,you will get more points -up to 7 per question,for a perfect score of 42. The test is 2 days long: 4.5 hours each day,3 problems. The 12 people with the highest scores will get into MOSP,formerly,and still referred to by the qualifiers,as MOP. 18 other promising students will be invited. It is a summer training camp that lasts 3 weeks,after which there is a test called the TSTST. The top 6 scorers will represent the United States at the IMO.[/quote]
I think he knows after 8 years. But nonetheless good information


function
Euler
abstract algebra
algebra
functional equation

Hello everyone. This is a bizarre problem that Ramanulan gave:

$ 1\plus{}2\plus{}3\plus{}4\plus{}...\equal{}1/12$

All I've been able to deduce is that $ 1\plus{}2\plus{}3\plus{}4\plus{}...\equal{}(1/1)^\minus{}1/2\plus{}(1/2)^\minus{}1/2\plus{}(1/3)^\minus{}1/2\plus{}...$, and after this, somehow diophantine equations would come into play. However, I can't make any headway, so could someone explain it to me?

$ \infty \equal{} 1/12$? Are you sure?

I don't really get your question.

That makes no sense; it does not even converge!  :huh:

Sorry, -1/12. And Ramanujan got this not me. All I know is that somehow, diophantine equations must be involved.

Then Ramunujan got it wrong.

Explain your source. Explain your problem more in-depth. You've corrected me from saying $ \infty \equal{} 1/12$ to saying $ \infty \equal{} \minus{} 1/12$? How likely.

EDIT: Me being a #@#^@&. I apologize.

What is the problem anyway? You just gave us an equation.

I don't know the problem. Source is from a book titled "The Indian Clerk", though it will be a great coincidence if any of you have it. The problem crept out of one of Ramanulan's letters to Hardy, and even Hardy was puzzled so as to how he got it. I hope I haven't wasted everyone's time.

No, you haven't.

What exactly are you trying to show? That $ \infty \equal{} 1 \plus{} 2 \plus{} 3 \plus{} \ldots \plus{} \equal{} \frac { \minus{} 1}{12}$?

If it is, I bid you good luck.

EDIT: Me being a #@#^@&. I apologize.

This is one of those [b]fantastic[/b] results that seems obviously false but ties into a lot of deep mathematics; I believe it is originally due to Euler.

This sum occurs in the computation of the zero-point energy of the vacuum, and the mathematics that makes it precise is known as [url=http://en.wikipedia.org/wiki/Zeta_function_regularization]zeta function regularization[/url].  A "proof" of the sum is given [url=http://cornellmath.wordpress.com/2007/07/30/sum-divergent-series-ii/]here[/url], and an explanation of how it relates to physics can be found [url=http://math.ucr.edu/home/baez/week126.html]here[/url].

Here's a question that may sound stupid at first, but it is the key to understanding what Ramanujan meant by this statement: What does "=" mean for an infinite series? Are there any other possible interpretations of what "=" can mean?

[hide="Direct Answer:"] Ramanujan did not mean that this series equals $ \frac { \minus{} 1}{12}$ in the sense of "the limit of the partial sums" (as we usually mean when we assert that an infinite series equals some number). He meant it in the sense of [url=http://en.wikipedia.org/wiki/Analytic_continuation]Analytic Continuation[/url] of the [url=http://en.wikipedia.org/wiki/Riemann_zeta_function]Zeta Function[/url] via the [url=http://en.wikipedia.org/wiki/Riemann_zeta_function#The_functional_equation]Functional Equation[/url].

Now, how "realistic" is this interpretation of "="s? Well that is another question, but there is some motivation for it.[/hide]

Edit: Nevermind.. I see t0rajir0u beat me to it.

I find this whole phenomenon rather fascinating.  When students are first exposed to series summation, they are taught to stress rigor and to avoid making "nonsensical" statements like

$ 1 \minus{} 1 \plus{} 1 \minus{} 1 \pm ... \equal{} \frac {1}{2},$

even though Euler, one of our greatest mathematical minds, made such statements all the time and used them to give quite beautiful, but by modern standards unrigorous, proofs (such as the solution to the Basel problem).  It can take them a lot more time before they realize that there are alternate ways to think about infinite series:  [url=http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation]Cesaro summation[/url] turns the above into an honest statement, and if you want to be convinced that Cesaro summation is a legitimate way to think about infinite series you only have to look at [url=http://en.wikipedia.org/wiki/Fej%C3%A9r_kernel]how nice it makes the theory of Fourier series[/url].  One might call this the "physicist's" approach to mathematics (not least because it is directly relevant to physics), and I think it contains an interesting lesson for pure mathematicians.

This particular result is also fascinating:  $ 12$ and its companion $ 24$ are numbers that show up in a surprising number of places, including but not limited to

- this result,
- the [url=http://en.wikipedia.org/wiki/Dedekind_eta_function]Dedekind eta function[/url] (as explained in John Baez's post) and the [url=http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions#Modular_discriminant]modular discriminant[/url],
- the [url=http://en.wikipedia.org/wiki/Leech_lattice]Leech lattice[/url],
- an elementary theorem on [url=http://www.math.utexas.edu/~villegas/publications/polygons.ps]lattice polygons[/url],
- the second [url=http://en.wikipedia.org/wiki/Bernoulli_numbers]Bernoulli number[/url],

and it appears that all of these facts are intimately related!

[quote="Zhero"]Then Ramunujan got it wrong.

Explain your source. Explain your problem more in-depth. You've corrected me from saying $ \infty \equal{} 1/12$ to saying $ \infty \equal{} \minus{} 1/12$? How likely.[/quote]
And I don't know why you're being so rude!  Even in the unlikely event that this was some kind of mistranslation of something Ramanujan wrote, I think it is perfectly legitimate (especially in hindsight) to ask if Ramanujan was onto something even if at first the result doesn't appear to make sense.  That is, after all, the original meaning of [b]paradox.[/b]

As far as explaining the source, this is unfortunately one of those folklore identities that gets thrown around a lot without proper context, so I'm not surprised that the OP wasn't able to provide more details.

I apologize for my remark. I'm unfamiliar with this part of mathematics, and the literal notation confused me. I had no clue what I was talking about. Or perhaps I did, and had an inkling that this was something similar to $ \minus{}1 \equal{} 1 \plus{} 2 \plus{} \ldots$, and the inner no-that's-not-rigorous <text> came out of me. Despite this, I should've at least thought that something was a bit too fishy to go on (why on earth would I reasonably expect someone to show that $ \infty \equal{} \minus{} 1/12$?,) and shouldn't have said it so arrogantly...

On that note, the links provided are quite interesting. I may be going off on a tangent, but the "rigorous evaluation" for $ 1 \plus{} 2 \plus{} 4 \plus{} \ldots \equal{} \minus{} 1$ seems reminiscent of the "proof" that if $ x \equal{} 1 \plus{} 2 \plus{} \ldots$, then $ x \equal{} 1 \plus{} 2x$, so $ x \equal{} \minus{} 1$. Is there any connection?

[quote="Zhero"]the "rigorous evaluation" for $ 1 \plus{} 2 \plus{} 4 \plus{} \ldots \equal{} \minus{} 1$ seems reminiscent of the "proof" that if $ x \equal{} 1 \plus{} 2 \plus{} \ldots$, then $ x \equal{} 1 \plus{} 2x$, so $ x \equal{} \minus{} 1$. Is there any connection?[/quote]
Yes; this argument works fine in the [url=http://en.wikipedia.org/wiki/P-adic_number]$ 2$-adic numbers[/url] and I suppose you can also think of it as a simple case of analytic continuation whereby $ 1 \plus{} z \plus{} z^2 \plus{} ...$ is analytically continued from the open disk $ |z| < 1$ to $ \mathbb{C} \minus{} \{1 \}$ as $ \frac{1}{1 \minus{} z}$.  One of the reasons people like analytic continuation is that it makes functional equations like $ (1 \plus{} z \plus{} z^2 \plus{} ...) \equal{} 1 \plus{} z(1 \plus{} z \plus{} z^2 \plus{} ...)$ true in essentially the largest possible setting.

[quote="t0rajir0u"]As far as explaining the source, this is unfortunately one of those folklore identities that gets thrown around a lot without proper context, so I'm not surprised that the OP wasn't able to provide more details.[/quote]

This equation was stated by Ramanujan in his famous letter to Hardy (and other British mathematicians) in 1913. See page 29-30 [url=http://books.google.com/books?id=eu3F_1J-waUC&printsec=frontcover&dq=Ramanujan+%22Letters+and+Commentary%22#PPA29,M1]here[/url].

It's a good thing Hardy didn't use it to dismiss him :P

Anyway, one of the comments in the posts that I linked to reminded me of another example:  any reasonable summation method will tell you that

$ ... \plus{} \frac {1}{z^2} \plus{} \frac {1}{z} \plus{} 1 \plus{} z \plus{} z^2 \plus{} ... \equal{} 0,$

for example, if the sum is written as $ \frac {1}{1 \minus{} \frac {1}{z}} \plus{} \frac {z}{1 \minus{} z}$.  But the actual summation itself diverges for every value of $ z$; if $ |z| > 1$ the right part diverges, if $ |z| < 1$ the left part diverges, and if $ |z| \equal{} 1$ both parts oscillate.  However, when $ |z| \equal{} 1$ it's possible to compute a Cesaro sum of the sequence (if you interpret a partial sum as $ \sum_{k \equal{} \minus{} N}^{N} z^k$) which will again turn out to be $ 0$ in every case except $ z \equal{} 1$, where both parts of the summation diverge in the most straightforward way possible.  When $ z \neq 1$, you can think about this in terms of the "average value" of $ z^k$, i.e. as an [url=http://en.wikipedia.org/wiki/Equidistribution]equidistribution[/url] result if $ z$ isn't a root of unity and as a simple geometric statement about the centroid of an $ n$-gon if $ z$ is an $ n^{th}$ root of unity.

One way to interpret this result is, in my opinion, quite nice:  if you substitute $ z \equal{} e^{ix}$, the summation is none other than the Fourier series of the [url=http://en.wikipedia.org/wiki/Dirac_delta_function]Dirac delta function[/url]!  And the theory of distributions is the final rigorous setting in which everything makes sense.

Thank you. I understood most of the site though not all (I'm just an 8th grader!). What does O.P. mean?

"Original Poster" or "Original Post," depending on context.


What is the largest cude that would fit in a sphere. or u could say wat proportion of the volume of the sphere does the cube take.

cube and sphere are regular shapes, have same origin

let sphere radius 1, cube sidelength 2x, then by pythagoras from center to corner, sqrt(3x^2) = 1 implies x is 1/sqrt(3), so edgelength of cube is 2/sqrt(3)

do your own homework


Problem: #29 in Strang Linear Algebra

Prove that the identity matrix cannot be the product of 3 row exchanges (or five). It can be the product of 2 exchanges (or 4).

Now, to start, we try to count the number of rows that are different from the identity matrix. For the first row exchange, it's 2. Second, it becomes either 0, 3, or 4 [both can change to identity, two more might be different, or a different row may switch with a same row - resulting in one extra different row]. Third, it can go to 1,2,3,4,6. Four, 0,1,2,3,4,5,6,8... But I think I'm doing this incorrectly. Can anyone help?

Well, use the fact that det(I)=1 and that every row changing operation multiplies the determinant by -1.  So an odd number of row changing operation will leave you with a negative determinant. (while an even number will result in positive determinant).


Canceling out the [b]1/2[/b], and getting rid of parenthesis (is that how you call them in English?) yields:

[b][i]8x-8x^2+1+x<=11+11x-16+16x[/i][/b]

After doing necessary arithmetic you'll get:

[b][i]-8x^2+9x+1<=27x-5[/i][/b]

Then bring everything to the left side:

[b][i]-8x^2-18x+6<=0[/i][/b]

Multiplying everything by [b][i]-1[/i][/b] yields:

[b][i]8x^2+18x-6>=0[/i][/b]

Solving this quadratic inequality for positive [b][i]x[/i][/b] produces an ugly result:
[b][i]x>=0.29472708645...[/i][/b]  :(

PS Where can I find normal math signs? (not like x^2 or x>=y)
ty

[quote="dondigo"]I think that Neravo ment something different.
Possibly:
$x\sqrt{8(1-x)}+\sqrt{1+x}\leq\sqrt{11(1+x)}-\sqrt{16(1-x)}$
But I'm not sure.
To Neravo:If you want to write square roots write like that :
[b]\sqrt{something}[/b][/quote]

EXACTLY!!!

[quote="Hriundel"]
PS Where can I find normal math signs? (not like x^2 or x>=y)
ty[/quote]

You can use LaTeX: http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php


superior algebra
superior algebra theorems

Do there exist noncommutative rings with an element that has only the left inverse?

The classic example of something with a one-sided inverse: left and right shift operators, on a space of sequences.

Making this into a ring: let $ c_0$ be the space of sequences of real numbers which tend to zero, with the norm $ |x|\equal{}\max_n(|x_n|)$. Let $ A$ be the ring of bounded (continuous) linear operators from $ c_0$ to itself, with composition as the "multiplication" operation. The operator norm is $ |f|\equal{}\sup_{x\neq 0}\frac{|f(x)|}{|x|}$, and it has the property $ |f\circ g|\le |f|\cdot |g|$. We also have $ |f\plus{}g|\le |f|\cdot |g|$, and these operations are well-defined on $ A$.
Let $ L$ be defined by $ L(x_0,x_1,x_2,\cdots)\equal{}(x_1,x_2,x_3,\cdots)$ and $ R$ by $ R(x_0,x_1,x_2,\cdots)\equal{}(0,x_0,x_1,\cdots)$. We have $ |L|\equal{}|R|\equal{}1$, and $ L\circ R$ is the identity. Since $ L$ is not injective and $ R$ is not surjective, they don't have inverses on the other side.


geometry
induction
number theory
similar triangles

I just started Geometry, and I was really suprised at the difficulty of it.  My mom was really worried, becuase she said the Geometry was realy really hard, and even my cousin who's really smart had a tough time with it, and so on.  But When our class started, I was wondering - "Is this supossed to be a 10th grade class?" becuase everything is so easy.  We spent the first day doing algebra, and the teacher had to stop becuase some people were having trouble.  We even add to go over how to plot points, something that is like 3rd grade math.  We're going to spend 3 class periods, trying to learn the definitions of coplanar, colinnear, and how when two planes intersect, the intersection is a line.  And, our homework takes 5 minutes, if you really don't understand what the teacher is saying.

Is Geometry class going to be this easy all year, or is it going to change?  RIght now, we're using the Glencoe Geometry books.

American classroom geometry is always ridiculously slow and almost never taught well. I was in an "accelerated" program and it was still a joke. Back then, I didnt know about any of THIS, but looking back on it, I realize what a horribly taught class it was.

Unless you go to a good magnet school geometry will be a badly taught class.  You are better off do it on your own or taking the AoPS class for it.

Most math classes in schools(unless you go to TJ, Exeter, AAST) are generally badly taught (at least an A- if you memorize all the formulas/theorems  :(  no creativity and ingenuity needed )

due to me being a TJ kid, i will say something:

TJ math courses aren't exactly exciting either.
it's basically the same as a normal math class.
memorize formulas, do practice problems, listen in class.
do that and you'll be fine

All the talent comes from being together with other talented people day after day after day....

Nosepuppy is right. I go to TJ as well. Creativity is learned at math team, not in math class. Of course, we have a few exceptionally good teachers that make some classes better than others, but almost all schools have some special teachers. I do not know anything about the other two schools you mentioned though  :) .

[quote="beta"]Okay... I was just assuming TJ had better math classes since TJ does so well on AMC/AIME/USAMO.
So most math classes in any school is basically a waste of time. We should just all homeschool ourselves on math.[/quote]

Aren't most of us doing that, just by answering problems posted by other people?

As a former Stuy student, I would like to disagree somewhat with Nosepuppy and JGeneson.  Classes at schools like ours aren't "exciting," exactly, but they aren't just "normal," either.  Our schools attract smart, good teachers.  So while the courses may be limitted by the nature of a classroom, they are actually very good by any reasonable standard of public school classes.
It is true that all the [i]really[/i] fun stuff happens in math team, though, which is entirely a result of having a critical mass of people who are interested in math.

My school doesn't have any exceptional math teachers.  The closest thing told me today at math team that I could use Fermat's Little to prove that 5 divides 3^(2n+1)+2^(2n+1).

And, just for kicks, I finally bought Geometry Revisted, because my geometry skills are sooooo lacking.  Should be arriving from Amazon tomorrow.  But if you'd really like to be engaged in geometry, that's the book to do it.

[quote="Magnara"]My school doesn't have any exceptional math teachers.  The closest thing told me today at math team that I could use Fermat's Little to prove that 5 divides 3^(2n+1)+2^(2n+1).[/quote]

If you have someone in your school who can tell you that, you DO have an exceptional math teacher. Exceptional because most school mathematics teachers probably have no memory of Fermat's Little Theorem, or never learned it in the first place, and never talk about number theory to high school students.

To get an idea of how bad mathematics instruction is in the United States (compared specifically to China, at the elementary school level), see the book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0805829091/learninfreed]Knowing and Teaching Elementary Mathematics[/url] by Liping Ma. You probably haven't got time to read the book right now, but everyone on your school's board of directors ought to be reading it, and one of your parents might find it interesting--and dismaying.

When I left China in the middle of grade 8, we had finished congruent and similar triangles and were just about to start circles, I think.

That equates to grade 12 materials here in Canada. Yeah, it's rather dismaying. It's no surprise that the Canadian IMO team members came from 1) homeschooling 2) schools with gifted programs and 3) new immigrants.  :D

[quote="tokenadult"][quote="Magnara"]My school doesn't have any exceptional math teachers.  The closest thing told me today at math team that I could use Fermat's Little to prove that 5 divides 3^(2n+1)+2^(2n+1).[/quote]

If you have someone in your school who can tell you that, you DO have an exceptional math teacher. [/quote]

What if it's not true?  I can't see how it's true, at least.  But it is true that 3^(2n+1) == (-2)^(2n+1) mod 5, and the result follows.

[quote="Osiris"]Eh?

32n + 1 + 22n + 1
= (3 + 2)(32n - 32n - 12 + 32n - 222 + ... + 22n)
= 5 (something)[/quote]

Tom, I know that it's divisible by 5 (as I alluded to when I noted that 3^(2k+1) == (-2)^(2k+1) ).  But I don't see what it has to do with FLT - do you see an FLT solution?

As an aside, we also can use everyone's best friend, induction:

3^1 + 2^1 | 5

If 3^(2k+1) + 2^(2k+1) = 5q holds, then:

9*3^(2k+1) + 4*2^(2k+1) = 4*(5q) + 5*3^(2k+1), which is obviously divisible by 5.  Hence it is true by induction.

Sorry to revive this old thread.. but I just realized that Liping Ma is my Chinese school teacher! :D