Let be $ x,y,z>0$ such that $ xyz\equal{}1$ . Prove that :

$ \frac{(x^2\plus{}yz)(y^2\plus{}zx)(z^2\plus{}yz)\sqrt{x^2y\plus{}y^2z\plus{}z^2x}}{(x^2y\plus{}y^2z\plus{}z^2x)^2}\ge \frac{8\sqrt{3}}{9}$

Hello,

By AM-GM $ {(x^2y \plus{} y^2z \plus{} z^2x)^2}\geq 9$

Now its enought to prove that $ {(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} yz)\sqrt {x^2y \plus{} y^2z \plus{} z^2x}}\geq 8\sqrt {3}$

By AM-GM to $ x^2 \plus{} yz ,$ $ y^2 \plus{} zx$ and $ z^2 \plus{} yz$ we have that $ {(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} yz)}\geq 8$

and $ \sqrt {x^2y \plus{} y^2z \plus{} z^2x} \geq \sqrt {3}$.

$ \implies \frac {(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} yz)\sqrt {x^2y \plus{} y^2z \plus{} z^2x}}{(x^2y \plus{} y^2z \plus{} z^2x)^2}\ge \frac {8\sqrt {3}}{9}$

IGNORE THIS POST !

[quote="enndb0x"]Hello,

By AM-GM $ {(x^2y \plus{} y^2z \plus{} z^2x)^2}\geq 9$

Now its enought to prove that $ {(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} yz)\sqrt {x^2y \plus{} y^2z \plus{} z^2x}}\geq 8\sqrt {3}$

By AM-GM to $ x^2 \plus{} yz ,$ $ y^2 \plus{} zx$ and $ z^2 \plus{} yz$ we have that $ {(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} yz)}\geq 8$

and $ \sqrt {x^2y \plus{} y^2z \plus{} z^2x} \geq \sqrt {3}$.

$ \implies \frac {(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} yz)\sqrt {x^2y \plus{} y^2z \plus{} z^2x}}{(x^2y \plus{} y^2z \plus{} z^2x)^2}\ge \frac {8\sqrt {3}}{9}$[/quote]

You're wrong!  :blush:

yes the first step is wrong, your use AM-GM is true, but after no,


Let $ ABCDEF$ be a convex hexagon with all the sides of length 1. Prove that one of the radii of the circumcircles of triangles $ ACE$ or $ BDF$ is at least 1.

[quote="mto"]Lemma:

Let MNPQ be a convex quadrilateral where MN=NP=1 and  2.[MQP] =< [MNP]  then the radius of circumstance of MPQ is at least 1

Apply the lemma, if all the radius are less than 1, takes the sum on all the angles we come to a contradiction that

 2*2*180 >  (6-2)*180[/quote]

If your Lemma is not in "iff" form, but only "if" you can't use it like that. If it is iff provide a proof of your lemma.

[quote="Goblin"]This problem is from Kvant (M951).[/quote]

  Well, it was quite straightforward to discover the result in the lemma when I compared the angles ABC and AOC where O is the center of
circumstance circle of ACE

Let $ m(B)\plus{}m(D)\plus{}m(F)\geq360$ . $ O$ be the circumcenter of triangle $ AEC$. If circumradii of $ AEC$ is $ <1$ then $ FE>EO$ , $ DC>OC$ , $ AB>OB$ and               $ m(FOE)>m(EFO)$ , $ m(EOD)>m(EDO)$ , $ m(AOB)>m(ABO)$ and  $ \frac{m(B)\plus{}m(D)\plus{}M(F)}{2}<180$ and this is a contraction.


inequalities
search
absolute value
inequalities unsolved

Find the minimum of $ x^2\plus{}y^2\plus{}z^2$ where $ x,y,z \in \mathbb{R}$ and satisfy $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}1$

Enjoy

Condition of problem may be rewritten as:

$ (x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xy \minus{} yz \minus{} zx) \equal{} 1$ and since second bracket on $ LHS$ is nonnegative we have $ x \plus{} y \plus{} z > 0$.

Notice that from last equation we have:

$ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} \frac {1 \plus{} (xy \plus{} yz \plus{} zx)(x \plus{} y \plus{} z)}{x \plus{} y \plus{} z} \equal{} \frac {1}{x \plus{} y \plus{} z} \plus{} xy \plus{} yz \plus{} zx$

and since $ xy \plus{} yz \plus{} zx \equal{} \frac {(x \plus{} y \plus{} z)^{2} \minus{} x^{2} \minus{} y^{2} \minus{} z^{2}}{2}$ last equation implies:

$ \frac {3(x^{2} \plus{} y^{2} \plus{} z^{2})}{2} \equal{} \frac {1}{x \plus{} y \plus{} z} \plus{} \frac {(x \plus{} y \plus{} z)^2}{2} \equal{} \frac {1}{2(x \plus{} y \plus{} z)} \plus{} \frac {1}{2(x \plus{} y \plus{} z)}$ $ \plus{} \frac {(x \plus{} y \plus{} z)^2}{2}\geq \frac {3}{2}$

This inequality follows from $ AM \geq GM$ so $ x^{2} \plus{} y^{2} \plus{} z^{2}\geq 1$ so minimum of $ x^{2} \plus{} y^{2} \plus{} z^{2}$ is 1 and triple $ (1,0,0)$ shows that this value can be achieved.  q.e.d.

Actually,this inequality is the consequence of the inequality:
square root(a^2+b^2+c^2)^3>=/a^3+b^3+c^3-3bac/(a^3+b^3+c^3-3abc=1)
Applying this inequality and we're done!

ghjk, where did this inequality come from?
$ \sqrt {(x^2 \plus{} y^2 \plus{} z^2)^3}\ge x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$

It's very famous inequality! You even can search it on AOPS since I can't remmember specifically which post includes it! :)
PS:The RHS should have absolute value sign! :wink:

[url=http://www.artofproblemsolving.com/community/c6h247447p1358044]Indonesia TST 2009 First Stage Test 2 Problem 3[/url]


Let the operation * be defined by $ a * b \equal{} ab \plus{} a \minus{} b$. Which of the expression below is wrong, or are they all correct?

A. $ a * a \equal{} a^2$
B. $ a * b \equal{} (\minus{}b) * (\minus{}a)$
C. $ (a * b) * a \equal{} a^2 * b$
D. $ (a * b) * c \equal{} (a * c) * b$
E. All four are correct

Second one is clearly wrong.. I haven't had time to check the others.

All of them are correct, including the second one.
The second one:
$ (\minus{}b)*(\minus{}a)\equal{}(\minus{}b)(\minus{}a)\plus{}(\minus{}b)\minus{}(\minus{}a)\equal{}ba\minus{}b\plus{}a\equal{}ab\plus{}a\minus{}b\equal{}a*b$

Ah you're correct. I had read it as $ a * b \equal{} (\minus{}a) * (\minus{}b)$, which doesn't work, I think.

no that wouldnt work because when you simplify the equation you would get $ (\minus{}a)*(\minus{}b) \equal{} ab \plus{} b \minus{} a$  which doesnt equal $ ab \plus{} a \minus{} b$.

[hide="Solution"]

Let's consider all four of them.

$ a * b \equal{} a^2 \plus{} a \minus{} a \equal{} a^2$ so A is true.

$ a * b \equal{} ab \plus{} a \minus{} b$ and $ (\minus{}b) * (\minus{}a) \equal{} (\minus{}b)(\minus{}a) \plus{} (\minus{}b) \minus{} (\minus{}a) \equal{} ab \plus{} a \minus{} b$ so B is true.

$ (a * b) * a \equal{} (ab \plus{} a \minus{} b) * a \equal{} a(ab \plus{} a \minus{} b) \plus{} ab \plus{} a \minus{} b \minus{} a \equal{} a^2 b \plus{} a^2 \minus{} ab \plus{} ab \plus{} a \minus{} b \minus{} a\equal{} a^2 b \plus{} a^2  \minus{} b$

Now, compare that to $ (a^2 * b) \equal{} a^2 b \plus{} a^2 \minus{} b$ so they are equal. Thus, C is true.

$ (a * b) * c \equal{} (ab \plus{} a \minus{} b) * c \equal{} c(ab \plus{} a \minus{} b) \plus{} ab \plus{} a \minus{} b \minus{} c \equal{} abc \plus{} ac \minus{} bc \plus{} ab \plus{} a \minus{} b \minus{} c$

$ (a * c) * b \equal{} (ac \plus{} a \minus{} c) * b \equal{} b(ac \plus{} a \minus{} c) \plus{} ac \plus{} a \minus{} c \minus{} b \equal{} abc \plus{} ac \minus{} bc \plus{} ab \plus{} a \minus{} b \minus{} c$

So, D is true as well. Thus, the answer is E. [/hide]


I had some problem with this, but I think that is not an Olympiad level!!

[quote] Let $x,y,z,t\in\mathbb{R}^{*}_{+}$ such that: $\left\{\begin{array}{lll}x^{2}+y^{2}\leq 1 \\ z^{2}+t^{2}\leq4\end{array}\right.$

Proove that:   $xz+yt\leq 2.$

Thanks for help![/quote]

Tanks, it was was easy as I thought, but I had really problem with it! :lol:

This can be also proved in trigonometrical way I think.


I noticed the average AIME score in 2008 was 4.77, but in all the other years it's between 2 and 3. Was the exam much easier? Or did the AMC do a better job of weeding out the "underperformers" thus giving a better contestant pool?

I really felt that the easy problems on this year's AIME were much easier than the easy ones on previous AIME's, which probably eliminated a lot of the 1's and 2's.  Also, the AMC 12 was harder, so less people qualified to take the AIME.

I don't think the overall difficulty was significantly different - AIME15 has the reasons. It's possible some more sophisticated statistical analysis would show this.

Additionally, the floor score was actually a bit LOWER than some recent years.

Those are some interesting theories, also, I think that the removal of calculators eliminated a lot of brute forcing- number bashers from the AIME pool....

...you weren't allowed to use calculators on AIME for awhile...possibly since the start.  I know for a fact it was not allowed from 2001-2008.

Dude I'm talking about removing calculators from the AMCs....how would removing calculators from the AIME have any impact on the AIME participant pool?

yeah, that and the 1.5 for blanks from 2.5, would certainly make the it more difficult to make the AIME

They keep lowering the floor score partly because they are increasing the number of students who get to take the USAMO. Also, the difficulty of the AIME really depends on what level you are at. Some tests have problems of more even difficulty, which might be harder for people who can only solve the easier questions but easier for people who tend to get 10 or above. Since the 2007 AIME I was a particularly hard test, (some kids who went to IMO got less than 10), 2008 will seem easy by comparison.


Let $d_n$ be the last digit, distinct from 0, in the decimal expansion of $n!$. Prove that the sequence $d_1,d_2,d_3, \ldots$ is not periodic.


Is there a fairly simple way to convert a LaTeX file to a .gif, .jpg, or .jpeg format?

Export it as a .ps file then use command line Imagemagick to create any type of image file from that.

I'm not familiar with Imagemagick.  Could someone please elaborate?

The idea is that you install [url=http://www.cs.wisc.edu/~ghost/]Ghostscript[/url] and [url=http://www.imagemagick.org/script/index.php]Imagemagick[/url] and say [i]convert myfile.ps myfile.png[/i] on the command line. There's no problem installing them both on Windows.

I am assuming you want to convert a complete page of latex, rather than just a bit as in the TeXer. You can do this online for example at [url=http://www.sciencesoft.at/index.jsp?link=latex&lang=en]ScienceSoft[/url] without installing anything on your computer. There's a whole long list of image formats offered.

I have ghostscript from when I downloaded asymptote, and when I went to the download page in sourceforge.net, it gave me the list in the attached screenshot.  Which file(s) do I need to download?

Are you using Windows? If so, then you don't want any of those as you want a 'binary'. Go to http://www.imagemagick.org/script/binary-releases.php#windows and then select the top file ImageMagick-6.3.4-1-Q16-windows-dll.exe. This includes an installation program which will make life easier.

Incidentally, please don't upload Word files with images in as many people don't open unknown Word files as they could contain viruses etc. Just copy and paste the image into Paint, save as gif or jpg then upload that to your post.

I have Windows XP.  How do I find the Command Prompt?

Either Start, Accesories, Command Prompt [i]or[/i] Start, Run, type cmd and press Enter.
You will then have to navigate to the directory containing your file which if you aren't familiar with DOS is not easy so here is a simpler method.

Use Notepad (or even TeXnicCenter) to create a new file and put in it:
[code]convert image.ps image.jpg
pause[/code]but changing the names of the file to the right ones. Save the file as myconvert.bat (or any other name as long as it ends in .bat) in the [i]same[/i] directory as your image. Then in Windows Explorer double-click on it and it should work.

Is there any advantage to compiling to ps rather than pdf, for converting to another format?  If so, how do I tell it to do that (when I change the output profile to LaTeX=>PS, it still saves it as a pdf file, but disables view output)?

When I converted the pdf file to jpg (techinically jpeg), it made a separate file out of each page--not useful for my current purpose.  The only other file type I can use is gif.  But when I use that, it rapidly flips through all the pages.  Is there some workaround for this?

[quote="roadnottaken"]Is there any advantage to compiling to ps rather than pdf, for converting to another format?  If so, how do I tell it to do that (when I change the output profile to LaTeX=>PS, it still saves it as a pdf file, but disables view output)?[/quote]If you installed Ghostscript after TeXnicCenter then you need to tell TeXnicCenter where to find Ghostscript which should be in Build->Define Outut Profiles->LaTeX=>PS. Use the Wizard button to set it up.

[quote="roadnottaken"]When I converted the pdf file to jpg (techinically jpeg), it made a separate file out of each page--not useful for my current purpose. [/quote]That makes sense as how can more than one page fit in an image without making it too large to view sensibly? Of course you could use a graphics program to put the images together but really I can't understand what you want to achieve by this. So what exactly are you trying to do?

There is a website called zinch.com that allows students to create profiles that can be viewed by admissions officers.  One of the features is the z-folio, where you can display writing, images, audio, or videos.  Unfortunately, the writing section requires you to use their text editor, which, naturally, is incompatible with $\LaTeX$.  The image section only accepts files in .jpg, .jpeg, or .gif format, so I am stuck trying to funnel multiple-page documents including lots of equations through one of these file types if I want to display them.  Now, I would prefer to upload it as one file, but, obviously, treating a large document as an animation does not work!

So you certainly wouldn't want to upload a massive jpeg which could only be read by lots of scrolling. I suggest that each page of your document is converted into an image and you could label them as my_work_page_1, my_work_page_2 etc.
[quote="roadnottaken"]When I converted the pdf file to jpg (techinically jpeg), it made a separate file out of each page--not useful for my current purpose.[/quote]Isn't that precisely what you want?

I would prefer to at least keep individual solutions together, but some of them run more than a page.  I also have a tendency to prefer to not split files up whenever possible.

Try using the -append command when you use ImageMagick's convert and it ought to produce one image. In fact there's even an option to use +append when the images will be side-by-side rather than top to bottom.


Acacia Avenue has eight houses, numbers $ 1, 3, 5, 7$ on one side opposite numbers $ 2, 4, 6, 8$. [$ 1$ is opposite $ 2$, $ 3$ opposite $ 4$, etc.]

An eccentric postman sets himself some rules. he delivers alternately to odd and even number houses and never crosses the road to a house directly opposite one to which he has just delivered mail. He always starts at number $ 1$ and returns there at the end for a cup of tea.

Assuming he always has mail for all $ 8$ houses, how many different routes can he use?

Oh, sorry  :blush:

I've never been good in combinatorics... How have you solved the problem?

Thanks,
Gabriele  :)

[hide]
Just draw out the houses and then you will see he has 3 options for the second house, 2 for the third and 2 for the fourth. The problem says that he must come back to a 1, so that basically limits all of your other options (since he can't pick a 2 right before a 1).

I get 3*2*2*1*1*1 = 12

Hope I did that right
[/hide]

[hide="Solution?"] When he starts at house $ 1$ there are $ 3$ possible even houses he can go to. From there, he can go to any of $ 2$ odd houses (he can't go to house $ 1$ or the house directly across from the one he chose). He then must go to house $ 2$ (since if it were the last house, he'd have to cross over directly for his cup of tea, and he can't do that). From there he has $ 2$ possiblities for his next choice of odd houses, and everything else is forced to be $ 1$ choice. So, $ 3\times{2}\times{2}\equal{}\boxed{12}$.[/hide]


Well, every invertible matrix is row equivalent to the identity matrix (as can be proved algorithmically using Gaussian elimination). (b) is the identity matrix and (c) and (d) are obviously invertible. So the only actual computation we need to do is to row-reduce (a).

OK ... (a) is singular. It has rank 2.

You now know the answer to the multiple choice question. What do you say?


Adams plans a profit of $ 10\%$ on the selling price of an article and his expenses are $ 15\%$ of sales. The rate of markup on an article that sells for $ \$5.00$ is:
$ \textbf{(A)}\ 20\% \qquad\textbf{(B)}\ 25\% \qquad\textbf{(C)}\ 30\% \qquad\textbf{(D)}\ 33\frac {1}{3}\% \qquad\textbf{(E)}\ 35\%$

[hide="Click for solution"]
S=Selling price = cost + profit + expense = C + .10S + .15S, so $ S\equal{}\frac{4}{3}C$.  Therefore the answer is $ \boxed{\textbf{(D)}}$.
[/hide]


[i]Μιας και πλησιάζει ο εσωτερικός, βάζω ένα πρόβλημα από μεσογειάδα...[/i]

Να βρείτε όλους τους ακεραίους $ n$ ώστε το πολυώνυμο $ p(x)\equal{}x^5\minus{}nx\minus{}n\minus{}2$ να μπορεί να γραφεί ως γινόμενο 2 μη σταθερών πολυωνύμων με ακέραιους συντελεστές!

Την προσπαθησα την ασκηση και την εφτασα μεχρι ενα σημειο.

Εστω $ p(x)\equal{}f(x)g(x)$, οπου $ f(x),g(x)$ πολυωνυμα με ακεραιους συντελεστες.Διακρινω δυο περιπτωσεις.
1) Αν $ f(x)$:πρωτου βαθμου και $ g(x)$:τεταρτου βαθμου. Τοτε $ f(x)\equal{}x\minus{}r$. Τοτε r ακεραια ριζα του $ p(x)$. Τοτε ισχυει $ r^5\minus{}nr\minus{}n\minus{}2\equal{}0$.
Τοτε $ n\equal{}\frac{r^5\minus{}2}{r\plus{}1}$. Τοτε $ n\equal{}r^4\minus{}r^3\plus{}r^2\minus{}r\plus{}1\minus{}\frac{3}{r\plus{}1}$.
Αρα $ r\plus{}1\equal{}1,\minus{}1,3,\minus{}3$. Οποτε $ (r,n)\equal{}(0,\minus{}2),(\minus{}2,34),(2,10),(\minus{}4,342)$.

2)Αν $ f(x)$:δευτερου βαθμου και $ g(x)$:τριτου βαθμου. Εστω $ f(x)\equal{}x^2\plus{}kx\plus{}l,g(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}c$. Επειδη $ p(x)\equal{}f(x)g(x)$ μετα απο πραξεις προκυπτει
$ p(x)\equal{}x^5\plus{}(a\plus{}k)x^4\plus{}(l\plus{}ka\plus{}b)x^3\plus{}(c\plus{}kb\plus{}la)x^2\plus{}(kc\plus{}bl)x\plus{}cl$.
Τοτε προκυπτει το συστημα:

$ a\plus{}k\equal{}0$
$ l\plus{}ka\plus{}b\equal{}0$
$ c\plus{}kb\plus{}la\equal{}0$
$ kc\plus{}bl\equal{}\minus{}n$
$ cl\equal{}\minus{}n\minus{}2$

Με απαλοιφη των αγνωστων μεταξυ των εξισωσεων προκυπτει:
$ k^2(k^2\minus{}2b)\plus{}b(k^2\minus{}b)\equal{}\minus{}n$
$ k(k^2\minus{}b)(k^2\minus{}2b)\equal{}\minus{}n\minus{}2$

Αν απαλειψω τον $ n$ μεταξυ αυτων προκυπτει
$ k^2(k^2\minus{}2b)\plus{}b(k^2\minus{}b)\minus{}k(k^2\minus{}b)(k^2\minus{}2b)\equal{}2$
την οποια δεν μπορω να λυσω στο Z. Καμια βοηθεια :?:

έχω την εντύπωση πως τελειώσαμε
το Ρ θα εχει παράγοντα  (x-r) (5ου βαθμού)
Το r θα είναι η ακέραιος η άρρητος
Αν ήταν άρρητος τότε το πηλίκο δεν θα μπορούσε να έχει ακέραιους συντελεστές,άτοπο
Άρα ισχύουν τα προηγούμενα

Νομιζω οτι  με την πρωτη περιπτωση δεν τελειωνει το προβλημα.Οταν το P εχει παραγοντα x-r τοτε πρεπει r ακεραιος.

   Ομως στη δευτερη περιπτωση μπορει τα πολυωνυμα δευτερου και τριτου βαθμου να εχουν ακεραιους συντελεστες χωρις απαραιτητως να εχουν
   ακεραιες ριζες,ΠΧ μπορει το δευτερου βαθμου να μην εχει καν ριζες στο R.Αρα πρεπει να λυθει το συστημα αλλα εχω κολλησει.

   Η ασκηση ζηταει παραγοντοποιηση του Ρ,δεν αναφερει ακεραιες ριζες,μπορει ομως να χανω κατι.

αφού το Ρ είναι περιττού βαθμού με συντελεστές στο R έχει σίγουρα πραγματική ρίζα
Επειδή ο μεγιστοβάθμιος συντελεστής του P είναι το 1 και οι υπόλοιποι ακέραιοι τότε αν η ρίζα r είναι ρητή θα είναι και ακέραια (r=a/b τότε b διαιρεί τον μεγιστοβάθμιο)
Αν είναι άρρητη τότε το πηλίκο Q θα είχε έναν τουλάχιστον άρρητο συντελεστή, άτοπο
Άρα το Ρ έχρι ακέραια ρίζα ...

Εξακολουθω να πιστευω οτι το Ρ μπορει να γραφει σαν γινομενο ενος  δευτεροβαθμιου  και ενος τριτοβαθμιου με ακεραιους συντελεστες
  χωρις κανενα απο αυτα να εχει ακεραιες ριζες και αυτο ειναι που ζηταει η εκφωνηση.

  Θα δωσω ενα παραδειγμα.

  Να παραγοντοποιηθει το $ P(x)\equal{}x^5\plus{}x^4\plus{}3x^3\plus{}2x^2\plus{}2x\plus{}6$  σε πολυωνυμα με ακεραιους συντελεστες.Ευκολα προκυπτει

   $ P(x)\equal{}(x^3\plus{}2)(x^2\plus{}x\plus{}3)$

   Τοτε το Ρ εχει μοναδικη πραγματικη ριζα αρρητη αναμεσα στο-1 και στο -2 και γραφτηκε σαν γινομενο πολυωνυμων με ακεραιους συντελεστες.

   Αρα στην ασκηση ο τριτοβαθμιος παραγοντας του Ρ μπορει να εχει ακεραιους συντελεστες και οχι ρητες ριζες.

   Πρεπει ομως να λυθει το συστημα εκτος αν υπαρχει αλλος τροπος.

Σε όλα όσα είπες συμφωνώ. Μπορεί να κάνω λάθος (συχνά μου συμβαίνει)
Δοκίμασε να κανεις την διαίρεση του $ x^3\plus{}2$ με το $ x\plus{}\sqrt[3]{2}$
εννοώντας ότι ο παράγοντας x-r θα είναι παράγοντας πολυωνύμου 3 βαθμού

Κ Μπορη η ασκηση με εχει παιδεψει για αρκετο καιρο αλλα δυστυχως δεν μπορω να δωσω πληρη λυση.

   Αν θεωρησουμε οτι με την πρωτη περιπτωση δεν τελειωνουμε αλλα υπαρχει και η δευτερη οπου το Ρ γραφεται σαν γινομενο ενος τριτοβαθμιου

επι ενα δευτεροβαθμιο με ακεραιους συντελεστες χωρις ακεραιες λυσεις τοτε θα καταληξουμε στο συστημα που εχω καταληξει στο προηγουμενο post.Αφου λοιπον δεν μπορεσα να λυσω την τελικη εξισωση αρχισα να δινω ευκολες τιμες στο κ ,οπως 1,-1,0,2,-2 μηπως εχει καποια αλλη λυση.

  Τελικα βρηκα οτι υπαρχει μια τουλαχιστον επιπλεον λυση για $ k\equal{}\minus{}1$.Τοτε το Ρ παραγοντοποιειται και εχει την εξης μορφη

    $ P(x)\equal{}x^5\minus{}19x\minus{}21\equal{}(x^2\minus{}x\minus{}3)(x^3\plus{}x^2\plus{}4x\plus{}7)$  Τοτε  $ n\equal{}19$  και ελπιζω να μην εχω καποιο λαθος.Τελικα και αυτη η περιπτωση εδωσε λυση,πιθανον ομως να υπορχουν και αλλες.

  Θα παρακαλουσα τον bilot 4 να βαλει την πληρη λυση,αν βεβαιως την εχει.

Σακη καλησπερα.
Το προβλημα λυνεται οντως με την χρηση του συστηματος που αναφερεις. Δυστυχως ,ομως,  ουτε εγω καταφερα να λυσω την  εξισωση που βγαζεις στο τελος.
Το θεμα ειναι να το λυσεις αλλιως απο το συστημα και κατω.
 Ο Σιλουανος με μια δυσκολη και πολυ ωραια παραγοντοποιηση το εχει λυσει το προβλημα,οντως, λυνοντας το συστημα.
Μια αλλη λυση ομως μπορει να γινει, παντα μιλαμε για την δευτερη περιπτωση, χρησιμοποιωντας το $ P( \minus{} 1) \equal{} \minus{} 3$
Πιο συγκεκριμενα,
εχεις $ (x^2 \plus{} kx \plus{} l)(x^3 \plus{} ax^2 \plus{} bx \plus{} c) \equal{} p(x)$(1)
και απο το συστημα σου
$ a \equal{} \minus{} k$
$ b \equal{} k^2 \minus{} l$
$ c \equal{} \minus{} k^3 \plus{} 2kl$
αρα
η (1) για $ x \equal{} \minus{} 1$ δινει,
$ (l \minus{} k \plus{} 1)( \minus{} 1 \minus{} k \minus{} k^2 \plus{} l \plus{} 2lk \minus{} k^3) \equal{} \minus{} 3$
σχεση που χρησιμοποιωντας οτι ειναι ακεραιοι σε οδηγει στην λυση.  :)

Ευχαριστω γιατι με παιδευε η ασκηση για καιρο. Το υποψιαζομουνα οτι ηθελε καποιο trick και πραγματικα αυτη η ιδεα να το εφαρμοσω για χ=-1 ειναι πολυ εξυπνη γιατι φευγει το n.


induction
quadratics
trigonometry
national olympiad

4.For all integers n>=1 x_(n+1)=(x_1)^2+(x_2)^2+...+(x_n)^2 where x_1 is a positive integer .Find the least x_1 such that 2006 divides x_2006.

5.From a point Q on a circle with diameter AB different from A and B ,we draw a perpendicular to AB QH where H is on AB.The intersection points of the circle with diameter AB and with center Q and radius QH are C and D.Prove that CD bisects QH.

6.Each of 2006 students makes  a list with 12 schools among 2006.If we take any 6 students, there are two schools which at least one of them is included in each of 6 lists.A list which includes at least one school from all lists is a good list.
   a.Prove that we can always find a good list with 12 elements, whatever the lists are.
   b.Prove that students can make lists such that no shorter list is good.

For the problem 4:

We have that $x_{n+1}=x_n+x_n^2$ hence all $x_n$ are even. We will prove by a backward induction that $x_2$ must be divisible by $1003$.

$x_{n}\equiv 0 (\mod 1003)\Leftrightarrow x_{n-1}^2+x_{n-1}\equiv 0 (\mod 1003)\Leftrightarrow x_{n-1}\equiv -1$ or $x_{n-1}\equiv 0 (\mod 1003)$.

We will prove that $x_{n-1}\equiv -1$ cannot hold for $n-1\ge 2$:

$x_{n-1}=x_{n-2}^2+x_{n-2}\equiv -1 (\mod 1003)\Leftrightarrow (2x_{n-2}+1)^2\equiv -3 (\mod 1003)$.

But from quadratic reciprocity law we get that $3$ is quadratic residue $\mod 1003$, but since $1003=4\cdot 250+3$ we have that $-1$ is not a quadratic residue $(\mod 1003)$. Hence contradiction.

Since $1003\mid x_2$, we have that $x_1\equiv 0,-1 (\mod 1003)$. Hence the least value for $x_1$ is 1002, which works.

[quote="cosinerburc"]
5.From a point Q on a circle with diameter AB different from A and B ,we draw a perpendicular to AB QH where H is on AB.The intersection points of the circle with diameter AB and with center Q and radius QH are C and D.Prove that CD bisects QH. [/quote]

Don't understand the definition of C and D, could you please clarify? are there two circles, one with diameter AB and other with center in Q or what?

Daniel

Yes. This problem is very simple. Denote by $T=CD\cap QH$.

We have that : $\sin QOB=\frac{QH}{R}$ where $R$ is the radius of the circle with center in $O$.
After angle chasing we get that $\angle QTD=180^\circ-\angle QOB$.  Hence $\sin QTD=\frac{QH}{R}$.

On the other hand we have that $\sin CDQ=\frac{CQ}{2R}=\frac{HQ}{2R}$.

The conclusion follows from sine law in triangle $QTD$.

The last one:
[quote] If we take any 6 students, there are two schools which at least one of them is included in each of 6 lists
[/quote]
I assume that it means that any 6 lists have at least a school in common. Am I wrong? :?

a) Take five students $a_1,a_2,..,a_5$. Their lists have at most 12 universities in common. We will prove that the list $M$ of common universities is good. Since for every new student, the students $a_1,..a_5,x$ must have at least one universitiy in common, then student $x$ must have in his list one of the universities in $M$. Hence every student have in his list at least one of those universities.
b)In this case the problem is wrong because, If there are $5$ students with less than $12$ univ in common, the list formed by those less than 12 universities in good, contradiction.
Hence any $5$ students have all 12 universities in their list the same, hence every student have the same 12 universities which is definitely a contradiction.


calculus
integration
logarithms
calculus computations

hello folks.  :D

please do

$ \int_0^1\int_0^1\;\dfrac{(\minus{}\ln\,(xy))^3}{1\plus{}(x\,y)^2}\;\textbf dx\;\textbf dy$

[quote="misan"]hello folks.  :D

please do

$ \int_0^1\int_0^1\;\dfrac{( \minus{} \ln\,(xy))^3}{1 \plus{} (x\,y)^2}\;\textbf dx\;\textbf dy$[/quote]
simple just :

$ I(a) \equal{} \minus{} \frac {\partial^{3}\int \int \frac {(xy)^{a} d d }{1 \plus{} (xy)^{2}}} {\partial a^{3}}$

$ \equal{} \minus{} \frac {\partial ^{3}\int \int (xy)^{a}\sum_{i \equal{} 0}^{\infty}( \minus{} 1)^{i}(xy)^{2i} } {\partial a^{3}} \equal{}$

=
$ \minus{} \frac {\partial ^{3}\int \int \sum_{i \equal{} 0}^{\infty}( \minus{} 1)^{i}(xy)^{2i \plus{} a} } {\partial a^{3}} \equal{} \minus{} \frac {\partial ^{3}\int \int \sum_{i \equal{} 0}^{\infty}( \minus{} 1)^{i}\frac {1}{(2i \plus{} a \plus{} 1)^{2}} } {\partial a^{3}} \equal{} 4!\sum_{i \equal{} 1}^{\infty}( \minus{} 1)^{i}\frac {1}{(2i \plus{} a \plus{} 1)^{5}}$
and lets' $ a \to 0$
then $ ... \equal{} 4!\sum_{i \equal{} 1}^{\infty}( \minus{} 1)^{i}\frac {1}{(2i \plus{} 1)^{5}}$ and last can be easly calculated  :) try do it yourself. (or more generaly problem  find:
$ \int_0^1\int_0^1\;\dfrac{( \minus{} \ln\,(xy))^n}{1 \plus{} (x\,y)^2}\;\textbf dx\;\textbf dy$ )


Does anyone know of a [url=http://www.artofproblemsolving.com/Wiki/index.php/British_Flag_Theorem]British Flag Theorem[/url] for a cube?

Thanks in advance!

That depends on what you mean by "a British flag theorem."  If you just want to extend perpendiculars to every face, all you have to do is apply the 2-dimensional British flag theorem twice.

If I knew the lengths of $ AP$, $ BP$, $ CP$, and $ DP$, how would I find the side length of the cube?

My last post was worded slightly strangely.  Try dropping the perpendiculars to the faces and making right triangles and see what happens.

I drew the following diagram, but wasn't able to find a way to get the side length of the cube.  :oops:  Does anyone have a formula that might work?

Thanks!

Mathematica gave me a rather complex answer:
\[ \sqrt {\frac {a^2}{6} \plus{} \frac {b^2}{6} \plus{} \frac {c^2}{6} \plus{} \frac {d^2}{6} \plus{} \frac {1}{3} \sqrt {2} \sqrt { \minus{} a^4 \plus{} a^2 b^2 \minus{} b^4 \plus{} a^2 c^2 \plus{} b^2 c^2 \minus{} c^4 \plus{} a^2 d^2 \plus{} b^2 d^2 \plus{} c^2 d^2 \minus{} d^4}},
\]where $ a$, $ b$, $ c$, and $ d$ are the given lengths.

Use analytic geometry!

establish a coordinate system with the coordinates in terms of the side length of the cube and let $ (x,y,z)$ be the point O.  Write the equations for the distances of O to the vertices and solve the system of the 4 equations for the side length of the cube.


From page 50 [url]http://www.math.cornell.edu/~hatcher/AT/ATchapters.html[/url], it says :

Suppose we attach a collection of $ 2\minus{}$ cells $ e^{2}_{\alpha}$ to a path-connected space X via maps $ \phi_{\alpha}: S^{1}\to X$, producing a space Y . If $ s_{0}$ is a basepoint of $ S^{1}$ then $ \phi_{\alpha}$ determines a loop at $ \phi_{\alpha}(s_{0})$ that we shall call $ \phi_{\alpha}$, even though technically loops are maps $ I : \to X$ rather than $ S^{1}\to X$ . For different $ \alpha$’s the basepoints  $ \phi_{\alpha}(s_{0})$ of these loops $ \phi_{\alpha}$ may not all coincide. To remedy this, choose a basepoint $ x_{0}\in X$ and a path $ \gamma_{\alpha}$ in $ X$ from $ x_{0}$ to $ \phi_{\alpha}(s_{0})$ for each $ \alpha$. Then
$ \gamma_{\alpha}\phi_{\alpha}\overline{\gamma_{\alpha}}$  is a loop at $ x_{0}$ . [b]This loop may not be nullhomotopic in $ X$, but it will certainly be nullhomotopic after the cell $ e^{2}_{\alpha}$ is attached.[/b]

Can someone explain for me why does it say "[b]....., but it will certainly be nullhomotopic after the cell $ e^{2}_{\alpha}$ is attached.[/b] "? many thanks

The cell $ e_\alpha^2$ provides a homotopy from $ \phi_\alpha$ to the constant path.


We all know that the probability of heads coming up on one flip of a coin is .5, and that the probability that after 2n flips, exactly n of them result in heads is $(1/2){}^2{}^n*(2n!)/(n!)^2$. The probability that after 2m people each flip one coin exactly 2n times, that exactly $(1/2){}^2{}^n*(2n!)/(n!)^2*2m$ of them flip exactly n heads can also be similarily but more tediously calculated.

But what about if the chances of H and T are altered, so that 2/3 of the time, T will be flipped? How would we continue calculating these "continued" (for a lack of a more percise or [i]the[/i] formal name) probabilities?

Um I think these are called Bernoulli Trials.
Say the probability of an event $A$ is $n$.  Then the probability that $A$ happens exactly $m$ times out of $t$ trials is:

$\binom{t}{m}n^{m}(1-n)^{t-m}$.

The reasoning behind this:  Say we want 3 heads in a row then 2 tails in a row, where the probability of heads is $\frac{2}{3}$.  Then we have $\frac{2}{3}^{3}\cdot\frac{1}{3}^{2}$.  This corresponds to the $n^{m}(1-n)^{t-m}$ part of the formula.  But what we have so far only works with a specific sequence.  There are $\binom{5}{3}$ ways for the 3 heads and 2 tails to be arranged, and each of the separate events have the same probability as the 3 heads in a row then 2 tails in a row.  So we multiply $\frac{2}{3}^{3}\cdot\frac{1}{3}^{2}$ by $\binom{5}{3}$, and we get our probability.  Now compare this logic to the formula.

So going along that train of thought:

Given three events, the probability that you get a certain amount of each within $n$ trials would be a specific term in the expansions of

$(P(A)+P(B)+P(C))^n$

For example, say you wanted A to happen twice, B once, and C once, in four trials. Therefore, it would be the term

$\binom{n}{i,j,k}(P(A))^i(P(B))^j(P(C))^k$

And this can be extended to any number of probabilities ?

[quote="banblink182"][quote="Elemennop"]So it's just the expansion of $(P(A)+P^C(A))^n$ given $n$ trials, where the $m$th term is the probability it happens $m$ times out of the $t$ trials?[/quote]

Yes, it is.

So they're referred to the Bernoulli trials. Ok, thanks you're right - the probabilities don't really matter in that sense, you still calculate them the same way.[/quote]Banblink,
The distribution of concern here is a Binomial distribution, the probability of obtaining m successes in N independent "Bernoulli" trials: $\binom{N}{m}p^m(1-p)^{N-m}$ where $p$ is the probability of success in a single Bernoulli trial.

The Bernoulli distribution is the distribution that has only two possible outcomes, success or failure. One succeeds with probability $p$, and fails with probability $(1-p)$.

[quote="Elemennop"][quote="banblink182"]I'm not sure how $\displaystyle \binom{n}{i,j,k}$ is defined...

I think that it should be

$\binom{n}{i}*\binom{n-i}{j}$

But yea, extended to any number of probabilities. Interesting...[/quote]

$\displaystyle \binom{n}{i,j,k}$ is defined generally as $\frac{n!}{i!j!k!}$ when $i+j+k=n$. It's just used as an expansion of the binomial for trinomial, quadrinomial, and so on.[/quote]

The distribution for $\textstyle \binom{n}{i,j,k}$ is called the Hypergeometric distribution. It comes into play when one samples a population without replacing the samples.


Factor  $x^4+64$. I want to see a lot of solutions


[quote="cosinator"]Could someone explain to me how to solve problems asking for the minimum value of some expression such as this question from the mandelbrot sample test:

What is the minimum possible value of (a / 2b) + (b / 4c) + (c / 8a), where a, b, and c are positive real numbers?[/quote]

One of the ways, you should try to use one or another inequality and using it prove that given expression cannot have smaller value. You always must give exact values of variables then that minimum value is obtained. Sometimes you can prove that expression is larger than $N$, but there aren't such values then $N$ is get.

In your example, you can use AM-GM inequality to get minimum value:
$\frac{a}{2b}+\frac{b}{4c}+\frac{c}{8a}\geq3\left(\frac{abc}{64abc}\right)^\frac{1}{3}=\frac{3}{4}$

When you get this, you must find $a, b, c$ such that given expression is equal to $\frac{3}{4}$. In this case you must find then all term of LHS are equal. After some computation we find that $2a=b=c$.

[quote="cosinator"]thanks
but could you tell me what AM-GM is?[/quote]

The Inequality of Arithmetic and Geometric Means, in two variables, states that

$\frac{a + b}{2} \ge \sqrt{ab}$

For any positive reals $a, b$.  In three variables, it becomes

$\frac{a + b + c}{3} \ge \sqrt[3]{abc}$

And generally,

$\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1 a_2 ... a_n}$


What address did you guys get for sending the 3000 dollar check to awesome math.  I got one addressed directly to Titu Andreescu, but the e-mail was different, and my parents thought it was kind of suspicious that it would be paid directly to someone rather than an organization.  I'm kind of worried now that someone may have tapped my e-mail and tricked me into sending a check to him rather than awesome math.  Thanks

eh, i got that too

that's his personal email that the utdallas one reroutes to i think, i emailed him earlier with some lodging/meal questions, and his response came from the txa one


Just curious: is there some language where the word for "chocolate" is not just a simple variation of chocolate?

Well, it seems that the only counterexample is

[quote]Cherokee/tsalagi: uwodige kalisetsi, wodigekalsetsi (brown sweet)
Lakota: pejútašazi
Zapotec: dxuladi
[/quote]

if it can be called "a counterexample" at all  :lol:

in phillip pullmans, [u]the Golden Compass [/u] there is a drink called chocatl i think which is hot chocolate


induction
ratio
calculus
integration
algorithm
continued fraction
irrational number

Prove that $n^2+(n+1)^2$ is a perfect square for infinitely many $n \in \mathbb{N} \cup \{ 0 \}$.

[quote="JesusFreak197"]Umm... it isn't.[/quote]

a proof would be appreciated

[quote="JesusFreak197"]$n^2+(n+1)^2=n^2+n^2+2n+1=2n^2+2n+1$ is not a perfect square, so it cannot work for every case. In fact, the only case I can think of offhand that actually works is $n=3$.[/quote]

it doesn't have to work for $\bf{every}$ case. you have to prove that for every natural $n_0$, there is a $n \geq n_0$ for which it works (you must have $2n^2+2n+1=k^2$ for some natural $k$)

[quote="JBL"]Sum of two perfect squares equals a perfect square?  Sounds familiar to me ...

(You can cheat by [hide="spoiler for sure"]using induction and the [url=http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000129]Pell numbers[/url])[/hide][/quote]

sorry, but i don't get your hints :(

[quote="Gyan"]Perhaps  it reminded JBL of the famous problem: Find a Pythogorian  triangle where one   leg is one unit more than the other
(Or an almost isosceles right-angle triangle) .
(And  thus the  ratio of  this side to the hypotenuse  of the triangle is 'almost'  $\sqrt 2$ . (3/2 .. (7/5)    17/12.. etc .)

Like   $3^2+4^2 = 5^2 $   or $20^2+21^2 = 29^2$  or $119^2+120^2 = 169^2$   and   so on...

Hint: [hide]Continued fraction of  $\sqrt 2 $ is  $1+ \frac {1}{2+} \frac  1 {2+} \frac 1  {2+}  \dots $[/hide][/quote]

it is not helping me. i can't see the connection     :(

Gyan or JBL, please explain Pell solution to perfect_radio, this is not an easy trick to know. ty

More Hints  and some tools which can be used for problems like this.

[hide]Pell's equation  in  the simplified form  is  used to find a solution of the equation of the type:
(Note: Please look at any standard math book or google it for more info )

$kx^2 -  y^2 = 1$

For our particual case  we need the solution of the type:
$2x^2 - y^2 = 1$

(Your equation  $ m^2 = n^2+(n+1)^2 $
  is  nothing but  $2 m^2 - (2n+1)^2 =1 $  )

Trick is to use continued fraction of $\sqrt k $ in this case $\sqrt 2 $

[hide]we get  approximate values  (by taking few terms from  $1 \ + \frac 1 {2+} \ \frac 1 {2+} \ \dots $ )

we see   $\frac 3 2 $ , $ \frac 7 5 $ , $ \frac  {17} {12} $ , $\frac {41}{29} $,  $\frac {99}{70} $, $\frac {239}{169}$

etc ..

(the next value  $\frac  {a_n} {b_n} $ is given by
$a_n = 2 a_{n-1} + a_{n-2} $ and $b_n =2 b_{n-1} + b_{n-2} $

In fact  each value is pretty close to  $\sqrt 2$ , in fact   $a_n^2  = 2 b_n^2  +- 1$
If we take only $+1$ type values . We skip every other term,  that is, for   $(2n+1)$  we restrict out values to  $7, 41, 239$ you get all the values...

You can always get the next value from the last two values. .. There are infinite such values...

Hope this helps.[/hide][/hide]
(

ty Gyan because im not good explaining Pell solution, i only know the way to solve it  :blush:
i hope perfect_radio gets the way to solve his problem ... :)

I suppose if you go to infinity, there will be a set of a smaller infinity of numbers, but the question doesn't exactly seem right.

It's obviously Pythagorean triples, so you just have to find every case where the two legs are only one apart, but I'm not under the impression that all that many exist, although there may be quite a few if you use an extremely large set.

[quote]but I'm not under the impression that all that many exist[/quote]

Yes they do, you can keep going and will not run out of the values of  $n$ .... :)

As said before .. here  is the metod if you want to find  'next value' ..

[hide]Well start with $0$  and $3$
and each time  you get a new number as $6 (Previous  number) - (previous to previous numner) + 2 $

so next number is   $6 \cdot 3 -0  +2  = 20  $
and next is     $6  \cdot 20  - 3  + 2   = 119  $
and next is   $ 6 \cdot 119  -20 +2  = 696$

You get the pattern .. it can go as high as you wish ....

Now  for every such n ( $ 0 ,  3,  20,  119 , 696 $

   $n^2 + (n+1)^2 $ is a perfect square.. check it, and you can have as many  different values as you want ....

Hope this makes it clear. [/hide]

Can't resist a "show off" proof  !

[hide]Let   $ p_k = \frac {(2+ \sqrt 2) (3+ 2 \sqrt 2 )^ k + (2 - \sqrt 2)(3 - 2 \sqrt 2 )^ k } 4  $
and  $ q_k =  \frac {(1+ \sqrt 2) (3+ 2 \sqrt 2 )^ k + (1 - \sqrt 2)(3 - 2 \sqrt 2 )^ k } 2 $

direct substitution yields $p_0 =  1 , p_1 = 5,  q_0 = 1 , q_1 = 7 $

One can prove that  $p_k $ and $q_k$ both are positive integers when $k$ is a positive integer.

[hide](If  $(3+2 \sqrt 2)^k  = A +B \sqrt 2$   then  $ (3 - 2 \sqrt 2)^ k = A - B \sqrt 2 $   where $A$ and $B$ are integers)  then $p_k = A+B $  and $q_k = A+2B$  (Note that A is always an odd number ..etc ..[/hide]

Also as shown in above hint:    $q_k $ is  an odd number.

Also one can show easily  that for any $n =  \frac  {q_k   - 1 } 2 $
$n^2 + (n+1)^2 =  \frac  {{q_k}^2 + 1 } 2  =  p_k^2 $

This is true for all values of k so there are infinitely many solutions.,,

Not only that plug in values of k and you find all of them ... :D[/hide]

lol Gyan nice "show off" proof  :showoff:

[quote="JesusFreak197"]My point is, it will be so much smaller than the infinity of the set that it's using that it practically isn't infinity. I know, I know, it technically is, but it just doesn't seem like there should be an infinite number of values.[/quote]

JesusFreak197, i have the following analogy for you: let $\mathbb{Z}^+$ is the set of positive integers. now take out all the odd positive integers, let set of odd positive integers be $\mathbb{O}$. then what is remaining is the set of even integers $\mathbb{Z}^+-\mathbb{O}=\mathbb{E}$. now $\mathbb{O}$ is an infinite set and how come $\mathbb{E}$ is infinite then? we took out a lot of numbers and it just doesn't seem like there should be an infinite number of values but $\mathbb{E}$ does have an infinite number of values!!!

just wanted to lead you out of your intuitions.

the same thing happens when you take out the rational numbers out of the real number set $\mathbb{R}$. believe it or not, there are more irrational numbers than rational numbers!!! and yet, between any two irrational numbers, you can find a rational number; AND between any two rational numbers, you can always find an irrational number!!!  :huh:  :what?:  :huuh:  :lol:  :whistling:

edit: after reading a couple of times on what you wrote before, you will have to think of infinity in another sense. like $\aleph_0\pm c=\aleph_0$ or $\aleph_0+\aleph_0=\aleph_0$

[quote="Bictor717"][

That shows there are infinitely many $n$ that satisfy $n^2+k^2=(n+1)^2$, for integral $k$, instead of $n^2+(n+1)^2=k^2$.
Nice to know, though.[/quote]

 ... Yes, you can also notice that  if $n = 2m(m+1) $   and  $k = 2m+1 $  then for any $m$
 $(n+1)^2 = n^2+k^2$  :D

BTW - Just realized that  the  solution to the original problem, if you have a calculator, can be  found by this interesting result  -

$n =  \frac  {[(1+\sqrt 2)^{2k+1}]  -2 }  4  $

where [x] means integer value etc .(k is any positive integer)   interesting!

May be this can be an interesting  problem by itself  .  (prove that for such n ,  n^2+(n+1)^2 is a perfect square !)

[quote="Bictor717"][quote="Psycho"]Look at the set $A={0,1,4,9,16,25,36,49,64,81,100,121,...}$ which contains all perfect squares.
As you can see, every term is the preceding term +an odd number. Starting with $1,3,5,7,...$ you see that every odd number occurs exactly once.
In the set of odd numbers, there are infinitly many perfect squares, so there are infinitly many squares of the form $n^2+(n+1)^2$.
$\text{QED}$[/quote]

That shows there are infinitely many $n$ that satisfy $n^2+k^2=(n+1)^2$, for integral $k$, instead of $n^2+(n+1)^2=k^2$.
Nice to know, though.[/quote]
yeah, sorry  :blush:
realised i made a mistake after i posted it, and then i couldn't come on the computer till now  :?

[quote="Gyan"]

Well start with $0$  and $3$
and each time  you get a new number as $6 (Previous  number) - (previous to previous numner) + 2 $
[/quote]

How did you come up with this formula?... i'm sure it's correct but i'm having trouble proving it with induction

[quote="perfect_radio"]how can i find the continued fraction of $\sqrt k$? the only thing i understood is that it is periodical[/quote]

I'm not sure how to find it in general, but here are a couple interesting results concerning continued fractions (arrows mean periodic):

$\sqrt{n+1}=[\sqrt n,<2\sqrt n>]$
$\sqrt{n-1}=[\sqrt{n}-1,<1,2\sqrt{n}-2>]$

This gives you continued fractions if n is a perfect square, but otherwise you just get nested continued fractions. A simple algorithm, though, is to take the floor, then take the reciprocal of the fractional part and repeat. With a calculator, this should be sufficient.


geometry
circumcircle
parallelogram
ratio
geometric transformation
reflection
trigonometry

Two circles intersect at two points $A$ and $B$. A line $\ell$ which passes through the point $A$ meets the two circles again at the points $C$ and $D$, respectively. Let $M$ and $N$ be the midpoints of the arcs $BC$ and $BD$ (which do not contain the point $A$) on the respective circles. Let $K$ be the midpoint of the segment $CD$. Prove that $\measuredangle MKN = 90^{\circ}$.

I know I once gave a pretty neat solution, but I definitely can't remember it now so I'll try to post the outline of a not-so-nice solution (it has no ingenious ideas :().

(1) Try to find the locus of $K$, the midpoint of $CD$, as th line $CD$ varies. You'll see that it's the circumcircle of $OAB$, where $O$ is the midpt of $O_1O_2$ (the centers of the 2 circles).

(2) Find the locus of $P$, the midpoint of $MN$, as $CD$ varies. It's the circle with diameter $O_1O_2$, so it also has center $O$.

(3) Show that $PO\perp CD$ and we're done, because $O$ is on the perpndicular bisector of $AK$ (from (1)), and from (3) we find that the line $OP$ is the perpendicular bisector of $AK$, so $PA=PK$, so $K$ is on th circle centered at $P$ which passes through $A$, and that's precisely what we want, because $\angle MKN=90\Leftrightarrow K\in C(P,PA)$ (this is because $\angle MAN=90$).

O1O2 is center of two circle meet A,B O1M cut CB at P , O2N cut BD at Q then
   Then  <NAM=90 then  KP/NQ=BP/QK,then <PKM=<QNK => KNQ=90+<CPQ+<QDK=>MKN=90 :(

A different solution:

Let the mid-points of BC and BD be M' and N' respectively. Then M'KN'B is a parallelogram, $MM' \bot BC$ and  $NN' \bot BD$.

$\angle BNN'=\frac{1}{2} \angle BND=\frac{1}{2} \angle BAC=\angle MAC=\angle MBM'$. Thus the two right triangles BNN' and MBM' are similar.

Then, $\frac{N'N}{M'K}=\frac{N'N}{BN'}=\frac{BM'}{M'M}=\frac{KN'}{M'M}$, giving $\frac{N'N}{KN'}=\frac{M'K}{MM'}$. In addition,

$\angle NN'K=90^o +\angle DN'K =90 ^o +\angle CM'K =\angle KM'M$
Thus $\triangle NN'K \sim \triangle KM'M$.

So,
$\angle MKN=\angle MKM'+\angle M'KN'+\angle N'KN=\angle MKM'+\angle CM'K+\angle M'MK=180^o -\angle CM'M=90^o$

Although we already have 3 solutions for the original problem now, let me show a rather uncommon solution using inversion:

[color=blue][b]Problem.[/b] Two circles c and d intersect each other at two points A and B, and a line l passing through the point A meets these two circles at the points C and D (apart from the point A). Let M and N be the midpoints of the arcs BC and BD which do not contain A on the circles c and d, respectively. Let K be the midpoint of the segment CD. Prove that < MKN = 90°.[/color]

[i]Solution.[/i] The crucial point of the solution will be an inversion with center A. But before we undertake this inversion, we transform the problem into a more convenient form:

The point M is the midpoint of the arc BC on the circle c. Thus, the two chords BM and CM of the circle c have equal length. Hence, the chordal angles of these two chords must also be equal; in other words, we have < BAM = < CAM. Hence, the point M lies on the angle bisector of the angle BAC. Similarly, the point N lies on the angle bisector of the angle BAD. But, of course, the angle bisector of the angle BAD is the exterior angle bisector of the angle BAC; hence, since the interior and the exterior angle bisector of an angle are always perpendicular to each other, the angle bisector of the angle BAC and the angle bisector of the angle BAD are perpendicular to each other. Hence, < MAN = 90°.

The problem requires to prove < MKN = 90°. Since we know that < MAN = 90°, this will trivially follow if we show < MKN = < MAN. But this is equivalent to proving that the points M, N, A and K lie on one circle. Hence, in order to solve the problem, it is enough to prove that the points M, N, A and K lie on one circle.

Now, perform an inversion with respect to an arbitrary circle centered at the point A. Denote by B', C', D', M', N' and K' the images of the points B, C, D, M, N and K under this inversion. What properties do these new points B', C', D', M', N' and K' have?

- We know that the circle c passes through the points A, B, C and M. Since this circle passes through the point A, which is the center of the inversion, it is mapped by the inversion to a line; this line must pass through the images B', C' and M' of the points B, C and M under our inversion. Thus, the points B', C' and M' lie on one line. Similarly, the points B', D' and N' lie on one line.

- We know that the point M lies on the angle bisector of the angle BAC. Since our inversion has the center A, the images B', C' and M' of the points B, C and M under this inversion must lie on the rays AB, AC and AM, respectively, and thus we can state that the point M' lies on the angle bisector of the angle B'AC'. Similarly, the point N' lies on the angle bisector of the angle B'AD'.

- The point K is the midpoint of the segment CD. This seems to be a property hard to deal with using inversion. However, there is a workaround: We denote by $\infty_{l}$ the infinite point of the line l (this is the point of intersection of the line l with the line at infinity). Then, the image of this point $\infty_{l}$ under our inversion is the point A (since an inversion maps infinite points to the center of the inversion).

Also, since every segment is divided harmonically by its midpoint and the infinite point of the line containing the segment, the points K and $\infty_{l}$ divide the segment CD harmonically (since the point K is the midpoint of the segment CD, and the point $\infty_{l}$ is the infinite point of the line l = CD). In other words, the points C, D, K and $\infty_{l}$ on the line l form a harmonic quadruple. Now, it is well-known that if four points lying on a line through the center of an inversion form a harmonic quadruple, then so do their images under the inversion. Since the line l passes through the center A of our inversion, we thus conclude that the points C', D', K' and A, being the images of the points C, D, K and $\infty_{l}$ under our inversion, must also form a harmonic quadruple. Thus, the point K' is the harmonic conjugate of the point A with respect to the segment C'D'.

- Finally, let's see how the inversion transforms the assertion that we have to prove. We have already seen that, in order to solve the problem, it suffices to show that the points M, N, A and K lie on one circle. In other words, it suffices to show that the points M, N and K lie on one circle through the point A. Since the point A is the center of our inversion, and since an inversion maps circles through the inversion center to lines and conversely, it is clear that proving that the points M, N and K lie on one circle through the point A is equivalent to showing that their inversive images M', N' and K' lie on one line.

Hence, in order to solve the problem, it will be enough to show that the points M', N' and K' lie on one line.

Now, everything is simple: Since the points B', C' and M' lie on one line, and the point M' lies on the angle bisector of the angle B'AC', the point M' is the point of intersection of the angle bisector of the angle B'AC' with the line B'C'. Since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we have $\frac{B^{\prime}M^{\prime}}{M^{\prime}C^{\prime}}=\frac{AB^{\prime}}{C^{\prime}A}$. Similarly, the point N' is the point of intersection of the angle bisector of the angle B'AD' with the line B'D', and we conclude that $\frac{D^{\prime}N^{\prime}}{N^{\prime}B^{\prime}}=\frac{AD^{\prime}}{AB^{\prime}}$. Finally, since the point K' is the harmonic conjugate of the point A with respect to the segment C'D', we have $\frac{C^{\prime}K^{\prime}}{K^{\prime}D^{\prime}}=-\frac{C^{\prime}A}{AD^{\prime}}$. Thus,

$\frac{D^{\prime}N^{\prime}}{N^{\prime}B^{\prime}}\cdot\frac{B^{\prime}M^{\prime}}{M^{\prime}C^{\prime}}\cdot\frac{C^{\prime}K^{\prime}}{K^{\prime}D^{\prime}}=\frac{AD^{\prime}}{AB^{\prime}}\cdot\frac{AB^{\prime}}{C^{\prime}A}\cdot\left(-\frac{C^{\prime}A}{AD^{\prime}}\right)=-1$.

Hence, by the Menelaos theorem, the points M', N' and K' lie on one line. The problem is solved.

[b]EDIT:[/b] This problem comes from the 24th Russian Mathematical Olympiad 1997.

  Darij

As I remember a more general problem is G8 in the shortlist of IMO2002. I can post the original solution if it is needed.

[color=red][Moderator edit: Thanks for noticing! Now I have found various solutions for this problem G8 in the following sources:
http://www.kalva.demon.co.uk/short/soln/sh02g8.html
http://www.mathlinks.ro/Forum/viewtopic.php?t=22165
http://www.mathlinks.ro/Forum/viewtopic.php?t=22201
http://www.mathlinks.ro/Forum/viewtopic.php?t=17322
http://www.mathlinks.ro/Forum/viewtopic.php?t=15588 post #2 file IMO_2002_shortlist.pdf page 21 .][/color]

Let $M'$ and $N'$ be the midpoints of $BC$ and $BD$ respectively. It is obvious that $\triangle MM'K\sim\triangle KN'N$. Then $\angle MKC+\angle NKD=\angle BDC+\angle BCD+(\angle MKM'+\angle NKN')=\angle BDC+\angle BCD+\pi-\angle KM'M=\frac{\pi}{2}$. The result follows.

Please clarify your solution . I think It's not obvious and I believe there is a simpler solution

Basically my solution is the same as the one posted by E. Leung (even the notation!), so I'm not going to explain it in details.

Anyway, although it's not really that obvious, I personally believe that anyone associated with Math Olympiad should be able to solve it.

[b]A equivalent enunciation of the proposed problem:[/b]

Let the triangle $ABC$, the midpoint $D$ of the side $[BC]$ and two isosceles triangles $ABM,\ ACN\ (MA=MB,\ NA=NC)$
such that the line $AB$ separes the points $D,M$, the line $AC$ separates the points $D,N$ and $m(\widehat {AMB})+m(\widehat {ANC})=180^{\circ}.$
Prove that $DM\perp DN$ and $m(\widehat {DMN})=\frac 12 m(\widehat {AMB})$.

[b]Remark.[/b] After at least a hour I will prove with complex numbers !

My solution was somewhat different, but I also used spiral similarity. [I am just solving the problem the way levi stated it.]

Let $C'$ and $D'$ be the reflections of $C$ and $D$ in $M$ and $N$ respectively. It is easy to see that the right-angled triangles $BCC'$ and $BD'D$ are similar (this follows from $\angle{BMC} + \angle{BND} = 180^{\circ}$). Hence, triangles $BC'D$ and $BCD'$ are similar, and there is a spiral similarity (angle $90^{\circ}$) mapping the $BC'D$ to $BCD'$. This spiral similarity maps $C'D$ to $CD'$, so $C'D$ and $CD'$ are perpendicular. Since $MK \parallel C'D$, $NK \parallel CD'$ we are done.

([b]See the last my reply[/b]). I note:

$ u=m(\widehat {AMB}),\ v=m(\widehat {ANC})\ (u+v=180^{\circ}),\ x=\cos u+i\cdot \sin u,\ y $ $ =\cos v+i\cdot \sin v\ (xy=-1). $

$ A(a),\ B(b),\ C(c),\ D(d),\ M(m),\ N(n)\Longrightarrow 2d=b+c,\ m= $ $ \frac {a-bx}{1-x},\ n=\frac {c-ay}{1-y}\Longrightarrow $

$ m-d=\frac {2a-b(1+x)-c(1-x)}{2(1-x)},\ n-d=\frac {-2ay-b(1-y)+c(1+y)}{2(1-y)} $ $ ,\ xy=-1\Longrightarrow $

$ n-d=\frac {2a\cdot \frac 1x -b(1+\frac 1x )+c(1-\frac 1x)}{2(1+\frac 1x)}=\frac {2a-b(1+x)-c(1-x)}{2(1+x)}\Longrightarrow $

$ \frac {n-d}{m-d}=\frac {1-x}{1+x}=\frac {1-\cos u-i\cdot \sin u}{1+\cos u +i\cdot \sin u}=\frac {2\sin^2 \frac u2 -2i\sin \frac u2\cos \frac u2}{2\cos^2 \frac u2 +2i\sin \frac u2 \cos \frac u2}=-i\cdot \tan \frac u2 \Longrightarrow $

$ (m-d)=i\cdot cot \frac u2 \cdot (n-d)\Longrightarrow DM\perp DN, \tan \frac u2=\frac {DN}{DM} $, i.e. $ m(\widehat {DMN})=\frac 12 \cdot m(\widehat {AMB}). $

[b]1. Remark.[/b] Let $ P $ be the point of the line $ BC $ for which $ m(\widehat {APC})=u,\ m(\widehat {APB})=v\ (u+v=360^{\circ}) $.

Then the $ AMBM,\ ANCP $ are cyclic quadrilaterals and thus, we obtain the initial proposed problem !.

[b]2. Remark.[/b] This problem admits a generalization (see http://www.mathlinks.ro/Forum/viewtopic.php?p=289313#p289313 ):

$ \blacksquare $. Let $ ABC $ be a triangle and three isosceles triangles $ BCL,\ CAN,\ ABM\ (LB=LC,NC=NA,MA=MB) $

such that the lines $ BC,CA,AB $ separe the pointa $ A,B,C $ from the points $ L,N,M $ respectively and

$ m(\widehat {BLC})+m(\widehat {CNA})+m(\widehat {AMB})=360^{\circ} $. Then

$ m(\widehat {MLN})=\frac 12\cdot m(\widehat {BLC}),\ m(\widehat {LNM})=\frac 12 \cdot m(\widehat {CNA}),\ m(\widehat {NML})=\frac 12\cdot m(\widehat {AMB}).\blacksquare $.

[b]For[/b] $ L: =D $ [b]we obtain the previous problem !.[/b]

[hide="My Proof:(I have used directed angles to avoid cases)"]Extend $DN$ to $C'$ such that $DN=NC'$.As we know $MB=MC;NB=NC'$ (from properties of $M$ and $N$) and $\angle BMC=\angle BAD=\angle BNC'$, we can say $\triangle BMC\sim \triangle BNC'$ and so $CB/BC'=MB/BN$.
Also $\angle CBM=\angle C'BN$ implies $\angle CBC'=\angle MBN$.So $\triangle CBC'\sim \triangle MBN$.Then $\angle C'CB=\angle NMB$.
Also notice that in $\triangle CC'D$, $K,N$ are the mid-points of $CD$ and $C'D$.So $KN||CC'$.Then $\angle NKD=\angle C'CD=\angle C'CB+\angle BCA=\angle NMB+\angle BMA=\angle NMA$.So points $M,N,K,A$ are con-cyclic.Hence $\angle MKN=\angle MAN=\frac {1}{2}(\angle CAB+\angle BAD)=\frac {\pi}{2}$[/hide]

Dear Mathlinkers,

1. M' the symmetric of M wrt K
2. the triangles BMN and DM'N are equal
and we are done

Sincerely
Jean-Louis

Hi J.L.

[quote="jayme"]
2. the triangles BMN and DM'N are equal
[/quote]

I only can see that BN=ND and MN=M'N, but somewhat I can't go any further.

[quote="jayme"]
and we are done
[/quote]

why?

If you would be so kindly to explain a little bit longer... I would be even more grateful.

Dear Mathlinkers,

<MBN = <MBA + <ABN ;
<MBN = <ACM + <NDA ;
<ACM = <CDM' = <ADM' ;
<MBN = <ADM' + <NDA ;
<MBJN = <NDM'.

Sincerely
Jean-Louis

I have a different one using transformations. Let $\angle CAB = A$ and $\angle DAB = 180 - A$. Also, note that $CB/CM = 2 \cos A/2$(by using sine law and $\sin(A) = 2\sin A/2 \cos A/2$. Consider the sequence of transformations:

(1) A spiral similarity with centre $C$, ratio $2\cos A/2$ and angle $A/2$.

(2) A spiral similarity with centre $D$, ratio $\dfrac{1}{2 \sin A/2}$ and angle $90-A/2$.

Note that this takes $M \to B, B \to N$, so $M \to N$. Apply the first transformation to $K$, let the image be $K'$. Now, contruct a right angled triangle $\triangle ECD$ with hypotenuse $CD$, such that $\angle ECD = A/2$.

Now, By the dilation properties of (1) $CK' = BC \cos A/2$. But by our right triangle, $CE = BC \cos A/2$, so $E = K'$. Then the second transformation takes $E$ to $E'$, where $E'$ lies on $CD$ (by our right triangle). Also by our right triangle, $DE = CD \sin A/2 = 2DK \sin A/2$, so in fact $K = E'$, meaning $K$ stays invariant under the transformations.

So, the resultant transformation is a spiral similarity about $K$ with net rotation $(90 - A/2) + A/2 = 90$ and dilation $\dfrac{1}{\tan A/2}$, which takes $M \to N$ i.e. $\angle MKN = 90$.

[i]Remark: [/i]There is a shorter solution given by Number1. Consider the composition $R_M \circ R_N$ taking $C \to B \to D$. This is equivalent to a half turn about $K$. But, then we know that $\angle (KM, MN) = \angle CMB/2$ and $\angle (KN, MN) = \angle DMB/2$. Easy to see $\angle CMB + \angle DMB = 180$, so $\angle MKN = 90$.


I know, I know, this was discuss before.
However, the topic was removed to GFF, and, then, get full with spams. What if we start the discussion before? I'd like to.
I want to know your opinion about:

*Choosing a girlfriend.
*Have a lot or a few girlfriends.
*Wait until being about 25 to have one or just know being a teenager.
*How to be a good boyfriend, and
*anything you want to add.

As I already said, althought this was discussed before, we couldn't finish because of spams.

How old are you exactly? And what are you studying? Is the institution co-ed or boys only? What kind of relationships are you looking for: short term or long term? These are questions we you should answer before seeking personal advice.

But I can tell you one thing Jose: always be honest. You can have multiple girlfriends - even multiple physical relationships are okay - as long as you don't deceive any of them into believing that she is [i]the[/i] one.

How to be a good boyfriend: Listen. Try to understand. If your relationship is physically intimate, pay some attention to what she wants! Make sure she knows you appreciate her.

Choosing: It depends on what you want. The more serious you hope your relationship will be, the more care you should take.

I know these are vague and fairly obvious, but it is sad how many people don't follow these basic principles, and relationships crash and burn in sadness and anger. :(

I'm not asking for me personally (to bubka, that asked), I just want everyone to post his opinion (I'd like girls to post too, to see what they think). I want to try to "make" the way of being of a good boyfriend, not for me, for all of us here, so post your opinion for everyone, not only for me.

I'm not asking, also, for a relationship exclusive for "me". I repeat, I want this to be profitable for all.

Thanks bubka and Xevarion that post, I want more and more!!!!

[b]Now that this topic has been moved to games and fun factory, please continue serious discussion and do not spam or make pointless posts![/b]

Anyways, I think choosing a girlfriend totally depends on your character. If you just want somebody that you can go out with, you can just quite quickly choose a girlfriend, and you will not have to go too far to find one. However, not many people would do that . If you want a long-term relationship where you would be happy, you need to choose very carefully and before even starting to go out with a girl you like, you must explore her character and other features, so that you do not get fooled by first appearances.
I think having a few girlfriends is better, beacuse then you will have a longer, more stable relationship(instead of getting a new gf every half a year or more often), and you will get more experience, which can come in very handy later in life.
I am not really sure about the best age to first have a girlfriend. This also totally depends on yourself: if having a gf as a teen des not mess with your religion, parents, etc., and if you think it would make your life better, I think you should go for it. And also you should not care if your friends have/do not have girlfriends, beacuse that does not mean anything.
I really have no idea of how to be agood boyfriend. I can just point out the obvious: be a nice, caring, loving, and honest person who always tries to do the best thing for his loved one.

true, but they also help us develop in another sense.

[quote="Ashegh"]i think girl friends stop us from developing,[/quote]

It what sense? freedom maybe? I don't think that having a girlfriend is loosing my freedom, because we can continue our lives. E.g.: I can go out at night with or without a girlfriend

[quote="drunner2007"]girlfriends [i]can[/i] stop you from developing certain parts of your life but they also help you develop in others.[/quote]

Nice sentence drunner. Can you give us examples?

We continue here: http://www.mathlinks.ro/Forum/viewtopic.php?t=100208&start=180


real analysis
limit
real analysis solved

I think this is pretty well known (I've seen it many times, but took me a long time to come up with a soln, which I don't even know is correct), but I'd like to see as many solns as possible:

What's the limit of (1^n+2^n+..+n^n)/n^n as n->infinity?

This is Cesaro Stolz!!! case: oo/oo!!!

tschuess!!!

Somehow I really doubt this :D

Remember that n is also the exponent of 1, 2, ... Could you post a soln using Cesaro?

somehow I saw it like so:

lim sum(k^p)/n^(p+1), when n tends to infinity.. the limit of this is 1/(p+1)

sorry about that!! :?
cheers! :D :D

[quote="grobber"] I'd like to see as many solns as possible:

What's the limit of (1^n+2^n+..+n^n)/n^n as n->infinity?[/quote]

Wn= (1/n)^n+....+((n-1)/n)^n =\sum_{k=1 to n-1} (1-k/n)^n

(1-k/n)^n  \leq e -k

Wn  \leq \sum_{k=1 to n-1}e -k

limsup Wn  \leq 1/(e-1)

Choose q in N for n \geq q => Wn  \geq \sum_{k=1 to q-1}(1-k/n)^n

now we tend n-->+oo, for any q in N, liminf Wn  \geq \sum_{k=1 to q-1}e -k

now we tend q-->+oo, 1/(e-1)  \leq liminf Wn  \leq limsup Wn  \leq 1/(e-1)

\lim Wn = 1/(e-1)

the limit of the sequence is 1/(e-1) + 1

It's time for a caf and croissant


Find the minimum and maximum values of this expression for $ (a, b, c, d, e)$ are chosen from $ (1, 2, 3, 4, 5)$ with no repeated values.

$ \frac {1}{a \plus{} \frac {1}{b \plus{} \frac {1}{c \plus{} \frac {1}{d \plus{} \frac {1}{e}}}}}$

[hide="Hint"]Every time there is a fraction sign, the value becomes either a lot larger, or a lot smaller.[/hide]

[hide="Yes."]
$ \frac {1}{a \plus{} \frac {1}{b \plus{} \frac {1}{c \plus{} \frac {1}{d \plus{} \frac {1}{e}}}}}$

To maximize a fraction $ \frac{1}{x}$, we would seek to [i]minimize[/i] the denominator $ x$. In this case, we would seek to minimize $ \frac{1}{a\plus{}\frac{1}{b\plus{}...}}$. Naturally, we would choose $ a\equal{}1$. We would also want to minimize $ \frac {1}{b \plus{} ...}$, so, by similar logic, we would want to [i]maximize[/i] $ b$. Thus we take $ b\equal{}5$.

Continuing in this fashion, we get unimpossible's answers for the maximization, and, by similar logic, his answers for the minimization.
[/hide]


plz could some1 help me with this mess. :maybe:

A cube consists of 12 resitances on its edges.
four 1 ohm vertical resistances
four 2 ohm horizontal resistances
four 3 ohm oblique resistances

Calculate the total equivalent resistance between 2 diagnolly opposite points of the cube

Thanks :)

ask our tyagi sir :rotfl:  :rotfl:  :rotfl:

I dont want to see his glare when i ask him :(

not tooo difficult..

just use top view first,then envelope method and finally star delta conversion (yay..wen my sir taught these,i thought it useless)

is ans 1.6 ohms?? :P

the answer is somewhere near 1.4 ohms :maybe:

Ritu i know about star delta  But what is envelope method ?
plz tell :oops:

just ovelap the two symmetrical halves,through a middle line of symmetry..its difficult 2 exp in words.i'll draw a fig soon

u'll find that pairs of resistence come in parallel

and the calculations r really long :oops:


Anyone here know about virtual high school or VHS? Would you recommend taking an AP class there?

taking AP classes that have to do with science no way. ALOT of science classes have associated lab stuff. even the physics 1-2 honors i'm going to take has an lab at the university.
however online stuff is good for math, and those stupid required class  credits that you have to do.

I am currntly taking number theory on VHS, but it is completely useless as 1)I already know everything covered, having gone to PROMYS last summer, 2) The course doesn't offer any challenges. There are no proofs, the hardest part is cryptograms and puzzles which take a half-hour.  None of the other math courses looked challenging-- in short, it's a very easy class. Take one of the math classes if you like, but don't expect to learn very much if you have read the any math books.


[hide]
19 women are single
12 women don't have cell phones
28 women don't have their own car
24 women don't have their own house.

Let women #1 - 19 be single
#20 - 31 be deprived of a cell phone
#32 - 59 not have their own car
#60 - 83 not have their own house

#84 through 100 must have all 4, for a minimum of $\boxed{17\;\text{women}}$
[/hide]

[quote="13375P34K43V312"]For each 100 women in town, 81 are married, 88 have a cell phone, 72 own a car, and 76 own their own house. What is the least possible number of these 100 women who are married, have a cell phone, own a car, and own a house?[/quote]
I've seen this before in mathcounts...
[hide]$81+88+72+76-300=17$ Count the number of women then subtract what it would be if all of them had three of those things.[/hide]


in a triangle ABC the following relation is given:
$2a^{4}+b^{4}+c^{4}+18b^{2}c^{2}=2a^{2}(4bc+b^{2}+c^{2})$
. Find the measure of the triangle angles.

i tried to complete the square.. but I am stuck. I am too tired, I guess I'll try again tomorrow.


geometry
3D geometry
number theory
prime factorization

I will open up this topic with the following tricks:

1.  For finding squares notice the following:
$ 1^2$ = 1
$ 2^2$ = 4    $ 1^2$ + 1 + 2 = 4  (the sum of the current base and the previous base)
$ 3^2$ = 9    $ 2^2$ + 3 + 2 = 9
$ 4^2$ = 16  $ 3^2$ + 4 + 3 = 16
$ 5^2$ = 25  $ 4^2$ + 5 + 4 =  25
...
so, $ 25^2$ = 625, so $ 26^2$ = ?

Formula:  $ n^2 = (n - 1)^2 + n + (n - 1)$

2.  How many squares are there in a figure?
┌─┬─┬─┐
├─┼─┼─┤
├─┼─┼─┤
└─┴─┴─┘
say you had a 3 x 3 square
   How many 3 x 3 squares?     1  ($ 1^2$)
   How many 2 x 2 squares?     4  ($ 2^2$)
   How many 1 x 1 squares?     9  ($ 3^2$)

┌─┬─┬─┬─┐
├─┼─┼─┼─┤
├─┼─┼─┼─┤
├─┼─┼─┼─┤
└─┴─┴─┴─┘
And for a 4 x 4 square?
   How many 4 x 4 squares?     1  ($ 1^2$)
   How many 3 x 3 squares?     4  ($ 2^2$)
   How many 2 x 2 squares?     9  ($ 3^2$)
   How many 1 x 1 squares?     16 ($ 4^2$)

So, How many in an 8 x 8 square?
Well, we start with $ 8^2$ and go backward
   $ 8^2$ + $ 7^2$ + $ 6^2$ + $ 5^2$ + $ 4^2$ + $ 3^2$ + $ 2^2$ + $ 1^2$ =  ?

3.  The sum of the first 50 numbers:
By grouping the first and the last and the second and second to last:
$ 1 + 2 + 3 + 4 + ... + 47 + 48 + 49 + 50$
we get
$ 51 + 51 + 51 + 51....... + 51 + 51 + 51 + 51$   but this is twice as many as we need
so we have $ \frac {50}{2} * 51$
Writing as a formula we get:
[hide]$ \frac {(lastnumber + firstnumber) * lastnumber}{2}$[/hide]

If you only wanted to add the even (or odd) numbers:
[hide]$ \frac {(lastnumber + firstnumber) * (thenumberofeven/oddnumbersinrange)}{2}$
Note:  the number of even numbers in the range 1 to 50 is 25
the number of odd numbers in the range 1 to 49 is 25[/hide]

If you start at a number other than one, subtract the total from 1 to the starting number using the formula above.

4.  Find the units digit in the number $ 23^{2007}$
Well, to find out what the units digit is, we are only interest in the last digit to the power, so lets look at the powers of three:
$ 3^1 = 3$
$ 3^2 = 9$
$ 3^3 = 27$
$ 3^4 = 81$
$ 3^5 = 243$
$ 3^6 = 729$
$ 3^7 = 2187$
$ 3^8 = 6561$
$ 3^9 = 19683$
Have you noticed the pattern?
[hide="click to see pattern"]Every four powers the last number repeats![/hide]
So, all we have to do is find the remainder of 2007/4
which is 3.  So we count down the list to find the last digit of the 3rd power=>7
So the last digit of $ 23^{2007}$ is 7
(No, you don't have to find as many powers as I did, only to where it repeats, I did the rest to emphasize the pattern)

5.  Notice:
$ \frac {1}{2} + \frac {1}{3} = \frac {5}{6}$
$ \frac {1}{3} + \frac {1}{4} = \frac {7}{12}$
$ \frac {1}{5} + \frac {1}{8} = \frac {13}{40}$
Are you noticing a pattern?
[hide="click to see pattern"]$ \frac {1}{2} + \frac {1}{3} = \frac {5}{6}$:  2+3=5; 2*3=6
$ \frac {1}{3} + \frac {1}{4} = \frac {7}{12}$: 3+4=7; 3*4=12
$ \frac {1}{5} + \frac {1}{8} = \frac {13}{40}$:  5+8=13; 5*8=40[/hide]

[hide="So the generalized formula is"]$ \frac {1}{a} + \frac {1}{b} = \frac {a + b}{a*b}$
[/hide]
Do you see how it could be used on this problem?

The sum of two numbers is 32.  The product of these two numbers is 48.  What iis the sum of the the reciprocals of the two numbers?  Express your answer as a common fraction.  (Warm-up 6, problem 4, 2007-08)

6.  Do you know your decimal equivalents of 9ths?
$ \frac {1}{9} = 0.\overline{1}$
$ \frac {2}{9} = 0.\overline{2}$
$ \frac {3}{9} = 0.\overline{3}$
$ \frac {4}{9} = 0.\overline{4}$
.
.
.
$ \frac {1}{99} = 0.\overline{01}$
$ \frac {2}{99} = 0.\overline{02}$
$ \frac {13}{99} = 0.\overline{13}$

Special thanks to Cheri and Barbara for these tips.

To easily find the total number of factors of a number,
1.  Find the prime factorization
2.  Take the exponents of each prime factor and add 1 to each
3.  Multiply the resulting numbers together to get the answer.

Example:
Find the total number of factors of 56
1.  $ 56\equal{}2^3*7$
2.  2's exponent is 3 and 7's exponent is 1.  3+1=4; 1+1=2
3.  4*2=8
Check
1, 2, 4, 7, 8, 14, 28, 56 => 8 factors!

[b]Finding the sum of all the positive factors of a number:[/b]

First, find the prime factorization in exponent form.

Say we want to find the sum of all the positive factors of 240.  The prime factorization is:
$ 2^3 * 3^1 * 5^1$

We take the base number and add the powers of each prime number up to its exponent (start with 0) and multiply the resulting sums together to get your answer.
$ 2^0 \plus{} 2^1 \plus{} 2^2 \plus{} 2^3)(3^0 \plus{} 3^1)(5^0 \plus{} 5^1)\equal{}$
$ (1 \plus{} 2 \plus{} 4 \plus{} 8)(1 \plus{} 3)(1 \plus{} 5) \equal{} 360$

[quote="danb80"][b]Finding the sum of all the positive factors of a number:[/b]

First, find the prime factorization in exponent form.

Say we want to find the sum of all the positive factors of 240.  The prime factorization is:
$ 2^3 * 3^1 * 5^1$

We take the base number and add the powers of each prime number up to its exponent (start with 0) and multiply the resulting sums together to get your answer.
$ 2^0 \plus{} 2^1 \plus{} 2^2 \plus{} 2^3)(3^0 \plus{} 3^1)(5^0 \plus{} 5^1) \equal{}$
$ (1 \plus{} 2 \plus{} 4 \plus{} 8)(1 \plus{} 3)(1 \plus{} 5) \equal{} 360$[/quote]
Or if you know the geometry series, you can use this:

Let's say that the positive integer $ n$ has the prime factorization of $ a^x b^y c^z$

The sum of positive factors of $ n\equal{} \frac{a^{x\plus{}1} \minus{}1}{a\minus{}1}\times\frac{b^{y\plus{}1} \minus{}1}{b\minus{}1}\times\frac{c^{z\plus{}1} \minus{}1}{c\minus{}1}$

[hide="Reason"]The reason behind this is simple. Since the formula for finding the sum of geometry series, which is

$ \frac{a_1(1\minus{}r^n)}{1\minus{}r}$can be also $ \frac{a_1(r^n \minus{}1)}{r\minus{}1}$

In this equation the value of variable are these:

$ a_1\equal{}$First number in the sequence

$ r\equal{}$Rate between numbers

$ n\equal{}$The number of terms

Why we add 1 to our exponents is because we want the number of factors which would be $ (x\plus{}1)(y\plus{}1)(z\plus{}1)$

Now if when we apply to this formula, $ a$ is going to be always 1.
So we can leave that out since it's not going to matter..

That yields this main formula:

$ n\equal{} \frac{a^{x\plus{}1} \minus{}1}{a\minus{}1}\times\frac{b^{y\plus{}1} \minus{}1}{b\minus{}1}\times\frac{c^{z\plus{}1} \minus{}1}{c\minus{}1}$

Let's see if it works:

The question was to find the sum of all positive factors of 240.

Now we know that $ 240\equal{}2^3\cdot 3 \cdot 5$

We apply that to our formula:

$ n\equal{} \frac{2^4 \minus{}1}{2\minus{}1}\times\frac{3^2 \minus{}1}{3\minus{}1}\times\frac{5^2 \minus{}1}{5\minus{}1}$

This can be simplied to this:

$ n\equal{} \frac{15}{1}\times\frac{8}{2}\times\frac{24}{4}$

Which I believe it simplifies to $ 360$[/hide]

Better get it right! We went over that just 2 weeks ago.  :ninja:

[quote="alanchou"]I suppose it's more important to know why these rules work.

Like, say, find the number of perfect square divisors in $ 810000$.[/quote]

[hide="I Think"]$ 810000\equal{}2^4\cdot3^4\cdot5^4$

$ 2^4\cdot3^4\cdot5^4\equal{}(2^2\cdot3^2\cdot5^2)^2$

$ 3 \cdot 3\cdot 3\equal{} \boxed{27}$[/hide]

[quote="alanchou"]Ok, I suppose it was easier since 810000 is a perfect square.

Now, how about the number of perfect cube divisors of 810000?

Because these types of problems appear all the time

[size=59]I don't even live in Oklahoma, why am I here?[/size][/quote]

[hide]That doesn't make any harder, I think..

$ 810000\equal{}2^4\cdot3^4\cdot5^4$

$ 2^4\cdot3^4\cdot5^4\equal{}(2^1\cdot3^1\cdot5^1)^3$

$ 2\cdot2\cdot2\equal{}\boxed{8}$[/hide]

You are always welcome in Oklahoma Forum.
It's good to have someone really active in this forum...

Umm, $ (2\times 3\times 5)^3\neq 2^4\times 3^4\times 5^4$.

[hide="although the answer is correct this probably makes more sense"]
The prime factorization of $ 810000$ is $ 2^4\times 3^4\times 5^4$.  For a number to be a perfect cube, all of the exponents of it's prime factorization must be multiples of 3. (if you don't know why, figure it out yourself)  So the exponent of the 2 can be 0 or 3, so that's 2 possible values for the exponent of 2.  We would use similar reasoning for 3 and 5, and come up with 2 values for the exponents of 3 and 5.  Therefore, the answer is $ 2\times 2\times 2\equal{}8$.
[/hide]

Plus, if I said perfect cube factors of $ 135000$, you wouldn't be able to express it as a perfect power.

[quote="alanchou"]Umm, $ (2\times 3\times 5)^3\neq 2^4\times 3^4\times 5^4$.

[hide="although the answer is correct this probably makes more sense"]
The prime factorization of $ 810000$ is $ 2^4\times 3^4\times 5^4$.  For a number to be a perfect cube, all of the exponents of it's prime factorization must be multiples of 3. (if you don't know why, figure it out yourself)  So the exponent of the 2 can be 0 or 3, so that's 2 possible values for the exponent of 2.  We would use similar reasoning for 3 and 5, and come up with 2 values for the exponents of 3 and 5.  Therefore, the answer is $ 2\times 2\times 2 \equal{} 8$.
[/hide]

Plus, if I said perfect cube factors of $ 135000$, you wouldn't be able to express it as a perfect power.[/quote]

Sorry. I think the valid way to express it should be $ (2\times 3\times 5)^3\times (2\cdot 3\cdot 5)$

Allow me to try $ 135000$

$ 13500\equal{}2^3 \cdot3^3\cdot 5^4$

$ 2^3 \cdot3^3\cdot 5^4\equal{}(2 \cdot3\cdot 5)^3 \times 5$

$ 2\times 2\times 2 \equal{} \boxed{8}$.

Ah, now I see where that method comes from

And should I mention that it should say "sum of smallest 50 [u]positive integers[/u]" instead of first 50 numbers, cause we don't know how numbers is defined in this.

Plus,

[hide="formula for the sum of first n odd postive integers"]
Is $ n^2$
[/hide]
[hide="formula for the sum of first n even positive integers"]
Is $ n(n\plus{}1)$
[/hide]

There's the proof without words here on AoPS for the first one, and the second one, just factor out the 2 and you get the sum of the first $ n$ positive integers

Also, in #3, in fact, for any number, the last digit repeats after 4.

[hide="proof"]
We are trying to prove that $ a^5\equiv a\mbox{ (mod 10)}$.  Subtracting a from both sides and factoring, we get $ a(a\minus{}1)(a\plus{}1)(a^2\plus{}1)\equiv 0\mbox{ (mod 10)}$.  All even $ a$ work because there's an even number and a multiple of 5, and if we used the zero product property, we'd get all odd $ a$ as numbers that satisfy this congruence.  If all even $ a$ and all odd $ a$ work, then all values of $ a$ work.
[/hide]

A related problem is to find the last 2 digits of a number to some large power.  In this case, it repeats after every 20.  Unfortunately I don't have a proof yet.

Principle of Inclusion and Exclusion (PIE)

The principle of inclusion and exclusion (PIE) gives us a way to relate the size of the union of two or more sets to the sizes of the sets and their intersections. For two sets this is simply the following: |AUB| = |A| + |B| - |A∩B|

This relationship can be easily seen by examining the Venn diagram. If we count up all the elements in A and all the elements in B, we will count each element of A∩B twice, so we subtract by the size of A∩B to correct for our over-counting. We first “include” everything, and then “exclude” the things we over-included.

Note that the formula above is equivalent to: |A∩B| = |A| + |B| - |AUB|
The principle of inclusion/exclusion is useful whenever we know or can easily count three of the sets in the expression above in order to compute the size of the fourth. Usually we are trying to find either AUB or A∩B.

Some examples:

Sample Problem #0: There are 100 students in a school, 50 don’t play sports, 70 don’t play music, and 40 don’t do either. How many students do both?

Solution: Let U be the set of all students in the school, A be the subset who play sports and let B be the subset in the school band. We are given that |U| = 100, |Ac| = 50, |Bc| = 30, and |(AUB)c| = 40. We want to compute A∩B. Counting complements gives |A| = 100-50 = 50, |B| = 100-70 = 30, and |AUB| = 100 – 40 = 60. Applying the principle of inclusion/exclusion (PIE) we have: |A∩B| = |A| + |B| - |AUB| = 50 + 30 – 60 = 20.

see [url]http://ocw.mit.edu/NR/rdonlyres/hs/com/898FF3EB-CF44-4A4F-8B1C-417395517A3F/0/lecturenotes_3.pdf[/url] for further coverage of this subject.

Here's an extension of PIE.  If you can get this one, then you understand the idea of PIE pretty well

Given three sets $ A$, $ B$, and $ C$, find a formula for the number of elements in at least 2 of the sets.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be an abelian, transitive subgroup of $ S_A$, the symmetric group on the set $ A$. Prove that $ \sigma(a)\neq a$ for all $ \sigma\in G \minus{} 1$ and all $ a\in A$. Deduce that $ |G|\equal{}|A|$.

The stabilizer subgroups $ G_a$ are conjugated and for G being abelian they are equal. So $ \bigcap_{a\in A}G_a\equal{}G_a$ is the kernel of the operation. Because G is a subgroup of $ S_A$ the kernel is trivial. $ \Rightarrow G_a\equal{}\{1\}$. Because this holds for all a, G operates regulary. Hence there is a bijection between G and A.


How do I write the delta x of two points? For example I have a diagram with many points, I want to write a formula the contains the delta x and delta y of two specific points. Imagine A and B are points on a diagram with many other points.

$ h = \sqrt{{\Delta{x}_{BA}}^2 + {\Delta{y}_{BA}}^2} $ is this correct?

I was thinking that just writing $ \Delta{x} $ was a bit ambiguos given there are many points on the diagram.

By "delta" of two points, do you mean distance?  Then the standard notation for distance between two vectors x and y is just |x - y|.

[quote="xxreddevilzxx"][quote="paladin8"]I'm not sure, but personally I would probably write $x_A-x_B$ and $y_A-y_B$.[/quote]

or $x_B-x_A$ and $y_B-y_A$[/quote]

In that case its probably best to do  $|x_B-x_A|$ and $|y_B-y_A|$ and not worry about the sign (as polymorphic said).

I just thought their might be one term I could use for the distance (yep, Im using delta here) between two points. So referring to my original sample, would this make sense if you saw it.

$h = \sqrt{\Delta{x}^2 + \Delta{y}^2}\\
h = \sqrt{|x_B - x_A |^2 + |y_B - y_A|^2}$

I mean make sense in terms of notation (not theory).

Looks to be correct, although I believe the absolute values are optional in this case.  You may or may not need them later though.


How does Ohio pick candidates for ARML? Is it based strictly on the OCTM?

I think it's based on the OHMIO, which is the second level of the OCTM, although it might be by combined score.


Also post on your grade level. Just for information. Btw, 8 and I'm a freshman. It does make a big difference for freshmen.

woot, I'm the highest voter right now!!

That'll change though.

I got a 10 and I'm a freshman (9th grade).

I got a 9... I think it's sort of hard to judge scores without seeing the tests... when do they get shipped?

And my score matches my grade  :) .  Make sure to include grades because it makes a big deal to freshmen!

So the papers are returned back to contestants?

[quote="matt276eagles"]8, freshman.  Good luck, chuckzeduck.   I'm nervous: :([/quote]

Yeah, me too. Anyone have any guesses on the freshman floor for mosp

I guess 8.

EDIT: No, it's going to be 7.

EDIT #2: It's going to be $n$, where $n\le 8$.

9 :P  :rotfl: You guys with 8 will probably be in, seeing how last year's cutoff was a 7 and the average was 4 points higher.

[quote="Phelpedo"]9  :rotfl:[/quote]

Ouch.  :rotfl:

From the AMC website, 432 people took the USAMO and 222 scored 7 or below.  Assuming that freshmen on average are slightly worse at USAMO than other grades, the cutoff will be at most 8.

But are freshmen actually worse at USAMO?  I think so, or at least hope so.

I got a 14. Which means I wildly underestimated my score(predicted a 9-12). Hoping that'll make Freshman MOP

Sirorange freshman or non-freshman?

And matteagles I'd say that if freshmen are "good" at USAMO then they'd be REALLY good... like eric larson last year.  So hopefully the really good freshmen will qualify in blue mop so the next 30 can fill in.

What I'm worried about is a tie situation... like what almost happened getting into USAMO.

I'm a freshman (the one who posted above.)

Yeah, jrshoch. Do you think they'll go to mosp index, as from another topic in this forum (sry, too lazy to link it here)?

14

[hide]Don't worry freshmen, I'm a junior.  I won't be taking one of your spots.   :P [/hide]

JB

I got 9 as a freshman, which probably means that the cutoff will be 10.  I would say a higher cutoff, but 10 would be, for my purposes, more ironically appropriate, especially as I have no earthly idea how I could have scored as low as I did, and my earlier display of hubris at school.  Also, this was a cutoff for score distributions on the AMC site.  Honestly, wasn't everybody underestimating the cutoff last year?

According to the phone thread, probably nobody has yet been notified save HMs/winners.

18, sophmore. If the cutoff is < 21... :wallbash:

15 Sophomore
 :)
I'm pretty happy, since I did pretty poorly the second day. I'm hoping I'll make MOP next year

13, 8th grade.  I have a question about the scoring:  Hypothetically, if a solution is complete but awkward, will points be deducted?  (No, that wasn't really hypothetical... not even close  :P )

[quote="orl"][quote="Ravi12346"]13, 8th grade.[/quote]

Then I know with probability of 1.01 what you will get next year.  :D[/quote]
Are you trying to say he's going to own everybody with a 41? :D

You should make a 43 option, just to see how many liars there are at AoPS.

Awkward solutions are a gamble. At best, you lose any tiebreakers they have. At worst, they decide your solution is crap and deduct points off it. Plus, there is a greater chance of you missing a non-rigorous part of your proof if your solution is awkward.

f(x) = ((((1/((3/4)^2-1/2))-((sin(pi/4))^-2)(15^2+11^2-(2^3+1)^(1/2))^(1/3))^(i(2i)^3)-(3(8^(2/3)))^2)/7)^(1/(1/(1/3-1/12)))-1 + e^2*sin(pi*x)
My score next year will be (with probably >= 1.01) equal to f(infinity).
Yes, it will own everyone with a score that is actually defined.

P.S. Put that large expression (without the e^2*sin(pi*x) part) into Google, and it will tell you that it equals one.  :lol:

[quote="JellyWorld"]I;m not from the USA so I didn't take the test, but I'll try it and get someone to mark it, then I'll post my score :P[/quote]

That score wouldn't really be relevant, as your grader would probably not have the official graders' rubric.

You all did better than me... I only got 5 on the AIME, but I'm an 8th grader though.

Congrats to all.


On: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21188[/url]
Armo asserts: [quote="Armo"]Using that $\cos{x}+\cos{y}=2\cos{\frac{x+y}{2}}\cos{\frac{x-y}{2}}$, $\cos{x}=1-2\sin^{2}{\frac{x}{2}}$
we get $ \cos{A} +\cos{B}+ \cos{C}=4\sin{\frac{A}{2}}\sin{\frac{B}{2}}
\sin{\frac{C}{2}}+1$.[/quote]
I couldn't prove this identity using only those two, but I did find a [hide="solution."]I say $A,B,C = A_1,A_2,A_3$ respectivly and $A_i=A_{i+3}$ or the indices are cyclic mod 3. So, we get:
$\displaystyle \sum_{i=1}^{3}\left(\cos A_{i} + \cos A_{i+1}\right) = \displaystyle \sum_{i=1}^{3}\left(2\cos\frac{A_{i} + A_{i+1}}{2}\cos\frac{A_{i} - A_{i+1}}{2}\right)             \\
\displaystyle2 \sum_{i=1}^{3}\left(\cos A_{i}\right) = \displaystyle \sum_{i=1}^{3}\left(2\cos\frac{180 - A_{i+2}}{2}\cos\frac{A_{i} - A_{i+1}}{2}\right)             \\
\displaystyle \sum_{i=1}^{3}\left(\cos A_{i}\right) = \displaystyle \sum_{i=1}^{3}\cos\left(90-\frac{A_{i+2}}{2}\right)\left(\cos\frac{A_{i}}{2}\cos\frac{A_{i+1}}{2}+\sin\frac{A_{i}}{2}\sin\frac{A_{i+1}}{2}\right)             \\
\displaystyle S = \displaystyle \sum_{i=1}^{3}\sin\frac{A_{i+2}}{2}\left(\cos\frac{A_{i}}{2}\cos\frac{A_{i+1}}{2}+\sin\frac{A_{i}}{2}\sin\frac{A_{i+1}}{2}\right)             \\
\displaystyle S = \displaystyle \sum_{i=1}^{3}\left(\sin\frac{A_{i+2}}{2}\cos\frac{A_{i}}{2}\cos\frac{A_{i+1}}{2}+\sin\frac{A_{i}}{2}\sin\frac{A_{i+1}}{2}\sin\frac{A_{i+2}}{2}\right)             \\
\displaystyle S = \displaystyle \sum_{i=1}^{3}\left(\sin\frac{A_{i+2}}{2}\cos\frac{A_{i}}{2}\cos\frac{A_{i+1}}{2}\right)+\displaystyle \sum_{i=1}^{3}\left(\sin\frac{A_{i}}{2}\sin\frac{A_{i+1}}{2}\sin\frac{A_{i+2}}{2}\right)
\\
\displaystyle S = \displaystyle T+\displaystyle3\sin\frac{A_{1}}{2}\sin\frac{A_{2}}{2}\sin\frac{A_{3}}{2}             \\$
Now, to simplify T:
$\displaystyle \sum_{i=1}^{3}\left(\sin\frac{A_{i+2}}{2}\cos\frac{A_{i}}{2}\cos\frac{A_{i+1}}{2}+ \sin\frac{A_{i}}{2}\cos\frac{A_{i+1}}{2}\cos\frac{A_{i+2}}{2}\right) = \displaystyle \sum_{i=1}^{3}\cos\frac{A_{i+1}}{2}\left(\sin\frac{A_{i+2}}{2}\cos\frac{A_{i}}{2}+ \sin\frac{A_{i}}{2}\cos\frac{A_{i+2}}{2}\right)                               \\
2T = \displaystyle \sum_{i=1}^{3}\cos\frac{A_{i+1}}{2}\sin\frac{A_{i+2}+A_i}{2}                               \\
2T = \displaystyle \sum_{i=1}^{3}\cos\frac{A_{i+1}}{2}\sin\frac{180-A_{i+1}}{2}
\\
2T = \displaystyle \sum_{i=1}^{3}\cos\frac{A_{i+1}}{2}\cos\frac{A_{i+1}}{2}
\\
4T = \displaystyle \sum_{i=1}^{3}2\cos^2\frac{A_{i+1}}{2}                 \\
4T - 3 = \displaystyle \sum_{i=1}^{3}\left(2\cos^2\frac{A_{i+1}}{2}-1\right)               \\
4T - 3 = \displaystyle \sum_{i=1}^{3}\cos A_{i+1}                                 \\
4T - 3 = S                                                 \\$
Now, we set P:
$P = \sin\frac{A_{1}}{2}\sin\frac{A_{2}}{2}\sin\frac{A_{3}}{2}                              \\
S = T + 3P                                 \\
4T - 3 = T + 3P                         \\
3T  = 3P + 3                         \\
T  = P + 1                         \\
\Rightarrow  S = (P+1) + 3P                   \\
S = 4P+1                  \\
\cos A_{1} + \cos A_{2} + \cos A_{3} = 4\sin\frac{A_{1}}{2}\sin\frac{A_{2}}{2}\sin\frac{A_{3}}{2}+1                  \\$
[/hide] Other solutions are welcome and Armo if your out there please at least outline your solution.

$\cos A + \cos B = \cos{\left( \frac{A + B}{2} + \frac{A - B}{2} \right)} + \cos{\left( \frac{A + B}{2} - \frac{A - B}{2} \right)} =$

$= 2 \cos{\frac{A + B}{2}} \cos{\frac{A - B}{2}}$

Since $A + B + C = 180^o$, $\sin C = \sin{(A + B)}$ and $\cos C = -\cos{(A + B)}$.

Since $\frac{A + B}{2} + \frac C 2 = 90^o$, $\sin{\frac C 2} = \cos{\frac{A + B}{2}}$ and $\cos{\frac C 2} = \sin{\frac{A + B}{2}}$ .

$\cos A + \cos B + \cos C = \cos A + \cos B - \cos{(A + B)} =$

$= 2 \cos{\frac{A + B}{2}} \cos{\frac{A - B}{2}} - \left( \cos^2{\frac{A + B}{2}} - \sin^2{\frac{A + B}{2}} \right) =$

$2 \cos{\frac{A + B}{2}} \cos{\frac{A - B}{2}} - \left( 2 \cos^2{\frac{A + B}{2}} - 1 \right) =$

$2 \cos{\frac{A + B}{2}} \left( \cos{\frac{A - B}{2}} - \cos{\frac{A + B}{2}} \right) + 1$

Similarly as factoring $\cos A + \cos B$, we can factor $\cos y - \cos x$:

$\cos y - \cos x = 2 \sin{\frac{x + y}{2} \sin{\frac{x - y}{2}}}$

Put $x = \frac{A + B}{2}$, $y = \frac{A - B}{2}$. Then $x + y = A$ and $x - y = B$ and

$\cos{\frac{A - B}{2}} - \cos{\frac{A + B}{2}} = 2 \sin{\frac A 2} \sin{\frac B 2}$

Hence,

$\cos A + \cos B + \cos C = 2 \sin{\frac C 2} \left( 2 \sin{\frac A 2} \sin{\frac B 2} \right) + 1 =$

$= 4 \sin{\frac A 2} \sin{\frac B 2} \sin{\frac C 2} + 1$


[quote="Maxim Bogdan"]$ a,b,c\in[\alpha;\beta],\alpha > 0.$Prove that:$ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\le\frac {(2\alpha \plus{} \beta)(2\beta \plus{} \alpha)}{\alpha\beta}$[/quote]
It is old.a special case of Kantorovis 's ineq

$ a,b,c\in\mathbb{R},and \displaystyle\sum\frac {1}{1 \plus{} a^2} \equal{} 2$. Prove that: $ abc(a \plus{} b \plus{} c \minus{} abc)\le\frac {5}{8}$


Show that $a$ and $b$ have the same parity if and only if there exist
integer $c$ and $d$ such that $\displaystyle a^2+b^2+c^2+1=d^2$.

$a^2+b^2+1$ can only have the form $4k+3,4k+1,4k+2$. The numbers $d+c,d-c$ have the same perity, so $(d+c)(d-c)=4k+3,\ 4k+1$, or $4k$. This shows that $a^2+b^2+1$ can't be $4k+2$, so $a,b$ have the same parity.

On the other hand, if they do have the same parity, then let $a^2+b^2+1=mn$, where $m,n$ must be odd. Then the system of equations $d-c=m, d+c=n$ has a solution and we're done.


Suppose that $ a\plus{}b\equal{}\sqrt{3\sqrt{3}\minus{}\sqrt{2}},\ a\minus{}b\equal{}\sqrt{3\sqrt{2}\minus{}\sqrt{3}}$, find the value of $ a^4\plus{}a^2b^2\plus{}b^4$.

$ (a\plus{}b)^2\equal{}a^2\plus{}2ab\plus{}b^2\equal{}3\sqrt{3}\minus{}\sqrt{2}$
$ (a\minus{}b)^2\equal{}a^2\minus{}2ab\plus{}b^2\equal{}3\sqrt{2}\minus{}\sqrt{3}$
Subtracting, $ 4ab\equal{}3\sqrt{3}\minus{}\sqrt{2}\minus{}3\sqrt{2}\plus{}\sqrt{3}\equal{}4\sqrt{3}\minus{}4\sqrt{2}$, $ ab\equal{}\sqrt{3}\minus{}\sqrt{2}$, $ a^2b^2\equal{}5\minus{}2\sqrt{6}$
Adding the first two, $ a^2\plus{}b^2\equal{}\sqrt{3}\plus{}\sqrt{2}$
Squaring that, $ a^4\plus{}b^4\plus{}2a^2b^2\equal{}5\plus{}2\sqrt{6}$
Then subtracting $ a^2b^2$ as found above gives:
$ a^4\plus{}a^2b^2\plus{}b^4\equal{}4\sqrt{6}$

That's right. :)

Let's learn by heart the following equations.

$ \boxed{(a \plus{} b)^2 \minus{} (a \minus{} b)^2 \equal{} 4ab}$

$ \boxed{(a \plus{} b)^2 \plus{} (a \minus{} b)^2 \equal{} 2(a^2 \plus{} b^2)}$


A bar of mass 0.5 kg lying on a horizontal plane with a friction coefficient b=0.1, is attached to the wall by means of a horizontal unstretched massless spring of spring constant k=[b]245 N/m[/b]. The bar is displaced so that the spring is stretched by x=3 cm and then released.

a) Find the period of oscillation of spring.

It is pretty much the same for determining the period of the oscillation with a friction nondependent of the velocity as if there is no friction, since the constant friction only changes the equilibrium, thus resulting the diminishing amplitude.
So the period will be $ 2\Pi\cdot\sqrt{\frac{m}{k}}$.


On page 7 of Engel, what is the second observation? (i.e. $S$ and $tS$ have the same life expectancy)

I think it is:

If $S$ is a quadruple $(a,b,c,d)$ which by a sequence of transformations as described in the question reaches the quadruple $(0,0,0,0)$ in $n$ transformations, then the quadruple $(at,bt,ct,dt)$ also reaches the quadruple $(0,0,0,0)$ in $n$ transformations.

i.e. $S$ and $tS$ require the same number of transformations to reach $(0,0,0,0)$. This follows because if $R=tS$ then $R_{1}=tS_{1}$ etc. (using the notation of the question).


$ \frac{2x^3 \minus{} 3x^2 \plus{}x\plus{}1}{2x^3\minus{}3x^2\minus{}x\minus{}1} \equal{} \frac{3x^3\minus{}x^2\plus{}5x\minus{}13}{3x^3\minus{}x^2\minus{}5x\plus{}13}$

I tried using compondedo and dividendo and got this:

$ \frac{2x^3\minus{}3x^2}{x\plus{}1} \equal{} \frac{3x^3\minus{}x^2}{5x\minus{}13}$

Can't use componendo and dividendo again because it'll go back to the original statement.

I tried cross multiplying the two expression and the expression  became messy.
Is there any other way?

It's not an equality. You're supposed to find the solutions. :huh:

[hide="Hint"]If there is a nice solution to this, synthetic division would probably find it. Start by observing
$ 2x^3 \minus{} 3x^2 \plus{} x \plus{} 1 \equal{} (2x^3 \minus{} 3x^2 \minus{} x \minus{} 1) \plus{} 2x \plus{} 2$
$ 3x^3 \minus{} x^2 \plus{} 5x \minus{} 13 \equal{} (3x^3 \minus{} x^2 \minus{} 5x \plus{} 13 ) \plus{} 10x \minus{} 26$[/hide]

oh sorry, i guess i didnt get it...

a attempt at redemption:

[hide="uses that idea"]$ \frac {2x^3 \minus{} 3x^2 \plus{} x \plus{} 1}{2x^3 \minus{} 3x^2 \minus{} x \minus{} 1} \equal{} \frac {3x^3 \minus{} x^2 \plus{} 5x \minus{} 13}{3x^3 \minus{} x^2 \minus{} 5x \plus{} 13}$
$ 1 \plus{} \frac {2x \plus{} 2}{2x^3 \minus{} 3x^2 \minus{} x \minus{} 1} \equal{} 1 \plus{} \frac {10x \minus{} 26}{3x^3 \minus{} x^2 \minus{} 5x \plus{} 13}$
$ \frac {2x \plus{} 2}{2x^3 \minus{} 3x^2 \minus{} x \minus{} 1} \equal{} \frac {10x \minus{} 26}{3x^3 \minus{} x^2 \minus{} 5x \plus{} 13}$
$ \frac {x \plus{} 1}{2x^3 \minus{} 3x^2 \minus{} x \minus{} 1} \equal{} \frac {5x \minus{} 13}{3x^3 \minus{} x^2 \minus{} 5x \plus{} 13}$
$ 3x^4 \plus{} 2x^3 \minus{} 6x^2 \plus{} 8x \plus{} 13 \equal{} 10x^4 \minus{} 41x^3 \plus{} 34x^2 \plus{} 8x \plus{} 13$
$ \minus{} 7x^4 \plus{} 43x^3 \minus{} 40x^2 \equal{} 0$
$ (x^2)(7x^2 \minus{} 43x \plus{} 40)$[/hide][hide="uses your last equation"]
$ \frac {2x^3 \minus{} 3x^2}{x \plus{} 1} \equal{} \frac {3x^3 \minus{} x^2}{5x \minus{} 13}$
$ 10x^4 \minus{} 41x^3 \plus{} 39x^2 \equal{} 3x^4 \plus{} 2x^3 \minus{} x^2$
$ 7x^4 \minus{} 43x \plus{} 40x^2$
$ \ldots$
[/hide]
(notice that the equation found by cross multiplication may have x4 coefficients, but it is indeed a quadratic)

from here, factor the quadratic and add 0 to the solutions of the quadratic


The letter T is formed by placing two
$ 2\: \text{inch}\!\times\!4\: \text{inch}$ rectangles next to each
other, as shown.  What is the perimeter of the T, in inches?
[asy]draw((1,0)--(3,0)--(3,4)--(4,4)--(4,6)--(0,6)--(0,4)--(1,4)--cycle);[/asy]

The perimeter is 2(2+4)=12, so the total is 24. We have to subtract the meeting point length twice from the total perimeter, so the perimeter of T is 24-4=20.


trigonometry
inequalities
calculus
calculus open

proove that $tg(\frac{5\cdot \pi}{18})>1+\frac{\pi}{18}$
how can i do this with lagrange
i already tried but i don't really know how to choose the interval
on wich i apply the theorem

apply arctan to your inequation and use the interval $[1;1+\Pi/18]$

rodolphe why apply on that exact interval?
there must be a reasoning
please tell me why?

$tg(\frac{5\cdot \pi}{18})>1+\frac{\pi}{18}$
$\Leftrightarrow \frac{5\cdot \pi}{18} > \textnormal{arctan}(1+\frac{\pi}{18})$
$\Leftrightarrow \frac{5\cdot \pi}{18} - \frac{\pi}{4} > \textnormal{arctan}(1+\frac{\pi}{18}) - \textnormal{arctan}(1)$
$\Leftrightarrow \frac{\pi}{36} > \frac{\pi}{18}\times\frac{1}{1+c^2}$ with $1<c<1+\pi/18$
$\Leftrightarrow \frac{\pi}{36} > \frac{\pi}{36} \times \frac{2}{1+c^2}$ true since $1+c^2 >2$

you didn't quite give me an solution
just prooved that the inequality is right using it as hypothesis

$\frac{\textnormal{arctan}(1+\frac{\pi}{18}) - \textnormal{arctan}(1)}{\frac{\pi}{18}}=\frac{1}{1+c^2}$ with $1<c<1+\frac{\pi}{18}$ using formulae $f(a)-f(b)=(a-b)f'(c)$ with $a\leq c\leq b$. Here we have $<$ since arctan' is strictly positive.
Now :
$c^2>1$
$\Leftrightarrow \frac{1}{1+c^2}<\frac{1}{2}$
$\Leftrightarrow \frac{\textnormal{arctan}(1+\frac{\pi}{18}) - \textnormal{arctan}(1)}{\frac{\pi}{18}} \times \frac{\pi}{18} < \frac{\pi}{36}$
$\Leftrightarrow \textnormal{arctan}(1+\frac{\pi}{18}) < \frac{\pi}{36}+ \textnormal{arctan}(1)=\frac{10\pi}{36}=\frac{5\pi}{18}$
$\Leftrightarrow 1+\frac{\pi}{18} < \textnormal{tan}(\frac{5\pi}{18})$

It's OK ?