trigonometry geometry MATHCOUNTS AMC AIME ARML algebra

Throughout the history of this site it has been difficult to keep the Getting Started Forum at the appropriate difficulty level.  For this reason I have decided to create a guide to help users understand what material is appropriate for this forum.

First, an explanation of suitable problems:

$ \bullet$ Most AoPS Volume 1 material.  Volume 2 nearly always belongs in the Intermediate Topics forum and sometimes the Pre-Olympiad forum (non-trivial proofs).
$ \bullet$ Topics from Algebra 1 which you generally do not see in Pre-Algebra
$ \bullet$ Topics from Algebra 2 - through *very* basic trig
$ \bullet$ Topics from High School Geoemtry
$ \bullet$ Most AMC 10 problems (between the first 5 and the last 3-5)
$ \bullet$ Harder MATHCOUNTS problems
$ \bullet$ Harder AMC 8 problems
$ \bullet$ The very simplest AIME problems are also reasonable
$ \bullet$ Most problems from most local competitions
$ \bullet$ The easier problems from current ARML (or about half of those from 10 to 25 year old ARML/NYSML contests)
$ \bullet$ Most math league problems, though harder problems such as the last 3 or so individual problems on a Mandelbrot competition should go in Intermediate.

[b]Note: If a problem involves a non-trivial step using more than one area of mathematics (algebra with number theory, geometry with serious algebra, etc.) then it does not belong in the Getting Started forum.  Remember that the average score on a typical AIME is 2-3 problems.  That's given the fact that AIME participants already score well on a qualifying exam.  At most 1 in 20 AIME problems belong in this forum.[/b]

It is always better to err on the side of putting a problem in a more advanced forum than in a too easy forum.

Please do not post that a message belongs in a different topic.  Moderators will move posts as appropriate.

To help clarify, here is a sample list of appropriate problems, though be aware that some of these are on the harder side of problems that should be posted in this forum.  If a problem is harder than these, or requires more steps, it probably belongs in the Intermediate forum.  [b]Please do not work any of these problems in this thread so that we can keep it clean for discussion of appropriate difficulty.  I will create a separate thread for each problem.[/b]

[u][b]Algebra[/b][/u]

$ \bullet$ Factoring (other than monomial factoring), including roots and zeros

$ \bullet$ Domain and range problems

$ \bullet$ Graphing non-linear equations

$ \bullet$ If $ f(x) \equal{} \frac {x^{4} \plus{} x^{2}}{x \plus{} 1}$, find $ f(i)$, where $ i^2=-1$.

$ \bullet$ If $ a \equal{} \log_{8}{225}$ and $ b \equal{} \log_{2}{15}$, then
\[ \text{(A)}\ a \equal{} b/2\ \ \ \ \text{(B)}\ a \equal{} 2b/3\ \ \ \ \text{(C)}\ a \equal{} b\ \ \ \ \text{(D)}\ b \equal{} a/2\ \ \ \ \text{(E)}\ a \equal{} 3b/2
\]
$ \bullet$ The sum of an infinite geometric series with common ratio $ r$ such that $ |r| < 1$ is 15, and the sum of the squares of the terms of this series is 45.  The first term of the series is
\[ \text{(A)}\ 12\qquad\text{(B)}\ 10\qquad\text{(C)}\ 3\qquad\text{(D)}\ 3\qquad\text{(E)}\ 2
\]
[u][b]Geometry[/b][/u]

$ \bullet$ A sphere is inside a circumscribing cube.  What fraction of the volume of the cube is inside the sphere?
\[ \text{(A)}\ \frac {1}{2}\qquad\text{(B)}\ \frac {\pi}{3}\qquad\text{(C)}\ \frac {3}{4}\qquad\text{(D)}\ \frac {\pi}{4}\qquad\text{(E)}\ \frac {\pi}{6}
\]
$ \bullet$ The point (1, 1, 1) is rotated $ 180^{\circ}$ about the $ y$-axis, then reflected through the $ y$-$ z$ plane, reflected through the $ x$-$ z$ plane, rotated $ 180^{\circ}$ abou the $ y$-axis, and reflected through the $ x$-$ z$ plane.  Find the coordinates of the point now.

$ \bullet$ Let $ ABC$ be an equilateral triangle of side $ s$.  Construct a circular triangle as follows.  Consider the circular arc from $ A$ to $ B$ with center of the arc at $ C$.  Another arc is from $ B$ to $ C$ with center at $ A$.  A third arc is from $ C$ to $ A$ with center at $ B$.  Find the formula for the area of the circular triangle, i.e., the area enclosed by the three circular arcs.
\[ \text{(A)}\ \sqrt {3}s^{2}\qquad\text{(B)}\ \frac {\pi}{3}s^{2}\qquad\text{(C)}\ \left(\frac {\pi}{2} \plus{} \frac {\sqrt {3}}{2}\right)s^{2}
\]

\[ \text{(D)}\ \left(\frac {\pi}{2} \minus{} \frac {\sqrt {3}}{2}\right)s^{2}\qquad\text{(E) None of these answers}
\]
[u][b]Counting[/b][/u]

$ \bullet$ How many positive integers less than 101 are multiples of either 5 or 7, but not both at once?

$ \bullet$ How many positive integers less than 1991 are of the form $ x^{n}$, where $ x\ge 1$ and $ n\ge 2$ are integers?

$ \bullet$ In how many ways can 3 identical algebra books and 3 identical geometry books be arranged on a shelf?

[u][b]Probability[/b][/u]

$ \bullet$ Find the probability of flipping at least 4 heads in 6 tosses of a fair coin.

$ \bullet$ A positive integer $ n$ not exceeding 100 is chosen in such a way that if $ n\le 50$, then the probability of choosing $ n$ is $ p$, and if $ n > 50$, then the probability of choosing $ n$ is $ 3p$.  The probability that a perfect square is chosen is
\[ \text{(A)}\ .05\qquad\text{(B)}\ .065\qquad\text{(C)}\ .08\qquad\text{(D)}\ .09\qquad\text{(E)}\ .1
\]
$ \bullet$ Rocky and Bullwinkle are playing Risk.  Rocky rolls one six-sided die, while Bullwinkle rolls two of them.  What is the probability that Rocky's roll is as high as Bullwinkle's largest number?

[u][b]Number Theory[/b][/u]

$ \bullet$ The number $ 10!$, when written in the base 12 system, ends with exactly how many zeroes?
\[ \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5
\]
$ \bullet$ Find the greatest integer that will divide 13511, 13903, and 14589 and leave the same remainder.

$ \bullet$ The product of any two of the positive integers 30, 72, and $ N$ is divisible by the third.  What is the smallest possible value of $ N$?

Recently I have moved a few proof-related posts from this forum.  Unless the proofs are extremely elementary, they do not belong here.

I have begun banning users from specific forums in which they post off-difficulty problems.  I will generally give warnings, but if a student is posting Olympiad level problems in general, I don't mind simply banning them and waiting for a request to return to this forum because they should be focusing on other forums to begin with.

Would it be possible to give any non-Americans an indication on the level of the education system in America? I am ignorant, for example of how old people in high school tend to be.  :ninja:

The first post is a good indication of the level and topics presented normally in a typical high school in America, and this same post is a good example what type of problems and their difficulty that fits in this subforum's category. The age can vary slightly but we all know the ages aren't much different from country to country, usually 13-18, excluding exceptional cases.

Please take the time to review to level of difficulty for the forum. We have had to move far too many posts to different forums lately.

[quote="moogra"]Actually the majority of problems posted here are above the level. The level for some meet intermediate topics level, but are not moved there since a higher level of expectation is in intermediate level now. I think we should just change the rules a bit and allow more difficult material.[/quote]I do not agree with changing the rules by allowing higher difficulty because:

1) There is a smooth transitioning from MATHCOUNTS to AMC, this forum serves such a purpose.
2) Users need to MOVE ON. Apparently, many users who has quite mastered a higher level of problem solving DO NOT move on to higher forums and post problems with the appropriate difficulty there. Look at the Olympiad Section. It is crowded by only the serious IMO-ers from around the world but... very few American users? The key is to move on.
3) Although we all have that habit of feeling like 'being at home', at least respect the forums' difficulty. It is NOT your home. Please do not post off difficulty problems.
4) We also know that as the year progressed, many students advanced in their problem solving skills. However, the flexibility of the forum's problem difficulty should not encompass the whole spectrum of problems. So what's the point of separating all these forums then?

 :spidy:

Bump because we are getting way too many intermediate level problems in the forum.

We do need a couple of extra moderators for this forum to help us keep the problem level where it should be, as described in the first couple of posts. If you would like to volunteer, please send me a PM.

Valentin: Should there be specific qualifications we should have before PM-ing you? I don't want to PM you (or anyone else to) who are a spammer and do not deserve to be a moderator.
Also, do you already have to be a moderator to sign up? Or can anyone PM you for this?

Thanks! :)

It's done for now, we don't need any more Moderators. When the time comes again that we would need any, I will make a post here, as before. So no more PMs please.

I would just like to bump this thread because recently there has been again a huge surge of users not getting that High School BASICS is for BASIC problems.

This needs to be bumped again. Just now I had to deal with [b]seven[/b] different threads that weren't in the appropriate forum, sadly by seven different users. From now on, if you don't possess enough judgment to decide if something like triple-variable manipulation problems is not what High School Basics is for, your thread might just be deleted instead of moved if you persist.

This was my first time of doing that and it got deleted instead of moved.  You said "If you persist on doing it."  It was my first time.  Why?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=378833]Your thread was not deleted.[/url] Use the search function.

Bump. [url=http://dictionary.reference.com/browse/basic]Here's a definition of Basic.[/url]

Bump. I don't want to re-instate a "State of Emergency" as I described four posts before in this thread.

The difficulty of the posts here in HSB is becoming way too difficult for its level once again. In fact, if you randomly select a topic in the first two pages, chances are that it's too hard for the forum. At the time of this post, the last few topics on my screen are:

[list]
[*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=447730]Need Graph Theory Problem[/url]: This thread is way out of place, as graph theory should be in the Pre-Olympiad Section, since it isn't used until the olympiad level. Furthermore, Turan's theorem may not even be appropriate for Pre-Olympiad, as it's actually olympiad level.
[*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=447702]Complex Roots[/url]: Like MCrawford mentioned in his first post, HSB should only have the simplest AIME problems. This is probably an AIME level 4-7, which is too hard for this forum. In general, problems that involve complex numbers should be reserved for HSI and higher.
[*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=447699]Excircle[/url]: While the solution to the posted problem only involves elementary techniques, they require nontrivial steps that would be deemed AIME level. Also, excircles are generally considered more advanced areas of geometry and thus are not covered in High School geometry. I'm more willing to let this one slide than the other ones, but it's still a bit too hard for this forum.
[*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=447392]system of equations[/url]: Nothing wrong with this one. (It just happens to be in the last five topics posted.)
[*][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=446919]Proof[/url]: Like the first example, this topic is completely out of place for its. To begin this rant, proofs are typically no-nos. Hard proofs, which include construction-type questions, are always no-nos. [/list]

Please read the first post of this thread again and make sure the difficulty levels of your posts in this forum are at the right level. Thanks.

Might I also take the time to request that the "Too Easy" and "Too Hard" buttons be used [b][i][u]whenever[/u][/i][/b] applicable.  They are supposed to align the difficulty of the fora with what they're supposed to be.

Yes, that's a great notion, but I'm pretty sure that people who have been doing that will continue doing so and people who haven't will continue not doing so upon notification, so we can only hope for an increased number of people who care.

How come its called High School Basics - shouldn't it be called something by skill instead of by age? Just wondering :O

post it in intermediate first, they would move it here if it was too easy.

[quote="hshiems"]Sorry I must bump this thread, but I have a question: would having a Number Theory Proof in this forum count as High School "Basics"? Or would a number Theory proof be too difficult for this forum?[/quote]
Generally speaking, it is not advised that proofs be posted in this forum. Most Number Theory proof problems are High School Pre-Olympiad level (or higher), so as SMOJ, post in a higher forum when in doubt.

[quote="snigy2000"]Would conics be a part of High School basics?[/quote]
Yeah, they teach it in 9th grade honors.

Are you allowed to post logarithm problems?

what about things from the intro to c & p class?

Edit: sorry for reviving the thread


Something happened, and when i was going to change my signature, the area had disappeared. HELP!!! What is happening

One topic per problem please! Continue at http://www.mathlinks.ro/Forum/viewtopic.php?t=46913


Για το πρωτο ερώτημα το ΜΝ ειναι ριζικος αξονας των δυο κυκλων  και ο ριζικος αξονας ειναι καθετος στη διακεντρο αρα KL καθετο με το ΜΝ.(Νομιζω)

[quote="kritikos"]Για το πρώτο ερώτημα το $ MN$ είναι ριζικός αξονας των δύο κύκλων και ο ριζικός άξονας είναι κάθετος στη διάκεντρο. Άρα $ KL$ κάθετο με το $ MN.$ (Νομίζω)[/quote]
Αυτό είναι σωστό και αποδεικνύεται εύκολα ότι τα σημεία $ M$ και $ N,$ ανήκουν στο ριζικό άξονα των δύο κύκλων.

Είναι προφανές ότι αυτό ισχύει για το σημείο $ M,$ γιατί έχουμε $ MC \equal{} MA \equal{} MB \equal{} MD$ $ ,(1)$

Για το σημείο $ N,$ αρκεί να αποδείξουμε ότι $ (NC)\cdot (NA) \equal{} (ND)\cdot (NB),$ ή ότι το τετράπλευρo $ ABDC$ είναι εγγράψιμο, που είναι προφανές από την $ (1).$

Ολοκληρώνεται έτσι η απόδειξη για το πρώτο ερώτημα, αλλά για το δεύτερο αναγκάστηκα να αναποδογυρίσω το σχήμα ... :)

Βρήκα μία ( σωστή ελπίζω ) λύση, αλλά είναι λίγο πολύπλοκη, γιατί χρησιμοποιώ στο τέλος το [b][size=100]θεώρημα Desarques[/size][/b] και δεν ξέρω αν ταιριάζει με το [b][size=100]Αρχιμήδης Level[/size][/b].

Θα περιμένω για μία απλούστερη προσέγγιση και θα βάλω αργότερα την απόδειξη για το δεύτερο ερώτημα, αν τελικά δεν υπάρχει κάποιο λάθος.

Κώστας Βήττας.

[quote="stergiu"][color=blue][i] Αυτή η άσκηση δεν πρέπει να λείπει από το Ελληνικό φόρουμ ![/i][/color]

Δύο κύκλοι  $ K,L$ εξωτερικοί μεταξύ τους εφάπτονται σε μια ευθεία στα σημεία $ A,B$.Από το μέσο $ M$ του $ AB$ φέρνουμε τα εφαπτόμενα τμήματα $ MC,MD$ προς τους δύο κύκλους $ K,L$ αντίστοιχα.Οι ευθείες $ AC,BD$ τέμνονται στο $ N$.Να αποδείξετε ότι :

  [b] α) [/b]Η   $ KL$ είναι κάθετη στην $ MN$

[b]   β) [/b]Αν οι   $ KL , MN$ τέμνονται στο $ R$ , τότε τα σημεία $ R,N,C,D$ είναι ομοκυκλικά.

 [u]Μπάμπης[/u][/quote]

[u][b]ΛΥΣΗ[/b][/u]

Το πρώτο ερώτημα το έκανα ακριβώς όπως και οι παραπάνω λύτες με το ριζικό άξονα.

Για το δεύτερο ερώτημα παρατηρούμε ότι αρκεί να δείξουμε ότι το τετράπλευρο $ MRDB$ είναι εγγράψιμο. Αλλά επειδή $ MR \perp KL$ έχουμε ότι το $ MRDL$ είναι εγγράψιμο. Επίσης είναι εύκολο να δείξουμε ότι το $ MDLB$ είναι εγγράψιμο, συνεπώς μόλις αποδείξαμε ότι τα σημεία $ M,R,D,L,B$ είναι ομοκυκλικά. Τώρα πλέον εύκολα με γωνιακές σχέσεις μπορούμε να δείξουμε το ζητούμενο.

Νίκος Ράπανος

[quote="Nick Rapanos"]....Για το δεύτερο ερώτημα παρατηρούμε ότι αρκεί να δείξουμε ότι το τετράπλευρο $ MRDB$ είναι εγγράψιμο. Αλλά επειδή $ MR \perp KL$ έχουμε ότι το $ MRDL$ είναι εγγράψιμο. Επίσης είναι εύκολο να δείξουμε ότι το $ MDLB$ είναι εγγράψιμο, συνεπώς μόλις αποδείξαμε ότι τα σημεία $ M,R,D,L,B$ είναι ομοκυκλικά. Τώρα πλέον εύκολα με γωνιακές σχέσεις μπορούμε να δείξουμε το ζητούμενο.

Νίκος Ράπανος[/quote]
Με τον ίδιο τρόπο αποδεικνύεται ότι και το τετράπλευρο $ MRCA,$ είναι εγγράψιμο και έτσι έχουμε :

Από εγράψιμο $ MRDB$ $ \Longrightarrow$ $ \angle MRD \equal{} 180^{o} \minus{} \angle MBN$ $ ,(1)$

Από εγράψιμο $ MRCA$ $ \Longrightarrow$ $ \angle MRC \equal{} 180^{o} \minus{} \angle MAN$ $ ,(2)$

Από $ (1),$ $ (2)$ $ \Longrightarrow$ $ \angle MRC \plus{} \angle MRD \equal{} 360^{o} \minus{} \angle CRD \equal{} 360^{o} \minus{} \angle MAN \plus{} \angle MBN$ $ ,(3)$

Από $ (3)$ $ \Longrightarrow$ $ \angle CRD \equal{} \angle MAN \plus{} \angle MBN \equal{} 180^{o} \minus{} \angle CND$ $ ,(4)$

Από την $ (4),$ συμπεραίνουμε ότι τα σημεία $ R,$ $ C,$ $ N,$ $ D,$ είναι ομοκυκλικά και ολοκληρώνεται η απόδειξη για το δεύτερο μέρος.

Αυτή η λύση, όπως την κατάλαβα μετά την υπόδειξη του Νίκου, είναι πολύ πιο απλή από αυτή στην οποία αναφέρθηκα πιο πάνω και δεν έχει νόημα να την παρουσιάσω, παρά μόνο σαν επέκταση του προβλήματος, για όποιον ενδιαφέρεται.

[b][size=100][color=DarkBlue]Έστω $ (O),$ ο περιγεγραμμένος κύκλος περί το τετράπλευρο $ RCND.$ Αποδείξτε ότι :

(a) – Το σημείο $ O\equiv KC\cap LD.$ είναι το κέντρο του κύκλου $ (O)$ και ότι $ NO\perp AB.$

(b) – Το σημείο $ N'\equiv KL\cap NO,$ είναι το αντιδιαμετρικό του $ N,$ στον κύκλο $ (O)$[/color][/size][/b].

Κώστας Βήττας.

[quote="vittasko"]

[b][size=100][color=DarkBlue]Έστω $ (O),$ ο περιγεγραμμένος κύκλος περί το τετράπλευρο $ RCND.$ Αποδείξτε ότι :

(a) – Το σημείο $ O\equiv KC\cap LD.$ είναι το κέντρο του κύκλου $ (O)$ και ότι $ NO\perp AB.$

(b) – Το σημείο $ N'\equiv KL\cap NO,$ είναι το αντιδιαμετρικό του $ N,$ στον κύκλο $ (O)$[/color][/size][/b].

Κώστας Βήττας.[/quote]

Πολύ καλά ερωτήματα κ. Βήττα :)

Για το πρώτο ερώτημα έχω:

$ \angle{ANO} \equal{} 90 \minus{} CDN \equal{} 90 \minus{} NAB\iff NO\perp AB$

Εστω τώρα $ NE$ το ύψος του τριγώνου $ ABN$. Τότε θα πρέπει το Ο να βρίσκεται επί της ΝΕ.
Εστω ότι η $ KC$ τέμνει την $ NE$ στο $ O'$. Τότε θα έχουμε:

$ NCO'$ όμοιο με $ ACK$ (afoy $ AK//NO$). Και επειδή $ CK \equal{} KA\Longleftrightarrow DO' \equal{} O'N$.

Αρα το $ O'$ ταυτίζεται με το $ O$ ως σημείο τομής της $ NE$ με τη μεσοκαθετο της $ DN$.

Συνεπώς $ K, C, O$ συνευθειακά. Ομοια $ L, D, O$ συνευθειακά κι άρα αποδείξαμε το πρώτο ερώτημα.

Το δεύτερο ερώτημα προς το παρόν δεν το έχω βρει αν και έχω ψάξει πολύ ώρα :oops: Προφανώς πρέπει να αποδιξουμε ότι $ N'$ ταυτίζεται με το ορθόκεντρο του τριγώνου $ ABN$. Βασικά προσπαθώ να αποδείξω ότι $ KL$ είναι η πολική το $ N$ w.r.t κύκλο $ ABCD$ ara arkei na dei3o oti to $ R$ einai to antistrofo toy $ N$ me polo to $ M$ i mia alli idea einai na apodei3o oti $ AD, BC, KL$ syntrexoun alla os ekei exo ftasei...


When the speed of a rear-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?

[quote="kabi"]Hello
Actually when we press accelerator the engine give a force in backward direction. So direction of friction is forward.
For more details,check this::
http://www.colchsfc.ac.uk/maths/cars.htm
Thank u.[/quote]
Wow, thank you very much for your article.They are so useful for me(I just think of the movement of a car as 4 wheels go forward, but the article says that just 2 wheels go forward,2 go backward)


in soal o man dar ye ketabe inglisi didam, soale assoniye,  dar nim safhe esbat kardam :)

sabet konid agar $n$ addadi taam bashad, ongah darim:

$\sum_{d|n}\frac{1}{d}=2$

(addadi taam ast ke mosaviye nesfe majmooe maghsoom alayh haye khodash bashad!)

khob ma midoonim ke  $\sum_{d|n}d=2n$ az tarafi agar  $\ d|n\Longrightarrow \frac{n}{d}|n$ pas darim $\sum_{d|n}\frac{n}{d}=2n$

dar danatije $\sum_{d|n}\frac{1}{d}=2$ game over and we are done.

rahe halle ghashangi bood amir jan,to vaghean khafani,ino midoonesti ya na?( ;) )

rahe hale nim safe ish chiye? aslan mage mishe jore dg i halesh kard


geometry
geometric transformation
rotation

[quote="Chinaboy"]I've always struggled with this type of problem.

Problems like, if you have n colors to color a octahedron/tetrahedron/cube, and you can/cannot reuse the colors, how many possible distinct coloring schemes are possible.

My main problem is with whether one is distinct with another because of rotations and all that stuff.

Any suggestions?

Thanks in adv.

(And yes, I've searched for it... although I'm not very good at searching for stuff, so any help is greatly appreciated.)[/quote]

[hide="0% Sure"] Supposing you have $n$ colours and the polyhedron have $k$ faces, we have that $n\geq k$. Then, I think the number of ways is $n*(n-1)*(n-2)*...*(n-k+1)$[/hide]

[hide="Something Important"]My last reply was if you can't reuse colours. If you can, the number of ways is $n^k$[/hide]

But is that taking into account rotations and certain schemes are not necessarily distinct by rotation?

What do I divide by to account for the rotation?


[quote="asdff"]If 17a+12b=211

list all positive integer solutions (a,b) that satisfy this equation

how about just all integer solutions (a,b-in terms of k)[/quote]

[hide]$17 \cdot 11+12 \cdot 2 = 211$. From the original equation, $17(a-11)+12(b-2)=0$. Then $17(a-11)=-12(b-2)$. Therefore, let $a=-12k+11$. Then $b=17k+2$. Then the solutions are $(-12k+11,17k+2)$ for any integer $k$.

For positive integer solutions, $-12k+11 \ge 0$, so $11 \ge 12k$ and $11/12 \ge k$. Also, $17k+2 \ge 0$, so $k \ge-2/17$. Only possible value of $k$ is $0$, which gives $(11,2)$.[/hide]

[hide]
All integer solutions wouldn't work- there would be infinite solutions.  So I'll just do the positive integer solutions.
Put everything mod 12.
Then the equation becomes
$5a\equiv7 \mod{12}$
So the lowest positive a that works is 11, which gives b=2.
Any lower, and a is negative. (e.g. -1,19)
The next lowest a that works is 23, which gives b=-15.
So the only positive integer solution is (11,2)

[/hide]

[quote="t0rajir0u"][quote="CircleSquared"]All integer solutions wouldn't work- there would be infinite solutions. [/quote]

So?  Characterize them.[/quote]

Okay, here goes.
[hide]
Based on the 3 examples in my previous posts:
Formula for $(x,y)$ such that $17x+12y=211$: $(11+12a,2+17a)$ where a is an integer.
[/hide]
But can anyone generalize to formula for $(x,y)$ such that $ax+by=c$?

[quote="usaha"][quote="kohjhsd"][quote="asdff"]If 17a+12b=211

list all positive integer solutions (a,b) that satisfy this equation

how about just all integer solutions (a,b-in terms of k)[/quote]

$17 \cdot 11+12 \cdot 2 = 211$. From the original equation, $17(a-11)+12(b-2)=0$. Then $17(a-11)=-12(b-2)$. Therefore, let $a=-12k+11$. Then $b=17k+2$. Then the solutions are $(-12k+11,17k+2)$ for any integer $k$.

For positive integer solutions, $-12k+11 \ge 0$, so $11 \ge 12k$ and $11/12 \ge k$. Also, $17k+2 \ge 0$, so $k \ge-2/17$. Only possible value of $k$ is $0$, which gives $(11,2)$.[/quote]

How did you go from 17(a-11)=-12(b-2) to a=-12k+11 and same with bs.[/quote]

I just let $17(a-11)=-12(b-2)=-12 \cdot17 k$ since $12$ and $17$ are relatively prime and $a,b$ are integers.

[quote="CircleSquared"]But can anyone generalize to formula for $(x,y)$ such that $ax+by=c$?[/quote]

if $(x_{0},y_{0})$ is a solution then $(x_{n},y_{n})=(x_{0}-nb,y_{0}+na)$ works.


The ones digit of the product of four consecutive positive integers
is 4. If this product is greater than 1000, what is the sum of the
four smallest such integers?

The 4th root of 100o is about 5.6, so the 2nd number should be around 6.

The product of 5, 6, 7, and 8 ends in a 0.
The product of 6, 7, 8, and 9 ends in a 4.

We have found the smallest possibility. The sum is 6+7+8+9=30.


number theory
relatively prime
combinatorics unsolved
combinatorics

Two runners start running along a circular track of unit length from the same starting point and int he same sense, with constant speeds $v_1$ and $v_2$ respectively, where $v_1$ and $v_2$ are two distinct relatively prime natural numbers. They continue running till they simultneously reach the starting point. Prove that

(a) at any given time $t$, at least one of the runners is at a distance not more than $\frac{[\frac{v_1 + v_2}{2}]}{v_1 + v_2}$ units from the starting point.
(b) there is a time $t$ such that both the runners are at least $\frac{[\frac{v_1 + v_2}{2}]}{v_1 + v_2}$ units away from the starting point. (All disstances are measured along the track). $[x]$ is the greatest integer function.

Note that both a) and b) can be summarized into: Let m(t) be the minimum of the 2 runners' distance to the starting point at time t. The maximum value of m(t) (0<=t<=1) is exactly \floor((v_1+v_2)/2)/(v_1+v_2).

- You can scale time and then scale distances so that WLOG, the length of the track is v_1+v_2, the speeds are still v_1 and v_2, and the runners meet at starting point again at time t=v_1+v_2. We need to show the maximum attained m(t) is exactly \floor((v_1+v_2)/2).

- The maximum value of m(t) is attained only when the two runners are equidistant from the starting point. (Otherwise you can either run time \epsilon forward or \epsilon backward and the distance to starting point of the closer runner would increase, and the other would still be further away.)

- In fact, if you label the portion of the track [0, (v_1+v_2)/2] "1st half," and [(v_1+v_2)/2,v_1+v_2] "2nd half" we get that m(t) can only be maximized when the runners are in opposite halves, because otherwise running time either forward or backward by \epsilon would increase both runners' distances from the starting point. Hence, to find maximum of m(t), it suffices to consider times when the 2 runners are in opposite halves and equidistant from the starting point.

- Consider a dummy runner 2' going with speed -v_1. (So 2' goes around in the opposite direction, starting at starting point at t=0, so her position always mirrors runner 1) By construction 2' is always equidistant to the starting point as runner 1. By the last point, we only have to consider times when 2 meets runner 2' (because exactly in these times runners 1 and 2 are in opposite halves and equidistant from starting point), and because their speeds add up to v_2+v_1, and the track has length v_1+v_2, this happens at t=0,1,2,...,v_1+v_2.

- If you consider the position of runner 1 at these times, you see that they exactly correspond to vertices of a regular (v_1+v_2)-gon inscribed in the circular track. (This is because v_1 is co-prime with v_1+v_2 so v_1*1,v_1*2,...v_1*(v_1+v_2) form a complete residue system mod v_1+v_2.) Hence, if v_1+v_2 is even the maximum distance is (v_1+v_2)/2 (because even regular polygon has diagonal points.) If v_1+v_2 is odd, the maximum distance is floor(v_1+v_2)/2. This is exactly what we needed to show.


algorithm
function
number theory
number theory theorems

Hey!

I've been interested in number theory for a long time, and with primes as my main interest.
The thing is that I found a kind of a nice algorithm, but I'm not sure whether it's very good or just average.

I know that there are functions and algorithms for finding primes, and I've heard that there are one function that generates all the primes between 0 and 100 but no ones that can generate more than that.

The thing is that I found out that it only takes around 5 very basic functions to generate the first 100 primes, 11 functions to generate the first 1000, 20 to generate the first 5000 and so on.
So this means that if I wan't to check if a number below 5000 is a prime I only have to do 20 simple calculations.

Now to my questions:
Is this a good algorithm? (If you look at number of calculations etc)
Have someone else already figured this out?

So if it has already been done I'm quite proud to have found it out myself =D, and if it isn't, well I'll post the formula here when it's done.

//Rasmus

If you mean the "basic" function I ment a function with one difference, one sum and two multiplications. Like 5x+1-2y.
(sorry for leaving out some details)

And what are the values setted into the function¿
More general: please describe exactly what you do...

Ok I'll do the best I can to explain.

Basically this is a project that I've been working on for a couple of weeks and isn't far from done yet. So the thing is that I don't want to release it in the wild if it's something never seen before until I feel that it's done and I've tested it properly etc.
Hope you understand  :oops:
So this is hardly a theorem yet, and I have no clue what it does for large n, but I havn't found anything like it so I'm hoping this is something that could optimize the time for finding primes (isn't that every man's dream?  :lol:).

So the theorem is something like this:

You can find if n is a prime (where n is a positive, natural number ofcourse) by using lets say q(n) number of functions f(n) where every function returns true if it is a prime number, and false if it isn't. q(n) < n/2 and shrinks to n/20 when n=5000 and continues. Every f(n) takes linear time.
The functions f(n) is calculated with another function, that is also linear.

So now I'm at the stage that I'm starting to think that Iäve discovered something really cool, and want to see if someone else beat me to it (Usually I found that goldbach have already found it etc).
And I also would like to find out how good/bad it is compared to other algorithms for finding primes. And I mean comparing the assymtoptic time, not memory consumtion etc.

As I suspected the algorithm is already known: http://www.ieeta.pt/~tos/software/prime_sieve.html#s


linear algebra
matrix
LaTeX
algebra
polynomial

A matrix $ B$ is said to be a square root of a matrix $ A$ if $ BB \equal{} A.$

(a) Find the square roots of
[2 2]
[2 2]

I couldn't type it using latex code; what's wrong with it?
[code]$A = \begin{bmatrix}
2&2\\2&2 \end$[/code]

(b) Find the square roots of
[5 0]
[0 9]

(c) Prove or disprove that every $ 2 \times 2$ matrix has at least one square root.

I tried to start with a general $ 2 \times 2$ matrix with entries $ a_{11} \equal{} a, a_{12} \equal{} b, a_{21} \equal{} c, a_{22} \equal{} d$ and then square it and equate the corresponding entries, but the system seems invincible.  But for (a), $ a \equal{} b \equal{} c \equal{} d \equal{} \pm 1$ by inspection.

It isn't enough to have \end : you need that to be \end{bmatrix}.

Consider a nilpotent matrix $ \begin{bmatrix}0&1\\0&0\end{bmatrix}$.

[quote="jmerry"]Consider a nilpotent matrix $ \begin{bmatrix}0 & 1 \\
0 & 0\end{bmatrix}$.[/quote]

Are you hinting that this matrix doesn't have a square root?  If so, how do we prove that?
What is the general technique for find square roots of any matrix?

A $ 2 \times 2$ nonzero nilpotent matrix has characteristic polynomial $ \lambda^2 \equal{} 0$.  But the "square root" of a nilpotent matrix must have characteristic polynomial $ \lambda^4 \equal{} 0$, and $ 2 \times 2$ matrices cannot have quartic characteristic polynomials.

On the other hand, over an algebraically closed field any diagonal matrix has a square root (take the square root of the diagonal entries); consequently, any diagonalizable matrix has a square root.

Concerning jmerry's hint: Are there any $ 2\times 2$ matrices $ A$ such that $ A^4 \equal{} 0$ but $ A^2\ne 0?$ (Key words: minimal polynomial, Cayley-Hamilton.)

The general technique for finding a square root of $ A:$ First, diagonalize $ A$ and write it as $ A \equal{} PDP^{ \minus{} 1}.$ (This step fails if $ A$ is not diagonalizable.) Then find square roots of the diagonal elements of $ D$ and write a matrix $ D^{1/2}$ whose diagonal elements are those square roots. (This step may fail in a field like $ \mathbb{R}$ in which not all elements have square roots. Even if it succeeds, the result may not be unique.) Then $ A^{1/2} \equal{} PD^{1/2}P^{ \minus{} 1}.$

[color=green](I had this mostly written when t0rajir0u posted.)[/color]


Find all positive integers $n$ such that $n^2+78n-2006$
is divisible by the product of digits of $n.$

See all the problems from Vol. 12(2006), #1-3 at[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=465391#p465391]Ukrainian journal  "In the World of Mathematics" problems contest[/url]

It's easily seen that the equation $x^2+78x-2006$ does not have a root modulo $3,4,5,7$.
So the only possible digits are $1,2$ and we can have at most one $2$.
Obviously numbers with all digits equal to $1$ work.
If we have one digit equal to $2$, then $n^2+78n-2006$ must be even so $n$ must be even, so the right most digit must be $2$ and also these kind of numbers work.
So answers are $(111\ldots 1)_{10},(111\ldots 12)_{10}$.


It's pretty straightfoward, I think...
1) Given that f, g and h are differentiable functions, and f(g(h(x))) = x, what is h'(x) in terms of f, f', g, g' and h?

2) Given the function f(x) = 2x^6 + 9x^5 + 10x^4 - 13x - 5, determine all intervals on which the graph of f is concave up, all intervals where it is concave down and find all inflection points for f.

Using the chain rule a few times on the first one gives:
f'(g(h(x)) . (g'(h(x)).h'(x)) = 1, or h'(x) = 1/(f'(g(h(x))).g'(h(x)))

For the second one we get f'' = 60x:^2:(x+2)(x+1).  This is positive when x<-2 or x>-1. It is negative when -2<x<-1, and is zero when x=0, -1 or -2. Testing on either side for these three points give these all as points of inflection,  I think.


One elevator leaves the 1st floor of a building at exactly 9 a.m.  It takes 5 seconds to travel between floors, stays 10 seconds at each floor and stops at each floor on its way to the top.  Another elevator leaves the 12th floor (top floor) at exactly 9 a.m., takes 5 seconds to travel between floors, stays 15 seconds at each floor and stops at each floor on its way to the bottom.  What is the first floor, which both elevators will be available at the same time?

[color=white]Please provide a solution.  Thanks.[/color]

[hide]My solution's a little long...I made a table of how long each elevator would stay on each floor in terms of the nth second that elapsed after 9 a.m. Elevator 1 stays on each floor (except the first) for ten seconds, then the interval for the next floor begins 5 seconds after the last second on the previous floor. The table for Elevator 1 started like this:
[u]Floors[/u]                                                               [u]Seconds after 9am[/u]
                            1     ,                                                         0
                            2      ,                                                        5-15
                            3     ,                                                        20-30   ...
Similarly, for elevator 2 except with the elevator staying 15 secs. per floor and starting on floor 12:
                           [u]Floors[/u]                                             [u]Seconds after 9am[/u]
                           12       ,                                                            0
                           11      ,                                                             5-20
                           10      ,                                                             25-40    ...
The first floor in which the seconds intervals overlap is the [b]7th[/b]. (please be right :) )[/hide]

[quote="robinhe"]Yes that is correct :lol: , but I was wondering if anyone had a better way of doing this problem (because making a table was also the way I did it).  Does anyone know how to do this problem using algebra or some other method that is short and concise?[/quote]

o.0 i completely doubt this is right
s=seconds past
f=floors past

$s=12.5f$
$s=180-12.5f$

$s=12.5f$
$-s=-180+12.5f$

$-180+25f=0$

$25f=180$

$f\approx7$

>.>

Sorry about the triple posts.  My computer suddenly stopped working so I clicked on the submit button several times.  I didn't know that it was going to submit them three times.  :oops:

Does anyone know how to delete posts?

About clicking on the submit button three times (and about elevators): Does pressing the elevator button more than once make the elevator arrive faster? Does clicking on the submit button three times make your computer work? ;) :P

About deleting posts: Yes, you can delete a post, but if and only if someone else has not already posted on top of it. Guess you'll have to leave this little task to the moderators. And deleting a post doesn't decrease your post count.

Haha. :laugh: No. My internet connection had a problem in which you could access it periodically.  While I was writing one of the posts, my internet failed again, so I tried a dumb thing and kept clicking the submit button.  Then I left.  I returned an hour later and my internet was working once again. :) So when I went back to check on my post(s), I saw this.


Let $x,$ $y$ and $z$ are positive numbers such that $x+y+z=1.$ Prove that
$\frac{1}{2-3xy}+\frac{1}{2-3yz}+\frac{1}{2-3zx}\le \frac{9}{5}$

$LHS=\frac{z}{2z-3xyz}+\frac{y}{2y-3xyz}+\frac{x}{2x-3xyz}\le\frac{z}{2z-\frac{1}{9}}+\frac{y}{2y-\frac{1}{9}}+\frac{x}{2x-\frac{1}{9}}$

with the substitution a=18x ; b=18y ; c=18z  we obtain

$9\left(\frac{a}{18}\frac{1}{a-1}+\frac{b}{18}\frac{1}{b-1}+\frac{c}{18}\frac{1}{c-1}\right)$

and now appliyng Jensen and GM-AM  we obtain

$9\left(\frac{1}{\frac{a^{2}+b^{2}+c^{2}}{18}-1}\right)\le\frac{9}{\frac{36}{6}-1}=\frac{9}{5}$

I think you may have applied Jensen's the wrong way...? $f(t) = \frac{1}{t-1}$ is convex.

i haven't more experience with Jensen, but $f(t)=\frac{1}{t-1}$ is concave in [0,1) , or not?

[quote="girino"]$\frac{z}{2z-3xyz}+\frac{y}{2y-3xyz}+\frac{x}{2x-3xyz}\le\frac{z}{2z-\frac{1}{9}}+\frac{y}{2y-\frac{1}{9}}+\frac{x}{2x-\frac{1}{9}}$

[/quote]
Why? I think, it's wrong. :wink:

yes yes,you are right. i'm just a beginner, sorry, i see the mistake   :blush:

Expanding and using  the fact that x+y+z=1 we have
$5(12-12(xy+yz+zx)+9xyz)\leq 9(8-12(xy+yz+zx)+18xyz-27x^{2}y^{2}z^{2})$

Then we use Schur that is $ab+bc+ca\leq \frac{1+9abc}{4}$
and we have to prove that $xyz\leq \frac{1}{27}$ which is true by AM-GM  :wink:

[quote="girino"]what is Schur? is there a solution without using that teorem? thaks   :wink:[/quote]
For any $a,b,c\in\mathbb{R}^{+}$ and $r\in\mathbb{R}$:
\[a^{r}(a-b)(a-c)+b^{r}(b-a)(b-c)+c^{r}(c-a)(c-b)\geqslant 0.\]

[quote="arqady"]My congratulations, silouan! :lol:[/quote]

Thank you Arqady . When I see a problem from you I try to solve it because it is beautiful and tricky. :)
Your solution is the same as mine?

[quote="silouan"]
Your solution is the same as mine?[/quote]
Yes!
And thank you. :)


[quote="Ignite168"]The volumes of two spheres are $972\pi$ and $288\pi$. Find the ratio of the area of their great circles.[/quote]
[hide]Find the radius of each sphere. You'll get that it is 9 and 6 respectively. Then it will have a 9 to 6 ratio os 3 to 2. Then square that to become 9 to 4 as the final answer.[/hide]

[hide]Express $\frac{972\pi}{288\pi}$ as a ratio. Simplify to $\frac{27}{8}$. Take the cube root, and then square. $ \left(\frac{27}{8}\right)^{\frac{2}{3}}$[/hide]


symmetry
inequalities proposed
inequalities

Let $ n$ be a positive integer and let $ a_1,a_2,\ldots,a_n$ be positive real numbers  such that:
\[ \sum^n_{i \equal{} 1} a_i \equal{} \sum^n_{i \equal{} 1} \frac {1}{a_i^2}.
\]
Prove that for every $ i \equal{} 1,2,\ldots,n$ we can find $ i$ numbers with sum at least $ i$.

Say $ k$ of the $ a_i's$ are greater than or equal to $ 1$: $ a_1, a_2, \ldots, a_k \geq 1$.  (Clearly, not all of them can be < 1).

Then for $ i \equal{} 1, 2, \ldots, k$, the assertion of the problem is obvious.  For $ i > k$, by AM-GM, $ \frac {a_1 \plus{} a_2 \plus{} \ldots \plus{} a_i}{i} \geq (a_1 \ldots a_i)^{1/i}$.  But also,

$ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_i \equal{} \frac {1}{a_1} \plus{} \ldots \plus{} \frac {1}{a_n} \minus{} a_{i \plus{} 1} \minus{} a_{i \plus{} 2} \minus{} \ldots \minus{} a_{n} \geq \frac {1}{a_1} \plus{} \ldots \plus{} \frac {1}{a_i}$, because $ \frac {1}{x} \minus{} x \equal{} \frac {1 \minus{} x^2}{x} \geq 0 \text{ for } x < 1$.  So by AM-GM, again,

$ \frac {a_1 \plus{} a_2 \plus{} \ldots a_i}{i} \geq \frac {\frac {1}{a_1} \plus{} \ldots \plus{} \frac {1}{a_i}}{i} \geq \frac {1}{(a_1 a_2 \ldots a_i)^{1/i}}$.

Since one of $ a_1 a_2 \ldots a_i$ or $ 1/(a_1 a_2 \ldots a_i)$ is bigger than or equal to $ 1$, we have the assertion.

The problem is very nice  :)

Due to the symmetry wlog assume that $ a_1\geq a_2\geq ...\geq a_k\geq 1\geq a_{k \plus{} 1}\geq...\geq a_n$

It is trivial that for $ j \equal{} 1,2,..,k$ we have that $ \sum_{i \equal{} 1}^j a_i\geq j$

Now observe that for $ j\geq k \plus{} 1$ we have that $ \sum_{i \equal{} 1}^j a_j\geq \sum_{i \equal{} 1}^j\frac {1}{a_j^2}$  (1)

By Holder we have $ (\sum_{i \equal{} 1}^j a_j)^2(\sum_{i \equal{} 1}^j\frac {1}{a_j})\geq j^3$

so from (1) we conclude that $ \sum_{i \equal{} 1}^j a_j\geq j$ and we are done . :)

[quote="Ahiles"]Sorry to everyone, I missed $ a_i^2$. Now problem is true...[/quote]

Fortunately the same proof I gave (and I think also the other proof using Cauchy-Schwartz) goes through.

[quote="jinjohn"]
Fortunately the same proof I gave (and I think also the other proof using Cauchy-Schwartz) goes through.[/quote]

You should change it to Holder  :wink:   . See above my solution . I think it is true , isn't it ?

We may focus only on the numbers $ a_i$ which are smaller than $ 1$, but here is another aproach:

It suffices to prove that $ a_1+a_2+\cdots a_n\geq n$. We shall first prove this assertion.

We have \[ \sum_{i=1}^{n} a_i +n=\sum_{i=1}^{n}(\frac{1}{a_{i}^{2}}+1)\geq 2\sum_{i=1}^{n} \frac{1}{a_{i}}.\]By adding $ 2\sum_{i=1}^{n} a_i$ in both terms, we obtain \[{ 3\sum_{i=1}^{n} a_i +n\geq 2\sum_{i=1}^{n}(a_i+\frac{1}{a_i}})\geq 4n\]
Therefore $ \sum_{i=1}^{n} a_i \geq n$. Now assume that for some $ k$, we have that \[ \displaystyle \sum_{m=1}^{k} a_{i_m}< k,\] for any $ \{i_1, i_2\dots i_k\}\subset\{1, 2\dots n\}$. Now we add all these possible sums together and we obtain \[ \ds \binom{n-1}{k-1}\cdot \sum_{i=1}^{n} a_i< k\cdot\binom{n}{k}.\]By simplifying we obtain $ \sum_{i=1}^{n} a_i<n$, which is a contradiction.

PS. This problem was proposed by me and Georgescu Flavian :P


algebra
polynomial
complex analysis
complex analysis unsolved

let $f$ be a polynomial of complex numbers,where if $z$ be on the unit circle,then $f(z)$ is so.
is it true that:$f(z)=z^{n}$

Something stronger is not more than 20 posts down on this same forum:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78814

Although there is probably a nicer solution for polynomials.

Being extremely nitpicky, the answer is of course "no" because of things like $-z^{2}$.


1)
The source of a sound wave has a power of 4.80 µW. Assume it is a point source.
    (a) What is the intensity 6.00 m away?
    (b) What is the sound level at that distance?

2)
You have 4 tuning forks that oscillate at different frequencies. What are the maximum number and the minimum number of different beat frequencies you can produce by using the forks two at a time, depending on how the frequencies differ?

[hide="my ideas"]
for the first one i use $I=\frac{P}{A}$ where P is the power $(4.8\cdot 10^{-6}W)$ and A is the area $6^{2}\pi$ but this does not seem to give the correct answer.
the second problem i have no idea how to start so have not gotten anywhere on it.[/hide]

1) First, a point source radiates a spherical wave, i.e. power is uniformly distributed over the surface of a sphere.

a) $I=\frac{P}{4\pi R^{2}}$

b) Sound level is the intensity (relative to the threshold of hearing) expressed in units of $dB.$

$\text{Sound level }= 10\log_{10}\left[\frac{I}{I_{0}}\right]\text{ dB.}$

The threshold of hearing is defined to be $I_{0}=10^{-12}\text{ W/m}^{2}$ @ $1,000\text{ Hz.}$

2) A beat frequency is frequency difference produced by the mixing of two tones together.

It comes from the fact that $\cos A\cos B = \frac12\cos\left(A-B\right)+\frac12\cos\left(A+B\right)$

If $A =2\pi f_{1}t$ and $B =2\pi f_{2}t$ then $\cos\left(A-B\right)$ represents a tone that oscillates at a "beat frequency" of $\left|f_{1}-f_{2}\right|$

Since you have four tuning forks, considering them pairwise you can produce $\binom{4}{2}$ beat frequncies.

Otherwise taking all four together means that you have:

$\cos A\cos B\cos C\cos D = \frac12\left(\cos\left(A-B\right)+\cos\left(A+B\right)\right)\cdot\frac12\left(\cos\left(C-D\right)+\cos\left(C+D\right)\right)$

Applying the FOIL method to this product and expanding the resulting cosine products results in four different beat frequencies.


Let's first convert the hard-to-manage mixed numbers to fractions.

$ 6 \frac{2}{3} \implies \frac{20}{3}$

$ 3 \frac{3}{7} \implies \frac{24}{7}$

$ 2 \frac{3}{16} \implies \frac{35}{16}$

Now, it's much easier to multiply them together. $ \frac{20}{3}\times\frac{24}{7}\times\frac{35}{16}$ comes out to be a perfectly round $ \fbox{50}$.


linear algebra
matrix
linear algebra unsolved

Hi everyone..
I have the following problems to solve :

1) Show that any invertible 2x2 matrix has a square root
2) For each n find a nxn matrix that has NO square root

Can anyone help?

idea for 2): The matrix $A$ with $1$ on the first side diagonal ($A_{i,i+1}$), rest $0$ has the property $A^{n-1} = 0, A^k \neq 0$ for $k < n-1$. It has rank $n-1$. If $A = B^2$, then the rank of $B$ must also be $n-1$. But by a problem posted in this week (http://www.mathlinks.ro/Forum/viewtopic.php?t=73922), this implies rank $A^k = n-1$ for all $k > 0$.


analytic geometry
geometry
trigonometry
vector
rotation
geometric transformation
complex numbers

Given $ O(0,0)$, $ P(4,3)$, $ Q(x,y)$, and $ R(1,0)$ such that $ PO\equal{}QO$ and $ \angle{QOP}\equal{}2\angle{POR}$.  Compute the ordered pair $ (x,y)$ if $ y>0$.

Thanks in advance!

(By the way, I don't know too much about trig/complex numbers, so if you could be detailed in those parts of your solution, I'd greatly appreciate it!)

Let $ \angle{PAO} \equal{} \theta$ such that $ \cos \theta \equal{} \frac {4}{5},\ \sin \theta \equal{} \frac {3}{5}$. Since two points $ P,\ Q$ are on the same circle $ x^2 \plus{} y^2 \equal{} 25\ (y > 0)$
we have $ x \equal{} 5\cos 3\theta ,\ y \equal{} 5\sin 3\theta$, now you need to use the formula of $ \cos \theta,\ \sin \theta$.
Please just do self study. Good luck. :)

Last edit.

I'm confused.  What do you mean by $ y\equal{}5\sin \theta$ was right?

[hide="Complex numbers (and explanation)"]
This problem is very easily solvable with complex numbers. But first, a little background to see why that is.

First of all, a complex number is a number of the form $ a\plus{}bi$ where $ i^2\equal{}\minus{}1$. We can plot these points in a complex plane that is in many ways analogous to the number line and to the Cartesian plane. In the complex plane, the $ y$ axis becomes the imaginary axis, and the $ x$ axis becomes the real axis. Then a number $ a\plus{}bi$ has real part $ a$ and imaginary part $ b$, so it becomes the point $ (a,b)$ in the complex plane.

But a point $ (a,b)$ in the complex plane can be treated as a vector in polar coordinates, with argument $ \theta$ and magnitude $ r$, where $ \theta\equal{}\arctan\frac{b}{a}$ and $ r\equal{}\sqrt{a^2\plus{}b^2}$. (Think about why that is, if you don't know polar coordinates). This is where things get interesting. Just as we can represent polar coordinates $ (r,\theta)$ in rectangular form, the point $ (r,\theta)$ in the complex plane becomes the complex number $ r\cos\theta\plus{}ri\sin\theta$. Euler's formula says that $ re^{i\theta}\equal{}r\cos\theta\plus{}ri\sin\theta$ (proven by considering series expansions).

Now, the most interesting part: If we take a point $ (a,b)$ in the complex plane (that is, the complex number $ a\plus{}bi$) and represent it as $ re^{i\theta}$, and then multiply it by, $ se^{i\gamma}$, we get $ rse^{i(\theta\plus{}\gamma)}$. Therefore, multiplying complex numbers adds their arguments and multiplies their magnitudes.

That means that if we take $ a\plus{}bi$ and multiply it by $ e^{i\theta}$, we've rotated it around the origin by $ \theta$.

Which brings us to our solution:

Project the diagram onto the complex plane. Let the affixes (complex number representations) of $ O,Q,R,P$ be $ o,q,r,p$. Then $ o\equal{}0$, $ q\equal{}x\plus{}yi$, $ r\equal{}1$, and $ p\equal{}4\plus{}3i$.

From the given information, $ |q|\equal{}|p|\equal{}\sqrt{4^2\plus{}3^2}\equal{}5$, and $ \arg q\equal{}2\arg p$. All we need to do is take $ p$ and rotate it by its own argument to end up at $ q$.

That is, letting $ \theta\equal{}\arg p$, we have $ p\equal{}5e^{i\theta}$. Then we multiply by $ e^{i\theta}$ to apply a rotation, and get $ q\equal{}5e^{2i\theta}\equal{}5\cos2\theta\plus{}5i\sin2\theta\equal{}5(2\cos^2\theta\minus{}1)\plus{}5i(2\sin\theta\cos\theta)$.

Drawing a right triangle gives $ \cos\theta\equal{}\frac{4}{5}$ and $ \sin\theta\equal{}\frac{3}{5}$.

Then $ q\equal{}5\left(\frac{32\minus{}25}{25}\right)\plus{}5i\left(\frac{2\cdot 3\cdot 4}{25}\right)\equal{}\frac{7}{5}\plus{}\frac{24}{5}i$.

Then $ Q\equal{}\boxed{\left(\frac{7}{5},\frac{24}{5}\right)}$.
[/hide]

You might also look into this problem I saw just the other day, which can be solved with complex numbers and rotations: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268996

[quote="kunny"][quote="Octahedr0n"][quote="kunny"]we have $ x \equal{} 5\cos 3\theta ,\ y \equal{} 5\sin \theta$[/quote]

How did you get that?  Thanks![/quote]

Sorry, $ y \equal{} 5\sin \theta$ was right. :oops:[/quote]

That was $ y \equal{} 5\sin 3\theta$. Sorry, I confused you.

kunny

[quote="grn_trtle"][hide="Complex numbers (and explanation)"]
This problem is very easily solvable with complex numbers. But first, a little background to see why that is.

First of all, a complex number is a number of the form $ a \plus{} bi$ where $ i^2 \equal{} \minus{} 1$. We can plot these points in a complex plane that is in many ways analogous to the number line and to the Cartesian plane. In the complex plane, the $ y$ axis becomes the imaginary axis, and the $ x$ axis becomes the real axis. Then a number $ a \plus{} bi$ has real part $ a$ and imaginary part $ b$, so it becomes the point $ (a,b)$ in the complex plane.

But a point $ (a,b)$ in the complex plane can be treated as a vector in polar coordinates, with argument $ \theta$ and magnitude $ r$, where $ \theta \equal{} \arctan\frac {b}{a}$ and $ r \equal{} \sqrt {a^2 \plus{} b^2}$. (Think about why that is, if you don't know polar coordinates). This is where things get interesting. Just as we can represent polar coordinates $ (r,\theta)$ in rectangular form, the point $ (r,\theta)$ in the complex plane becomes the complex number $ r\cos\theta \plus{} ri\sin\theta$. Euler's formula says that $ re^{i\theta} \equal{} r\cos\theta \plus{} ri\sin\theta$ (proven by considering series expansions).

Now, the most interesting part: If we take a point $ (a,b)$ in the complex plane (that is, the complex number $ a \plus{} bi$) and represent it as $ re^{i\theta}$, and then multiply it by, $ se^{i\gamma}$, we get $ rse^{i(\theta \plus{} \gamma)}$. Therefore, multiplying complex numbers adds their arguments and multiplies their magnitudes.

That means that if we take $ a \plus{} bi$ and multiply it by $ e^{i\theta}$, we've rotated it around the origin by $ \theta$.

Which brings us to our solution:

Project the diagram onto the complex plane. Let the affixes (complex number representations) of $ O,Q,R,P$ be $ o,q,r,p$. Then $ o \equal{} 0$, $ q \equal{} x \plus{} yi$, $ r \equal{} 1$, and $ p \equal{} 4 \plus{} 3i$.

From the given information, $ |q| \equal{} |p| \equal{} \sqrt {4^2 \plus{} 3^2} \equal{} 5$, and $ \arg q \equal{} 2\arg p$. All we need to do is take $ p$ and rotate it by its own argument to end up at $ q$.

That is, letting $ \theta \equal{} \arg p$, we have $ p \equal{} 5e^{i\theta}$. Then we multiply by $ e^{i\theta}$ to apply a rotation, and get $ q \equal{} 5e^{2i\theta} \equal{} 5\cos2\theta \plus{} 5i\sin2\theta \equal{} 5(2\cos^2\theta \minus{} 1) \plus{} 5i(2\sin\theta\cos\theta)$.

Drawing a right triangle gives $ \cos\theta \equal{} \frac {4}{5}$ and $ \sin\theta \equal{} \frac {3}{5}$.

Then $ q \equal{} 5\left(\frac {32 \minus{} 25}{25}\right) \plus{} 5i\left(\frac {2\cdot 3\cdot 4}{25}\right) \equal{} \frac {7}{5} \plus{} \frac {24}{5}i$.

Then $ Q \equal{} \boxed{\left(\frac {7}{5},\frac {24}{5}\right)}$.
[/hide]

You might also look into this problem I saw just the other day, which can be solved with complex numbers and rotations: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268996[/quote]

Except that, according to the problem statement, you must apply rotation three times, i.e. multiply by $ e^{2i\theta}$

[quote="kunny"]Since $ \cos \theta \equal{} \frac {4}{5},\ \sin \theta \equal{} \frac {3}{5}$, we have

$ x \equal{} 5\cos \theta \equal{} 5(4\cos ^ 3 \theta \minus{} 3\cos \theta) \equal{} \minus{} \frac {44}{25}$

$ y \equal{} 5\sin \theta \equal{} 5(3\sin \theta \minus{} 4\sin ^ 3 \theta) \equal{} \frac {117}{25}$[/quote]

Is that a continuation of your own solution?

[quote="Octahedr0n"]Thanks for your help!  However, I don't get this part:

[quote="grn_trtle"]Now, the most interesting part: If we take a point $ (a,b)$ in the complex plane (that is, the complex number $ a \plus{} bi$) and represent it as $ re^{i\theta}$, and then multiply it by, $ se^{i\gamma}$, we get $ rse^{i(\theta \plus{} \gamma)}$. Therefore, multiplying complex numbers adds their arguments and multiplies their magnitudes.[/quote]

Also does:

[quote="grn_trtle"]That means that if we take $ a \plus{} bi$ and multiply it by $ e^{i\theta}$, we've rotated it around the origin by $ \theta$.[/quote]

involve DeMoivre's?  Is that how we multiply something by $ e^{i\theta}$?[/quote]

About the first part: What's not to get? Multiplying $ re^{i\alpha}$ and $ se^{i\beta}$ gives $ rse^{i(\alpha\plus{}\beta)}$ purely from laws of exponents. Perhaps you're not familiar with the terminology (which comes from notions of vectors): If $ z\equal{}re^{i\theta}$, we define $ \arg z\equal{}\theta$ (the argument of $ z$), and $ |z|\equal{}r$ (the magnitude of $ z$).

About the second part: This doesn't exactly involve DeMoivre's (which is a consequence of Euler's formula anyway). We established previously that multiplying two complex numbers adds their arguments and multiplies their magnitudes. If we're multiplying by $ e^{i\theta}$, however, recognize that $ |e^{i\theta}|\equal{}1$. Therefore, whatever we're multiplying with doesn't see a change in magnitude; all that happens is we add $ \theta$ to it's magnitude—equivalently, we rotate it by $ \theta$.


If $ x\plus{}y\equal{}4$ and $ xy \equal{} 2$, then find $ x^{6}\plus{}y^{6}$.

[hide="different solution"]
$ (x\plus{}y)^{6}\equal{}x^{6}\plus{}6x^{5}y.....\equal{}4096$ work from there
[/hide]

[quote="7h3.D3m0n.117"]If $ x\plus{}y \equal{} 4$ and $ xy \equal{} 2$, then find $ x^{6}\plus{}y^{6}$.[/quote]
[hide="alternate solution"]$ x\equal{}4\minus{}y\Rightarrow (4\minus{}y)y\equal{}2\rightarrow y^{2}\minus{}4y\plus{}2\equal{}0\rightarrow (y^{2}\minus{}4y\plus{}4)\plus{}2\minus{}4\equal{}0\rightarrow (y\minus{}2)^{2}\minus{}2\equal{}0\rightarrow\boxed{y\equal{}2\pm\sqrt{2}}$

$ x\equal{}4\minus{}(2\pm\sqrt{2})\rightarrow\boxed{x\equal{}2\mp\sqrt{2}}$

Since one is $ \mp$ and the other is $ \pm$ it doesn't matter which one we use, so long as use the opposites. For simplicity, $ x$ will be $ \minus{}$ and $ y$ will be $ \plus{}$.

$ (2\minus{}\sqrt{2})^{6}\plus{}(2\plus{}\sqrt{2})^{6}\Rightarrow$

$ 2^{6}\minus{}6\cdot2^{5}\cdot\sqrt{2}\plus{}15\cdot2^{4}\cdot\sqrt{2}^{2}\minus{}20\cdot2^{3}\cdot\sqrt{2}^{3}\plus{}15\cdot2^{2}\cdot\sqrt{2}^{4}\minus{}6\cdot2\cdot\sqrt{2}^{5}\plus{}\sqrt{2}^{6}\plus{}2^{6}\plus{}6\cdot2^{5}\cdot\sqrt{2}\plus{}15\cdot2^{4}\cdot\sqrt{2}^{2}\plus{}20\cdot2^{3}\cdot\sqrt{2}^{3}\plus{}15\cdot2^{2}\cdot\sqrt{2}^{4}\plus{}6\cdot2\cdot\sqrt{2}^{5}\plus{}\sqrt{2}^{6}\Rightarrow$

$ 64\minus{}6\cdot32\cdot\sqrt{2}\plus{}15\cdot32\minus{}20\cdot16\cdot\sqrt{2}\plus{}15\cdot16\minus{}6\cdot8\cdot\sqrt{2}\plus{}8\plus{}64\plus{}6\cdot32\cdot\sqrt{2}\plus{}15\cdot32\plus{}20\cdot16\cdot\sqrt{2}\plus{}15\cdot16\plus{}6\cdot8\cdot\sqrt{2}\plus{}8\Rightarrow\boxed{1584}$

[/hide]

[quote="abacadaea"][hide="different solution"]
$ (x\plus{}y)^{6}\equal{} x^{6}\plus{}6x^{5}y..... \equal{} 4096$ work from there
[/hide][/quote]

This approach will just lead you to the well-known [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums]Newton sums[/url].


[url=http://home.cwru.edu/~jnt5/Planarity/]Planarity[/url]

Basically, a segment is drawn connecting every dot, but enough are removed to be able to move the points in such a way that no two segments overlap. Your goal is to move the points to that position. I'm just starting level 8 now; I finished level 7 in 8:12.

I just beat level 9 in 4:42, after beating level 8 in 10:12. :what?:

geez, how many levels are there?  Pretty soon the whole screen will be cluttered with dots.

Crazy game! I need a break

[Edit] Updated level 11, level 12 too many dots!
[Edit] Removed old score

Nice job. I got stuck on level 10, but I think that's because I only spent maybe six or seven minutes on it, then I got bored. :P

And I have no idea what the level limit is. It could go on forever, or it could stop at some set point.

Thanks for sharing, never knew such a cool game existed.  I stopped after level 11, at which point I had had enough.   :P

[quote="JesusFreak197"]And I have no idea what the level limit is. It could go on forever, or it could stop at some set point.[/quote]

[quote]How many levels are there?

There is no last level, but if you get past 10 or so then consider yourself in select company.
[/quote]

The second quote is from the Planarity website

thats an awesome game!  I made it to level 9, and now i have to go. Thanks!

i beat 11, the i tried 100, and it broke my computer there were so many dots!

Well, he's smart and he spends a fair amount of time playing it. :P He has kind of developed strategies for beating the levels, but it's starting to not work as well at level 18. ;)

[quote="G-UNIT"]Cant you just put all the vertices on the exact same spot, so none of them will intersect?  The problem is, I dont think a human hand has the percision to complete such a task.[/quote]

I just tried it; it worked. :D

And if you're decent with a mouse, that shouldn't be difficult at all.  I'll see how far I can get like this.

Edit:  Never mind, it didn't work for level 2.  Hmm...

Yea, I know, it sucks.  Now, Im going to try putting lines on top of each other, since a line is parallel to itself. :D

http://img.photobucket.com/albums/v494/dumpytrashbag/Inyourfacelevel17.jpg

Level 17.

wow thats crazyy

i spent like half an hour beating lvl 8  :D


geometry
rectangle
integration
calculus
logarithms
probability
expected value

Given a rectangle with side lengths $a$ and $b$. Pick randomly two points $x,y$ on it. What is the expected value of the length of the line segment $xy$?

[hide="I think this works:"] Assume that all points are equally likely to be chosen and that $x$ and $y$ are not necessarily distinct.  Then the lengths of the segments of any given $\overline{xy}$ are uniformly distributed from the shortest possible length to the largest possible length.  Since $x$ and $y$ are not necessarily distinct, the shortest length is $0$.  The largest length will be one of the diagonals of the rectangle, $\sqrt{a^{2}+b^{2}}$.  Then $xy$ is uniformly distributed on $(0,\sqrt{a^{2}+b^{2}})$.  The expected value of a (continuous) uniform distribution is $\frac{a+b}{2}$.  In our case $\frac{\sqrt{a^{2}+b^{2}}}{2}$.  [/hide]

[quote="mlok"]Have you tried to use some CAS on $\frac{1}{(ab)^{2}}\int_{0}^{a}\int_{0}^{b}\int_{0}^{a}\int_{0}^{b}\sqrt{(x-x')^{2}+(y-y')^{2}}dy'dx'dydx$?[/quote]
My university's version of Maple can't evaluate that in ten minutes.

[quote="illcrowflu"]Then the lengths of the segments of any given $\overline{xy}$ are uniformly distributed[/quote]
Why uniformly?

I guess it wouldn't be uniform  :blush: , my mistake.

I would rather run some symbolic integration program on
\[\frac 1{(ab)^{2}}\int_{0}^{a}\int_{0}^{b}\sqrt{x^{2}+y^{2}}(a-x)(b-y)\,dx\,dy\]
In principle, you can integrate most of the arising terms by hand. The only really unpleasant integral you'll encounter is $\int y^{2}\ln (a+\sqrt{a^{2}+y^{2}})\,dy$, which, after integration by parts, transforms into $\int\frac{y^{4}}{a\sqrt{a^{2}+y^{2}}+a^{2}+y^{2}}\,dy$, which you'd better handle with Wolfram Integrator. Still, the answer is an elementary function. :)


integration
trigonometry
calculus
calculus computations

Find
1.$ \int \frac{x.sinx}{cos^3x}dx$

2. $ \int \frac{cosx}{7\plus{}cos2x}dx$

Integrating by parts:
$ \int x\sec^2 x\tan x dx\equal{}\frac{x\sec^2}{2}\minus{}\int 1*\frac{\sec^2 x}{2} dx$
$ \int x\sec^2 x\tan x dx\equal{}\frac{x\sec^2\minus{}\tan x}{2}\plus{}C$


floor function
function
algebra
algebra unsolved

Find all tuples $ (x_1,x_2,x_3,x_4,x_5)$ of positive integers with $ x_1>x_2>x_3>x_4>x_5>0$ and
$ {\left \lfloor \frac{x_1+x_2}{3} \right \rfloor }^2 + {\left \lfloor \frac{x_2+x_3}{3} \right \rfloor }^2 + {\left \lfloor \frac{x_3+x_4}{3} \right \rfloor }^2 + {\left \lfloor \frac{x_4+x_5}{3} \right \rfloor }^2 = 38.$

L.H.S.=R.H.S.=6(mod 8), all squares=0,1,4 (mod 8),
So we get $ {\left \lfloor \frac{x_1+x_2}{3} \right \rfloor } , {\left \lfloor \frac{x_2+x_3}{3} \right \rfloor }, {\left \lfloor \frac{x_3+x_4}{3} \right \rfloor } ,{\left \lfloor \frac{x_4+x_5}{3} \right \rfloor }=0,(1 \;or \;3),(1 \;or \;3),2(mod4)$
Firstly we get one of them= $ 4$ as all of them $ \ge 1$, and one of the others $ =2$ as $ 6^2>38-4^2$, then easy to check that the rest two are both $ 3$.
So we get $ 9 \le x_2+x_3 \le 11$  and  $ 9 \le x_3+x_4 \le 11$
As $ x_2+x_3+2 \le x_3+x_4$ ,  we get $ x_2=4 ,x_3=5, x_4=6$ ,Substitute them into the the rest two floor function we get the tuples are $ (2,4,5,6,7)$ , $ (2,4,5,6,8)$ , $ (3,4,5,6,7)$ , $ (3,4,5,6,8)$


In triangle ABC, a,b,c be its sides. If the measure of angles A, B and C forms a geometrical sequence, and $b^{2}-a^{2}=ac$, then find angle B. (1985 China League)

$\measuredangle A=a$;$\measuredangle B=ar$;$\measuredangle C=ar^{2}$. We have $D$ in ray $CB$ such that $BD=DC$. Then triangles $ABC$ and $DAC$ are similars.

Hence:$\frac{ar}{2}=a$  :arrow:  $r=2$

$ar^{2}+ar+a=\pi$  :arrow: $7a=\pi$  :arrow:  $\measuredangle B=\frac{2\pi}{7}$

Another solution:
Through point $C$ construct $CD \parallel AB$ to meet the circumcircle of $\triangle{ABC}$ at $D$. Connect $AD$, then $ABCD$ is an isocelces trapzoid. By Ptolemy, we have: $b^{2}=a^{2}+c \cdot CD$.
From $b^{2}-a^{2}=ac$, we get $CD=a$, thus: $AD=DC=CB$, thus $\angle{B}=2 \angle{ABC}$.

In triangle $ABC$, since the mearuse of angles A, B and C forms a geometrical sequence, so the common ratio $q$ is $2$, thus $\angle{A}+\angle{B}+\angle{C}=7 \angle{A}= \pi$, thus we get $\angle{A}=\frac{\pi}{7}$, thus $\angle{B}= \frac{ 2 \pi }{7}$.


Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
$ {{x}_{m}\cdot{\prod_{i = m + 1}^{n}}{\frac {1}{{x}_{i}}}} = {\frac {\frac {\frac {{x}_{m}}{{x}_{m + 1}}}{\frac {{x}_{m + 2}}{\vdots}}}{\frac {{x}_{n - 2}}{\frac {{x}_{n - 1}}{{x}_{n}}}}}$

Is there a better way to format the $ \text{\LaTeX}$ formula above? Right now, it looks a little odd: with the vertical dots not quite centered between the "divided by" lines and with the vertical dots not aligned with the equals signs.

[b]2. Relevant equations.[/b]
Knowledge of $ {\text{\LaTeX}}$ formatting.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
Yea, I'm not really sure what to do here. I would think there is maybe a variation of the "\frac" command that will make the above $ {\text{\LaTeX}}$ formula look better.
[/hide]
Thanks,

-APSStudent

However you format it the RHS is never going to look good or even be understandable as division is not associative: 1/(2/3) doesn't equal (1/2)/3 so what would 1/2/3 mean?
What is wrong with:
$ x_m \cdot \prod_{i \equal{} m \plus{} 1}^n \frac {1}{x_i} \equal{} x_m \cdot \frac {1}{x_{m \plus{} 1}} \cdot \frac {1}{x_{m \plus{} 2}} \cdot \; \dots \; \cdot \frac {1}{x_n}$
Notice that I have removed unnecessary {...} so the code is easier to read.

Hey,
[quote="stevem"]
However you format it the RHS is never going to look good or even be understandable as division is not associative: 1/(2/3) doesn't equal (1/2)/3 so what would 1/2/3 mean?
[/quote]
Ahh, you're right. Yea, that did not occur to me when I was writing out the equation. Thank you for pointing that out.

[quote="stevem"]
What is wrong with:
$ x_m \cdot \prod_{i \equal{} m \plus{} 1}^n \frac {1}{x_i} \equal{} x_m \cdot \frac {1}{x_{m \plus{} 1}} \cdot \frac {1}{x_{m \plus{} 2}} \cdot \; \dots \; \cdot \frac {1}{x_n}$
Notice that I have removed unnecessary {...} so the code is easier to read.
[/quote]
Nice, that looks much better. Also, which is the correct convention: "$ {...}$" or "$ {\dots}$" [b]?[/b]

Thanks,

-APSStudent

Hey,
[quote="stevem"]
\dots is the standard. When the amsmath package is loaded \dots automatically adjusts the height depending on the surrounding symbols so $ a_1 \plus{} a_2 \plus{} \dots \plus{} a_n$ but $ a_1, a_2, \dots, a_n$.
[/quote]
Ok, so in general when [u]handwriting[/u] equations should the "dots" always be placed at the level of the operation? In other words, which of the following is correct?
Using "..."?
$ {{\sum_{i\equal{}m}^{n}}{{x}_{i}}} \equal{} {{x}_{m}}\plus{}{{x}_{m\plus{}1}}\plus{}{{x}_{m\plus{}2}}\plus{}...\plus{}{{x}_{n\minus{}2}}\plus{}{{x}_{n\minus{}1}}\plus{}{{x}_{n}}$
and
$ {{\prod_{i\equal{}m}^{n}}{{x}_{i}}} \equal{} {{x}_{m}\cdot{x}_{m\plus{}1}\cdot{x}_{m\plus{}2}\cdot...\cdot{x}_{n\minus{}2}\cdot{x}_{n\minus{}1}\cdot{x}_{n}}$

or

Using "\ldots"?
$ {{\sum_{i\equal{}m}^{n}}{{x}_{i}}} \equal{} {{x}_{m}}\plus{}{{x}_{m\plus{}1}}\plus{}{{x}_{m\plus{}2}}\plus{}\ldots\plus{}{{x}_{n\minus{}2}}\plus{}{{x}_{n\minus{}1}}\plus{}{{x}_{n}}$
and
$ {{\prod_{i\equal{}m}^{n}}{{x}_{i}}} \equal{} {{x}_{m}\cdot{x}_{m\plus{}1}\cdot{x}_{m\plus{}2}\cdot\ldots\cdot{x}_{n\minus{}2}\cdot{x}_{n\minus{}1}\cdot{x}_{n}}$

or

Using "\dots"?
$ {{\sum_{i\equal{}m}^{n}}{{x}_{i}}} \equal{} {{x}_{m}}\plus{}{{x}_{m\plus{}1}}\plus{}{{x}_{m\plus{}2}}\plus{}\dots\plus{}{{x}_{n\minus{}2}}\plus{}{{x}_{n\minus{}1}}\plus{}{{x}_{n}}$
and
$ {{\prod_{i\equal{}m}^{n}}{{x}_{i}}} \equal{} {{x}_{m}\cdot{x}_{m\plus{}1}\cdot{x}_{m\plus{}2}\cdot\dots\cdot{x}_{n\minus{}2}\cdot{x}_{n\minus{}1}\cdot{x}_{n}}$

Thanks,

-APSStudent

Usually the dots should where as \dots puts it which means it is at the same height as the surrounding symbols. So high up between + but on the line between , The problem comes when you use the multiplication dot as that merges with the continuation dots. In this case it is best to put a small space either side of the dots then use cdots.
$ \prod_{i\equal{}m}^n x_i \equal{} x_m \cdot x_{m\plus{}1} \cdot x_{m\plus{}2} \cdot \, \cdots \, \cdot x_{n\minus{}2} \cdot x_{n\minus{}1} \cdot x_n$

Incidentally, your code would be easier to read if you didn't put so many {...} in them. Separate parts of the code with spaces; LaTeX ignores these but it's easier to follow and less likely to end up with unmatched {}
Compare my
[code]$\prod_{i=m}^n x_i = x_m \cdot x_{m+1} \cdot x_{m+2} \cdot \, \cdots \, \cdot x_{n-2} \cdot x_{n-1} \cdot x_n$[/code]with your [code]{{\prod_{i=m}^{n}}{{x}_{i}}} = {{x}_{m}\cdot{x}_{m+1}\cdot{x}_{m+2}\cdot\dots\cdot{x}_{n-2}\cdot{x}_{n-1}\cdot{x}_{n}} [/code]

I have to say that as people's handwriting varies so much, can sometimes be barely legible, and is not used for publication, there is much more wide variation in use than in printed versions.

Hey,
[quote="stevem"]
Usually the dots should where as \dots puts it which means it is at the same height as the surrounding symbols. So high up between + but on the line between , The problem comes when you use the multiplication dot as that merges with the continuation dots. In this case it is best to put a small space either side of the dots then use cdots.
$ \prod_{i \equal{} m}^n x_i \equal{} x_m \cdot x_{m \plus{} 1} \cdot x_{m \plus{} 2} \cdot \, \cdots \, \cdot x_{n \minus{} 2} \cdot x_{n \minus{} 1} \cdot x_n$
[/quote]
Ok, thanks for the tip. However, seeing the use of: "\cdots" and "\dots"; what is the difference between the two? Since, in the following I am unable to see a difference.

Using "\cdots",
$ {{x}_{m}\cdot{\prod_{i \equal{} m \plus{} 1}^{n}}{\frac {1}{{x}_{i}}}} \equal{} {{x}_{m}\cdot\frac {1}{{x}_{m \plus{} 1}}\cdot\frac {1}{{x}_{m \plus{} 2}}\cdot\frac {1}{{x}_{m \plus{} 3}}\cdot\cdots\cdot\frac {1}{{x}_{n \minus{} 2}}\cdot\frac {1}{{x}_{n \minus{} 1}}\cdot\frac {1}{{x}_{n}}}$
$ {{x}_{m}\minus{}\sum_{i\equal{}m\plus{}1}^{n}{x}_{i}} \equal{} {{x}_{m}\minus{}{x}_{m\plus{}1}\minus{}{x}_{m\plus{}2}\minus{}\cdots\minus{}{x}_{n\minus{}2}\minus{}{x}_{n\minus{}1}\minus{}{x}_{n}}$

and

Using "\dots",
$ {{x}_{m}\cdot{\prod_{i \equal{} m \plus{} 1}^{n}}{\frac {1}{{x}_{i}}}} \equal{} {{x}_{m}\cdot\frac {1}{{x}_{m \plus{} 1}}\cdot\frac {1}{{x}_{m \plus{} 2}}\cdot\frac {1}{{x}_{m \plus{} 3}}\cdot\dots\cdot\frac {1}{{x}_{n \minus{} 2}}\cdot\frac {1}{{x}_{n \minus{} 1}}\cdot\frac {1}{{x}_{n}}}$
$ {{x}_{m}\minus{}\sum_{i\equal{}m\plus{}1}^{n}{x}_{i}} \equal{} {{x}_{m}\minus{}{x}_{m\plus{}1}\minus{}{x}_{m\plus{}2}\minus{}\dots\minus{}{x}_{n\minus{}2}\minus{}{x}_{n\minus{}1}\minus{}{x}_{n}}$

Thanks,

-APSStudent

Hey,
[quote="stevem"]
\dots behaves like \cdots in the examples you give but compare
$ a_1, a_2, \dots, a_n$ with $ a_1, a_2, \cdots, a_n$
\dots knows what height is correct and how to properly space (see the space before the last comma in the second example).
[/quote]
Got it, thanks for clearing that up.

Thanks,

-APSStudent


integration
algebra
polynomial
Putnam
calculus
college contests

I looked for a nicer way, but since the calculations weren't too mind bendingly horrid, I decided to plump for polynomial division

[hide]\begin{align*}
\int_0^1 \frac{x^4(1-x)^4}{1+x^2}\ dx &= \int_0^1 \left ( x^6 -4x^5+5x^4-4x^2+4 - \frac{4}{1+x^2} \right ) \ dx \\
& = \left [ \frac{x^7}{7} - \frac{2x^6}{3} + x^5 - \frac{4x^3}{3} +4x \right ]_0^1 - \pi \\
& = \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 - \pi \\
&= \boxed{\frac{22}{7} - \pi}
\end{align*}[/hide]

This is an old Putnam problem? I've assigned it in my Putnam problems class without knowing its source. When I do assign it, there's a part b):

Prove that $ 0 < \frac {22}7 \minus{} \pi < \frac1{630}.$

This result has been generalized by various authors, including if I remember corectly, the late prof. Lupas.

For part b)

$ \int\limits_{0}^{1}\frac{x^{4}(1\minus{}x)^{4}}{1\plus{}x^{2}}dx<\int\limits_{0}^{1}x^{4}(1\minus{}x)^{4}dx\equal{}\beta(5,5)\equal{}\frac{1}{630}$.

I have jut founded my old post.

Here is.[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=125923#125923[/url]

Out of interest, for someone who has only just started calculus, is that integral standard (the second one)?

I had to prove it for an old STEP paper (exam taken for uni entrance in the UK) but think that it was standard.

For reference, this integral is called [b]Dalzell's Integral[/b].

sorry for asking... in very old question
but is there any beautiful method other than polynomial division :( this question is beautiful... surely there is more insight rather than just dividin...

(From OTIS Integrals) We can use polynomial division. The expression simplifies to $x^6-4x^5+5x^4-4x^2+4-\frac{4}{x^2+1}$, which has antiderivative $\frac{1}{7}x^7-\frac{2}{3}x^6+x^5-\frac{4}{3}x^3+4x-4\tan^{-1}(x)$. Subtracting, we get $\left(\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\pi\right) - (0) = \frac{22}{7}-\pi$.


Two positive integer $ a$ and $ b$ are called [i]amicable[/i] iff $ \sigma(a) \equal{} \sigma(b) \equal{} a \plus{} b$ here $ \sigma(n)$ is sum of $ n$'s positive divisor
SHow that:
a.$ n$ and $ \phi(n)$ arenot [i]amicable[/i] for all natural $ n$
b.$ b$ and $ (2^a \minus{} 1)*b$ arenot [i]amicable[/i] too for natural $ a > 1$ and $ b$
c.Find all $ a$ such that :$ a$ and $ a \plus{} 1$ are [i]amicable[/i] .

Is it so insignificant everyguys????    :blush:

Some of your questions are probably unsolved problems in number theory, hence this post is in the wrong section.  For example as of 2000, it was not known if there exist amicable pairs $ a$ and $ a\plus{}1$.

b. is obviosly, because $ \sigma(mn)>\sigma(n) \ (m>1)$ numbers $ mn$ and $ n$ are not amicables.

[quote="keira_khtn"][quote="baz"]Some of your questions are probably unsolved problems in number theory, hence this post is in the wrong section.  For example as of 2000, it was not known if there exist amicable pairs $ a$ and $ a \plus{} 1$.[/quote]
   Are you sure this fact is true???I read these in AMM 1998 without solution ,so I think it couldnot br open!!! :!:[/quote]
Part (c) was open as of 2000, according to Richard Guy's book Unsolved Problems in Number Theory, 3rd edition.

[quote="Rust"]b. is obviosly, because $ \sigma(mn) > \sigma(n) \ (m > 1)$ numbers $ mn$ and $ n$ are not amicables.[/quote]
  Oh,sorry!It's my typo.We should plus one to the second number.The correct statement follows:
$ b$ and $ (2^a\minus{}1)b\plus{}1$ are not amicable


Right $ \Delta ABC$ has legs measuring 8 cm and 15 cm. The triangle is rotated about one of its legs. What is the number of cubic centimeters in the maximum possible volume of the resulting solid? Express your answer in terms of $ \pi$.

There are two possible areas. One is when the triangle is rotated about its length, $ 8$, and the other is when the triangle is rotated about its width, $ 15$. They each produce a cone.
$ V\equal{}\frac{8^215\pi}{3}\equal{}320\pi$
$ V\equal{}\frac{15^28\pi}{3}\equal{}600\pi$
Obviously, the $ 2^{\text{nd}}$ cone has a greater volume, and the answer is $ \boxed{600\pi}$


vector
complex numbers
linear algebra
linear algebra unsolved

1)Let $ A \in M_n{(\mathbb{C})},n \geq 5$ and $ 2n$ complex numbers $ \alpha_1,\alpha_2,...,\alpha_n,\beta_1,\beta_2,...,\beta_n \in \mathbb{C}$ such that :
  $ A \equal{} \left[\begin{array}{ccccc} |\alpha_1 \minus{} \beta_1|^2 & |\alpha_1 \minus{} \beta_2|^2 & |\alpha_1 \minus{} \beta_3|^2 & ... & |\alpha_1 \minus{} \beta_n|^2 \\
|\alpha_2 \minus{} \beta_1|^2 & |\alpha_2 \minus{} \beta_2|^2 & |\alpha_2 \minus{} \beta_3|^2 & ... & |\alpha_2 \minus{} \beta_n|^2 \\
|\alpha_3 \minus{} \beta_1|^2 & |\alpha_3 \minus{} \beta_2|^2 & |\alpha_3 \minus{} \beta_3|^2 & ... & |\alpha_3 \minus{} \beta_n|^2 \\
... & ... & ... & ... & ... \\
|\alpha_n \minus{} \beta_1|^2 & |\alpha_n \minus{} \beta_2|^2 & |\alpha_n \minus{} \beta_3|^2 & ... & |\alpha_n \minus{} \beta_n|^2\end{array}\right]$
Calculate $ \det(A)$
[hide="Answer of me"]  $ \det(A) \equal{} 0$[/hide]
2)Let $ B \in M_n{(\mathbb{C})},n \in \mathbb({N^{*}})$ and $ n$ Complex numbers $ z_1,z_2,z_3,..,z_n \in \mathbb{C}$ such that :
   $ |z_1| \equal{} |z_2| \equal{} ... \equal{} |z_n| \equal{} \phi > O$,$ B \equal{} [b_{ij}]_n ,b_{ij} \equal{} |z_i \minus{} z_j|$ for all $ 1 \leq i,j \leq n$
    $ B \equal{} \left[\begin{array}{ccccc} 0 & |z_1 \minus{} z_2| & |z_1 \minus{} z_3| & ... & |z_1 \minus{} z_n| \\
|z_2 \minus{} z_1| & 0 & |z_2 \minus{} z_3| & ... & |z_2 \minus{} z_n| \\
|z_3 \minus{} z_1| & |z_3 \minus{} z_2| & 0 & ... & |z_3 \minus{} z_n| \\
... & ... & ... & ... & ... \\
|z_n \minus{} z_1| & |z_n \minus{} z_2| & |z_n \minus{} z_3| & ... & 0\end{array}\right]$
Calculate $ \det(B)$
[hide="Answer of me"]  $ \det(B) \equal{} 0$ , $ n\geq 5$[/hide]

[quote="darij grinberg"]These are very nice! But you obviously have a solution, and nobody else has asked for one, so I didn't wish to spoil it for others.

I would like to add that 1) is not really about complex numbers. Here is how it generalizes to points in $ \mathbb{R}^m$:

[color=blue][b]Problem.[/b] Let $ m$ and $ n$ be positive integers such that $ n\geq m \plus{} 3$. Let $ \alpha_1$, $ \alpha_2$, ..., $ \alpha_n$, $ \beta_1$, $ \beta_2$, ..., $ \beta_n$ be vectors in the Euclidean space $ \mathbb{R}^m$. Let

$ A \equal{} \left(\left|\left|\alpha_i \minus{} \beta_j\right|\right|^2\right)_{1\leq i\leq n,\ 1\leq j\leq n}\in\mathrm{M}_n\left(\mathbb{R}\right)$.

Prove that $ \det A \equal{} 0$.[/color]

  darij[/quote]
[hide="Answer of me"]Let $ e_1 \equal{} \left[\begin{array}{ccccc} 1 & 0 & 0 & ... & 0 \end{array}\right]$,$ e_2 \equal{} \left[\begin{array}{ccccc} 0 & 1 & 0 & ... & 0 \end{array}\right]$,...,$ e_m \equal{} \left[\begin{array}{ccccc} 0 & 0 & 0 & ... & 1 \end{array}\right],$ is basis of $ R^{m}$
 $ \forall (k,j) \in \{1,2,...,n\}^2$
we have : $ \alpha_k \equal{} \sum_{i \equal{} 1}^{m}a_{ik}e_i,\beta_j \equal{} \sum_{i \equal{} 1}^{m}{b_{ij}e_i}$
   $ < \alpha_k,\beta_j > \equal{} \sum_{i \equal{} 1}^{m}{a_{ik}b_{ij}}$
 $ \forall p \in \{1,2,..,m\}$ put $ \delta_p \equal{} \left[\begin{array}{c}a_{p1} \\
a_{p2} \\
... \\
a_{pn}\end{array}\right]$
and $ \forall j \in \{1,2,..,n\}$ put $ C_j \equal{} \left[\begin{array}{c} \parallel{}\alpha_1 \minus{} \beta_j\parallel{}^2 \\
\parallel{}\alpha_2 \minus{} \beta_j\parallel{}^2 \\
... \\
\parallel{}\alpha_n \minus{} \beta_j\parallel{}^2\end{array}\right]$
  $ \alpha \equal{} \left[\begin{array}{c}\parallel{}\alpha_1\parallel{}^2 \\
\parallel{}\alpha_2\parallel{}^2 \\
... \\
\parallel{}\alpha_n\parallel{}^2\end{array}\right]$
  $ \beta \equal{} \left[\begin{array}{c}1 \\
1 \\
... \\
1\end{array}\right]$
We have $ \parallel{}\alpha_k \minus{} \beta_j\parallel{}^2 \equal{} \parallel{}\alpha_k\parallel{}^2 \minus{} 2 < \alpha_k,\beta_j > \plus{} \parallel{}\beta_j\parallel{}^2$
 Then $ C_j \equal{} \alpha \minus{}2 \sum_{i \equal{} 1}^{m}{b_{ij}\delta_i} \plus{} \parallel{}\beta_j\parallel{}^2\beta$
  $ Vect(C_1,C_2,..,C_n) \subset Vect(\alpha,\delta_1,\delta_2,..,\delta_m,\beta)$
And from here :
 $ rank(A) \equal{} \dim(Vect(C_1,C_2,..,C_n)) \leq \dim(Vect(\alpha,\delta_1,\delta_2,..,\delta_m,\beta)) \leq m \plus{} 2 < n$
(by $ n \geq m \plus{} 3$)
Therefore $ \det(A) \equal{} O$[/hide] Sorry for English of me   :P

This is pretty much the same as my solution, but somewhat shorter. Very nice!

  darij


trying to solve for r.

I get

1/r = 2*pi/C

but in my math book its

r = C/2*pi

What gives?  Im dividing r from the right hand side to get rid of it, then doing the same on the other side.  THen im getting rid of C by dividing by 1/C...help!

[quote="hunter34"]I don't think so, in your problem you solved for $\frac{1}{r}$ a.k.a the reciprocal of r. What superwizard188 is saying is that you take everything in your problem and set to the -1 power thus reciprocating it. But, I don't see how you got $\frac{1}{r}=\frac{2\pi}{C}$.
How did you get this?
The only way I see it is you just divide $C$ by $2\pi$ to get what $r$ equals.[/quote]

messycan, did you divide by r first and then by C?  Also, messycan's answer is a correct for of the equation... just cross multiply!

[quote="messycan"][quote="superwizard188"]just divide both sides by 2pi.

And yours is right; just take the reciprocal.[/quote]

so taking the recripical of both sides will equal the same?  such as:

1/2 = 3/4

will be the same as

2/1 = 4/3?[/quote]

The problem with that is $\frac{1}{2}\ne \frac{3}{4}$, but yeah. Look: $\frac{1}{2}= \frac{2}{4}\text{ and }\frac{2}{1}= \frac{4}{2}$

[quote="7h3.D3m0n.117"][quote="messycan"][quote="superwizard188"]just divide both sides by 2pi.

And yours is right; just take the reciprocal.[/quote]

so taking the recripical of both sides will equal the same?  such as:

1/2 = 3/4

will be the same as

2/1 = 4/3?[/quote]

Well think about $\frac{1}{100}$ and $\frac{100}{1}$. They are reciprocals of each other, but one is $0.01$ while the other is $100$. Completely different. You can do recipocals when dividing by complex fractions or in the presence of an equal sign. They don't equal their counterparts; reciprocals are just helpful in their peculiar ways.[/quote]

What he's asking, I think is if taking the reciprocal of both sides of an equation still makes it equal. The answer to that, obviously, is yes.

[hide]C=2*pi*r
Divide both sides by 2*pi
r=C/(2*pi)[/hide]


calculus
integration
limit
logarithms
real analysis
real analysis unsolved

how to prove:
$ \lim_{n\to{\infty}}\int_{\minus{}\infty}^{\plus{}\infty} e^{n\log(\frac{n\plus{}t}{n})\minus{}\frac{t(2n\minus{}t)}{2n}\minus{}t^2} dx\equal{}\sqrt{\pi}$

We have
\[ n\log\left(\frac {n \plus{} t}{n}\right) \equal{} n\log\left(1 \plus{} \frac {t}{n}\right) \equal{} n\left(\frac {t}{n} \plus{} O\left(\frac {t^2}{n^2}\right)\right) \equal{} t \plus{} O\left(\frac {t^2}{n}\right)
\]
and
\[ \frac {t(2n \minus{} t)}{2n} \equal{} t \plus{} O\left(\frac {t^2}{n}\right).
\]
So
\[ \int_{ \minus{} \infty}^{ \plus{} \infty} e^{n\log(\frac {n \plus{} t}{n}) \minus{} \frac {t(2n \minus{} t)}{2n} \minus{} t^2} dx \equal{} \int_{ \minus{} \infty}^{ \plus{} \infty} e^{ \minus{} t^2 \plus{} O\left(\frac {t^2}{n}\right)} dt.
\]
It is easy to show uniform conergence, so as $ n\to\infty$,
\[ \int_{ \minus{} \infty}^{ \plus{} \infty} e^{n\log(\frac {n \plus{} t}{n}) \minus{} \frac {t(2n \minus{} t)}{2n} \minus{} t^2} dt\to\int_{ \minus{} \infty}^{ \plus{} \infty} e^{ \minus{} t^2} dt \equal{} \sqrt\pi.
\]

EDIT: fixed dumbness

$ \int_{\minus{} \infty}^{\plus{} \infty}e^{\minus{}t^2}\,dx \neq \sqrt{\pi}$


function
induction
algebra unsolved
algebra

Find all functions from R to R that are continuous and
$f(f(x))+4x=4f(x)$

same idea http://www.mathlinks.ro/Forum/viewtopic.php?t=138856 or http://www.mathlinks.ro/Forum/viewtopic.php?t=128763

Almost.
To prove, that $f$ is bijective and increasing, is the same as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=128763]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=138856]here[/url].

But then $f^{(n)}(x)=(2x+n(f(x)-2x))2^{n-1}$ for all $n\in\mathbb{Z},x\in\mathbb{R}$ by induction,
and then for $x\leq y$ we have $f^{(n)}(x)\leq f^{(n)}(y)$ for all $n\in\mathbb{Z}$,
which gives $n[(f(x)-2x)-(f(y)-2y)]\leq 2y-2x$ for all $n\in\mathbb{Z}$,
and therefore $f(x)-2x=f(y)-2y$ for all $x,y$.

This gives $f(x)=2x+c$ for some $c\in\mathbb{R}$, and then
$8x+3c=f(f(x))+4x=4f(x)=8x+4c$ and $c=0$.
So $f(x)=2x$ is the only continuous solution.


EDIT [hide="HINT"]
The sum of a row of Pascal's Triangle
$2^{n}$ where $n$ is the row number.
For this problem $2^{2007}-\binom{2007}{2006}-\binom{2007}{2007}= 2^{2007}-2008$
That is a start.
[/hide]

[quote="AstroPhys"][hide]
The sum of a row of Pascal's Triangle
$2^{n}$ where $n$ is the row number.
For this problem $2^{2007}-\binom{2007}{2006}-\binom{2007}{2007}= 2^{2007}-2008$
[/hide][/quote]

Did you read the problem?

:|
I did not notice that it was every third integer.
Sorry.

Doesn't belong in HSB but this is well known and the nicest solution uses third roots of unity :wink:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1856077180&t=119683]Here[/url] is an old topic that describes a method which can be used to solve this problem. It "filters out" the terms that you don't want.


geometry
circumcircle
trigonometry
3D geometry
pyramid
algebra
polynomial

ABC is a triangle with A,B,C<=90°
H is the intersection of its heights.
and : [img]http://www.gnux.be/latex/data/fdb8a78f84c94665f247293ba4e649b8.png[/img]    [img]http://www.gnux.be/latex/data/5e852a08b2077c781b1621c53df824e6.png[/img]
Find the ray of the circle circumscribed of ABC.

PS:if u have a problem understanding this tell me abt it cuz I have some problems in translating math words
And thanks

In $\triangle ABC$ with orthocenter $H,$

$AH \cdot BH \cdot CH = 3,\quad AH^{2} + BH^{2} + CH^{2} = 7.$

Find radius of the circumcircle $R,$ I trust. All you need is $AH = 2R \cdot \cos A,$ etc. and

$\cos^{2}A + \cos^{2}B + \cos^{2}C = 1 - 2 \cdot  \cos A \cdot  \cos B  \cdot \cos C$ (look in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=66431]Paper pyramid[/url]).

$\sum_{cyc} \cos^{2}A = \frac{AH^{2} + BH^{2} + CH^{2}}{4R^2} = \frac{7}{4R^2},$

$1 - 2 \cdot \prod_{cyc} \cos A = 1 - 2 \cdot \frac{AH \cdot BH \cdot CH}{8R^3} = 1 - \frac{3}{4R^3}.$

Since the left sides are equal, so are the right sides.This leads to cubic equation

$\frac{7}{4R^2} = 1 - \frac{3}{4R^3}$  $\implies$ $\ 4R^3 - 7R - 3 = 0.$

By the rational root theorem, possible rational roots are $\pm 1,\ \pm \tfrac{1}{2},\ \pm \tfrac{1}{4},\ \pm 3,\ \pm \tfrac{3}{2},\ \pm \tfrac{3}{4}.$ It is sufficient to find only one rational root by trial and error, factor it out by long division (if it exists) , then solve the remaining quadratic equation. Rational root $-1$ is obvious and factoring it out yields

$(4R^3 - 7R - 3) : (R + 1) = 4R^2 - 4R - 3.$

The remaining quadratic equation is then

$ 4R^2 - 4R - 3 = 0$

with roots $-\tfrac{1}{2}$ and $\tfrac{3}{2}.$ Radius of the circumcircle is the only positive root, so that $R = \tfrac{3}{2}.$


How many pairs $ (x,y)$ of integers satisfy:

$ (1)$ $ 2x \ge 3y$

$ (2)$ $ 3x \ge 4y$

$ (3)$ $ 5x\minus{}7y \le 20$

[hide]
The intersections of the boundaries of the regions are $ (0,0),(60,40),( \minus{} 80, \minus{} 60)$, which encloses a triangle of area $ \pm \frac 12 \begin{vmatrix} 0 & 60 & \minus{} 80 \\
0 & 40 & \minus{} 60 \\
1 & 1 & 1 \end{vmatrix} \equal{} 200$. The lattice points lie in or on the boundaries of this triangle, and counting, we see that there are $ 19$ lattice points on each of the boundaries, with $ 60$ totally on the sides. By Pick’s Thereom, the number of lattice points inside the triangle is $ 200 \minus{} \frac {60}2 \plus{} 1 \equal{} 171$, and the answer is $ 171 \plus{} 60 \equal{} 231$.
[/hide]


When is the next problem set going to be released?  I like warning, and I'm getting bored with the other math I have to do.

It would be nice if next year there was a fixed schedule for problem releases.


I'd like to apply for PROMYS this year... The ten problems are quite challenging. I'm still working on them.

Just wondering, how many problems do I have to solve in order to qualify myself for PROMYS??

I wish to hear from those experienced participants:) How many problems had you usually worked out?

I had worked out almost all of them, but I realize now that it's not just about how many you work out.  They want to see how you approach problems and think, and whether PROMYS would work for you.  Based on the nature of the program, I'd say it's good not just to write solutions to what you've solved, but also make observations and try to generalize.  On the problems you can't solve, write lots of observations and say what you can.

So it's not just about complete solutions.  Get as many complete solutions as you can of course, but it's important to write what you can on the others too.

Does anyone know when we find out if we get in or not?  I turned in my application like 3 weeks ago, so I dunno if my school messed up, or something else happened.

I just got word a couple days ago, so you should be finding out soon.

Would an application for Financial Aid, or the ARML scholarship delay anything?

Should I be worried if I didn't get any new information about stuff like who's gonna pick me up at the airport at Boston or anything else about the camp (i.e. maps) other than the general stuff saying what to bring, housing, etc. that was supposed to be turned in on June 4th? It's getting [i]pretty[/i] close to July...

You should probably email them. Don't worry, I don't think it's a sign that they changed their mind about accepting you  :wink:


function
logarithms
real analysis
real analysis unsolved

Prove that all the zeros of the entire function $f(s)=1+2^s+3^s+4^s$ are simple.

Suppose we have a multiple root. Then $1^s+2^s+3^s+4^s=2^s\log 2+3^s\log 3+ 4^s\log 4=0$. Let $u=2^s$ and $v=3^s$. We get the system $u+u^2+v+1=0$, $u+2u^2+\alpha v=0$ where $\alpha=\frac{\log 3}{\log 2}\in (1,2)$. Hence $(2-\alpha)u^2-(\alpha-1)u-\alpha=0$. This equation has only real roots, so both $u$ and $v$ must be real. It means that both $\log 2\cdot\text{Im}\,s$ and $\log 3\cdot\text{Im}\,s$ are integer multiples of $\pi$. Since $\frac{\log 3}{\log 2}$ is irrational, this may happen only if $\text{Im}\,s=0$. But the original function has no real roots and we are done.


[color=darkblue]Let $ABC$ be an isosceles triangle for which $BC=1$ and $AB=AC=\sqrt{x^{2}-x+1}$, where $x\in [0,1]$. Prove that for any points $M\in (AB)$ and $N\in (AC)$ with the segments $[MN]$, $[BN]$, $[CM]$ we can construct a triangle.[/color]

Isn't interestingly this problem ? Even nobody ?


Let $ \Gamma_1$ and $ \Gamma_2$ be two circles with centers $ A$, $ B$ and radius $ a$, $ b$ respectively. $ MN$ is a chord of the circle $ \Gamma_1$ tangent to the circle $ \Gamma_2$ at the point $ P$. Prove that $ \overline{BM}\cdot\overline{BN}\equal{}2ab$ for any position of $ P$ in the small arc $ QR$ on $ \Gamma_2$.

Only now i see how this problem is easy. Sorry for posting in wrong forum.  :blush:

i wasn't able to prove for any point.....i could prove only for A,P,B collinear

When $ M \equal{} Q$ or $ N \equal{} R$ it's straightforward too. Good problem for Geogebra anyway.

Basically I did not notice a solution for this nice and extremely easy problem: take $BC$ a diameter of circle $\Gamma_1$, see that $\Delta BNP\sim\Delta BCM\implies \frac{BN}{BC}=\frac{BP}{BM}\iff BM\cdot BN=BP\cdot BC$, done.

Best regards,
sunken rock


What are the dimensions of a rectangle of a fixed perimeter P that result in the largest area?

NOTE: What exactly is a fixed perimeter?

it means it cant change

[hide]
it has to be a square[/hide]

[quote="Interval"]Your content did not reveal.[/quote]

it did.

SM4RT says it has to be a square

So, the answer is

"it has to be a square"?

[hide]
k so let $x$ an $y$ be the dimensions of the rectangle

$2x+2y=p\implies x=\frac{p}{2}-y$
$A=xy=\left(\frac{p}{2}-y\right)y=-y^{2}+\frac{py}{2}=-\left(y^{2}-\frac{py}{2}+\frac{p^{2}}{4}\right)+\frac{p^{2}}{4}$
$A=-\left(y-\frac{p}{4}\right)^{2}+\frac{p^{2}}{4}$

Therefore the maximum area $\frac{p^{2}}{4}$ occurs when $y=\frac{p}{4}$ then $x=\frac{p}{2}-\frac{p}{4}=\frac{p}{4}$

So you can see $x=y$ therefore its a square.
[/hide]

A fixed perimeter means that the perimeter will remain the same value even if the are changes.


If I can consistently get approximately the first 20 right on the AMC 12 then how many do you think I can get right on the AIME?

try a practice aime. amc 12 score is a horrible indicator of aime ability, although there is a positive correlation. some people can only get 15 right but can get 11 on the aime, while some can get a perfect but only get a 6 or a 7.

In AMC10 and AMC12 you have answer choices which can save you a lot of work and the like. In the AIME, it's a pure test of problem solving skill.

In the AMC exams, you can easily eliminate some of the answers.  While the AIME is technically a multiple choice test, with choices from $ 000\minus{}999$, it tests your problem solving abilities a lot more, in my opinion.

You would have to do a couple AIME tests yourself to see where you stand.


probability
arithmetic sequence
combinatorics unsolved
combinatorics

Five distinct numbers are drawn successively and at random from the set {1,2,...,n}.
Show that the probability of a draw which is such that the first three numbers drawn as well as all five can be arranged to form an arithmetic progression is greater than 6/(n-2)^3 !

I can't understand this problem, Orlando. Please, write it more clear.

I slightly try to reformulate the question:
Five distinct numbers are drawn successively and at random from the set {1,2,...,n}. At first you select three numbers. These three numbers have the property to form an arithmetic progression (AR). Then you select two further numbers. Now you have selected five numbers which have to satisfy the property of an AR again.
Show that the probability of this draw is greater than 6/(n-2)^3 !

Well, that's quite easy.
If we choose $a_1,a_2,a_3,a_4,a_5$ be arithmetical progression then the probability that $a_1,a_2,a_3$ also from arithmetical progression is, as easily seen $\frac 25$.
Now if we denote $a_n$ be the number of arithmetic progressions of $5$ members from $1,\cdots,n$ we must prove
$\frac 25 \cdot \frac{a_n}{\frac{n(n-1)(n-2)(n-3)(n-4)}{120}}\geq \frac 6{(n-2)^3}$ which leads to $a_n\geq \frac{n(n-1)(n-3)(n-4)}{8(n-2)^2}$
or it suffices to prove $a_n\geq \frac{(n-2)^2-1}{8}$.
However, counting by the smalest member of the progression we get $a_n=\sum [\frac{n-i}4]$ and using the easy identity
$[\frac{n}4]+[\frac{n+1}4]+[\frac{n+2}4]+[\frac{n+3}4]=n$ (remember Hermite) we deduce
$a_{n+4}-a_n=n$ and from now it's easy induction.

[i]Just edition:[/i]
Well, that's quite easy.
If we choose $a_1,a_2,a_3,a_4,a_5$ be arithmetical progression then the probability that $a_1,a_2,a_3$ also from arithmetical progression is, as easily seen $\frac 25$.
Now if we denote $a_n$ be the number of arithmetic progressions of $5$ members from $1,\cdots,n$ we must prove
$\frac 25 \cdot \frac{a_n}{\frac{n(n-1)(n-2)(n-3)(n-4)}{120}}\geq \frac 6{(n-2)^3}$ which leads to $a_n\geq \frac{n(n-1)(n-3)(n-4)}{8(n-2)^2}$
or it suffices to prove $a_n\geq \frac{(n-2)^2-1}{8}$.
However, counting by the smalest member of the progression we get $a_n=\sum \left[\frac{n-i}4\right]$ and using the easy identity
$[\frac{n}4]+[\frac{n+1}4]+[\frac{n+2}4]+[\frac{n+3}4]=n$ (remember Hermite) we deduce
$a_{n+4}-a_n=n$ and from now it's easy induction.

[quote="pbornsztein"]This one comes from ISL 1987.
I have to official solution so I won't look at it further....

Pierre.[/quote]

I have the official solution as Pierre and Myth does. But they seem to get lost in their huge file archives...  :P I could write it down if it is of interest for anybody ?  ;)

[quote="Myth"][quote="orl"]I have the official solution as Pierre and Myth does.[/quote]
I very rarely see official solutions.[/quote]

Once I send you some .pdf-files of the ISL/ILL project. Checking your files, allows you to see official solutions more often...  :P


inequalities
function
inequalities unsolved

[quote="Beginner"]If $a$,$b$ $c$ and $d$ are positive real numbers with $a+b+c+c=1$. Prove that \[ \frac{10-\frac{1}{a}}{1-a} + \frac{10-\frac{1}{b}}{1-b}+ \frac{10-\frac{1}{c}}{1-c}+ \frac{10-\frac{1}{d}}{1-d} \leq 32  \][/quote]

Let's transform this expression:
$= 9(\frac{1}{1-a} + \frac{1}{1-b}+ \frac{1}{1-c}+ \frac{1}{1-d})+(\frac{\frac{a-1}{a}}{1-a}+\frac{\frac{b-1}{b}}{1-b}+\frac{\frac{c-1}{c}}{1-c}+\frac{\frac{d-1}{d}}{1-d})=$
$9(\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{a+c+d}+\frac{a+b+c+d}{a+b+d}+\frac{a+b+c+d}{a+b+c}) -$
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=$

$32+9(\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}) -(\frac{b+c+d}{a}+\frac{a+c+d}{b}+\frac{a+b+d}{c}+\frac{a+b+c}{d}) \leq 32 \Leftrightarrow$

$9(\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}) \leq$
$(\frac{b+c+d}{a}+\frac{a+c+d}{b}+\frac{a+b+d}{c}+\frac{a+b+c}{d})$

And we get the proof of this inequality summarizing this ones:

$9\frac{d}{a+b+c} \leq \frac{d}{a}+\frac{d}{b}+\frac{d}{c}$
$9\frac{c}{a+b+d} \leq \frac{c}{a}+\frac{c}{b}+\frac{c}{d}$
$9\frac{b}{a+c+d} \leq \frac{b}{a}+\frac{b}{c}+\frac{b}{d}$
$9\frac{a}{b+c+d} \leq \frac{a}{b}+\frac{a}{c}+\frac{a}{d}$

That's all
V:)

[quote="Beginner"]If $a$,$b$ $c$ and $d$ are positive real numbers with $a+b+c+c=1$. Prove that \[ \frac{10-\frac{1}{a}}{1-a} + \frac{10-\frac{1}{b}}{1-b}+ \frac{10-\frac{1}{c}}{1-c}+ \frac{10-\frac{1}{d}}{1-d} \leq 32  \][/quote]
Also it is made with help Jensen's inequality.

[quote="Beginner"]now I'm curious, what is the function you use here?[/quote]
I have in mind the function $f(x)=\frac{10-\frac{1}{a}}{1-a}$

[quote="Beginner"]Are you sure that function  is always concave.
I've though that way too, but there exist subinterval in $[0,1]$ so that $f''(x)$ change sign , or we say that $f''(x)$ has solution on $[0,1]$.

Did you use case consider?

well maybe i'm wrong cause $f''(x)$ is too ugly so that i'm not sure  i differentiate it correctly.[/quote]
Yes, you are right. I wrongly thought, that if $a_1+...+a_n$ is a constant and $f(a_1)+...+f(a_n)\leq nf(\frac{a_1+...+a_n}{n})$, then always $f$ is concave.