Our coach Mrs. Ohlerich is trying to get together this year's upstate NY ARML team.  Still no funding: the estimated cost is 225 dollars per person.

I've copied and pasted the letter sent, and am attaching the application form.  For more information e-mail mathlady14@aol.com.  I can also post the info on fundraising and t-shirt design here if there's any demand.

UPSTATE NEW YORK MATH TEAM
American Regions Math League
4834 East Lake Rd.
Rushville, NY 14544
                                                                             		 Sandra Ohlerich
                                                    Executive Secretary
CONGRATULATIONS !!!!!                                                                                Tel 585-554-5207

We have recently received the list of scores on the 2005 AMC 10 and AMC 12 exams and you are to be congratulated for a fine performance. You have demonstrated that you have an excellent grasp of pre-calculus mathematics and that you are a good problem solver.

Because of your high achievement on the AMC, we are inviting you to apply for the 2005 Upstate New York Math Team that will compete in the American Regions Math League (ARML) on Saturday, June 4th at Penn State University.   We are planning to enter two teams (A and B) plus 5 alternates; this means 35 students will be participating. Since the number of invitations to apply is limited, your chances of being chosen are excellent although not guaranteed.

Enclosed please find the following:
    1.   An agenda of the activities of the ARML weekend.
2.	An application form
3.	Information about our T-shirt competition.
4.	How to make alternate arrangements if you have already registered to take the SAT on June 4th.
One of the reasons for our recent success is the increased participation of highly qualified math students from a broad geographic area all across Upstate New York.

We know that you would find this experience rewarding, challenging and fun. You would have the opportunity to meet and work with the best high school math students in Upstate New York. We would very much like to have you apply for the 2005 Upstate New York Math Team. We need enthusiastic math students who can solve tough math problems. Even more importantly, you would benefit academically from this experience.

From the enclosed Schedule of Events you can see that the team will have 0ne day of practice at Penn State before competing at the ARML competition on Saturday. We did one day at Penn State last year and everyone seemed happy with it.  If you are selected to the team it will be necessary to miss school on Thursday (6/2) and Friday (6/3) so that you can  meet our bus at one of our stops on Thursday morning or afternoon.  The weekend of June 2 – 4 is a busy one for many, with senior balls, music competitions and family activities. Please talk with your parents, check your calendar, and if the date is clear, send in your application by April 28.

We are excited about this year's team, but in order to compete at a high level we need to have many outstanding math students who are motivated to solve very challenging problems, and who want to work with other fine math students. We look forward to receiving your application to be part of the 2005 Upstate New York Math Team.

Younger students should not hesitate to apply. Over the years our most outstanding performers have started as 8th and  9th graders.  In order to select our team we use AMC, NYSML and local math league scores.  If you have been on the team in the past, you will get credit for your past participation.

Unfortunately we have not been able to secure funding as we have had in the past.  Therefore the cost per student will be  225 dollars, if we can get 35 people to come.  The fewer the number the more it would cost per student. This includes bus transportation (our biggest cost, over $3000), two nights room at  Penn State, two breakfasts and two lunches.  You will need to have money to pay for your lunch and dinner on Thursday, dinner on Friday and dinner on Saturday.  I have included on the schedule the same bus pickups as we used last year.  Part of the bus cost is determined on where the pick ups are and how many there are.  This seemed to work out pretty well last year so we will try it again.

Each student will need to provide his/her own financing. However, in the past about half the students have received financing from the home school which has been very willing to underwrite the cost, as it is an honor for both the school and the student to be selected. Please pursue this through your math team coach, your math supervisor or your principal. Regardless, we think that you would find the experience well worth the cost, even if you have to finance it yourself.

Applications must be received by April 28.  All applicants will be notified in mid May as to whether they have made the team. In that mailing details about transportation and fee payments will be included. Because you are being invited to apply, your chance of being selected is almost assured. Please understand that by sending in your application you are making a commitment to preparing for and participating in the two day practice session in Corning and the 2005 ARML competition as a member of the Upstate New York Math Team

If you are selected, you will be competing as a member of our team against over 1000 of the best high school math students in the country. By participating you will gain both skill and confidence in your problem solving skills, and you will be making new friends who are also interested in math. It is also a fun weekend. The Upstate New York Math Team has been entering a team in the ARML competition since 1982. During those years we have been privileged to have on our team some of the most outstanding high school math students from across Upstate New York. In recent years individuals on our team have distinguished themselves as among the best math problem solvers in the nation. Over the years we have been consistently among the top teams out of a total of 90, competing against All State Teams from California, Texas, Ohio, Pennsylvania and many other states. New York City and Chicago also field excellent teams which we are always ready to challenge.

The format at the ARML competition is identical to the format of the annual NYSML competition.  There are four types of questions at the ARML competition:

   	Team Question -  	A set of 10 difficult short answer questions. The team of 15 meets
	    (40 points)		together for 20 minutes to produce one set of answers.

  	 Power Question - 	One comprehensive question with multiple parts which is theoretical in
	    (40 points)		nature and may require proofs as well as computations. The team of
				15 meets together for 60 minutes to produce one comprehensive solution.

   	Individual Questions - 	Students work independently on 8 problems with each correct answer
	    (120 points)		counting one point for their team. It is in this category that students
				qualify for the individual awards at the meet.

   	Relay Questions - 	Each team is divided into five sub-teams of three with each sub-team
	    (40 points)		being a relay team. The answer to the problem for the #1 is passed to
				the #2 position who needs it to solve that question. The  #3 position will
				receive an answer from #2 to input into the #3 problem. The #3 position
				hands in the answer for credit.

We hope that those of you who have been on the team will continue to enjoy this wonderful experience with us and that new members will find the excitement and fun contagious. If you have any questions,  I would be happy to talk to you or your parents about this opportunity; so please feel free to call me. My number is on the front sheet.  Also, you could ask your math teacher, math team coach or the math supervisor in your school. Other coaches who can answer your questions are:   Mrs. Patty Pragel, Greece Arcadia H. S. (tel 585-621-4663),  Mrs. Stephanie Ladd, Nottingham H. S., Syracuse ( (tel 315-476-4180), and last but not by any means least, Mr. George Reuter, Canandaigua Academy (tel. 585-396-3313) We are all high school math teachers.

Since time is of the essence, if you are going to apply, please send your application to Mr. Reuter immediately.

Sandy Ohlerich

Note:  presumably this letter was sent out to high AMC/AIME scorers in Upstate NY.  If you don't get the letter in the mail, don't be deterred, you still have a shot at making the team (which is none too competitive).

Alison, I see that you've attached the cover letter.  Can you post the application form as well?


I have been loking for places to volunteer near cupertino, but can't find much.  If anyone has suggestions, please post here.  Thank you.

if ur ever near san francisco st anthony's is a good place

you can always volunteer at your local library! but if you want something more math, you can volunteer for the tech museum up in downtown san jose, its further though, many of my friends volunteer there

many people at my school go to the local veterans hospital for volunteering


Μήπως ξέρει κανείς την αλγεβρική απόδειξη της BCS (χωρίς χρήση αναλυτικής γεωμετρίας,ανάλησης κτλ,η επαγωγή επιτρέπεται)?

[quote="bilot4"]Kατ' αρχάς να πω ότι φυσικά επιτρέπεται η επαγωγή, η οποία μπορεί να αποδειχθεί πρώτα γαι την ανισότητα Andreescu κι απο εκει εύκολα να καταλήξεις στην BCS.
Ωστόσο η BCS έχει και μια άλλη απλή απόδειξη (αλγεβρική) η οποία βασίζεται απλά σε θεωρία τριωνύμου!
Συγκεκριμένα αν $ f(x) \equal{} (a_1 \minus{} b_1x)^2 \plus{} (a_2 \minus{} b_2x)^2 \plus{} ... \plus{} (a_n \minus{} b_nx)^2$, τότε $ f(x)\geq0$.
Αρα η διακρίνουσα $ D$ πρέπει να είναι μικρότερη ή ίση του μηδενός. Το σημαντικό στοιχείο τώρα που ολοκληρώνει την απόδειξη είναι να παρατηρήσει κανείς ότι:
$ D \equal{} 4(a_1b_1 \plus{} a_1b_2 \plus{} ... \plus{} a_nb_n)^2 \minus{} 4(a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2)(b_1^2 \plus{} b_2^2 \plus{} ... \plus{} b_n^2)$[/quote]
$ (\displaystyle\int^b_a f(t)g(t)\,dt)^2 \le \displaystyle\int^b_a f^2(t)\,dt \displaystyle\int^b_a g^2(t)\,dt$
Είναι η ΒCS για ολοκληρώσιμες (συνεχείς για το λύκειο) συναρτήσεις στο $ [a,b]$ Στο attachment δίνονται δυο τρόποι απόδειξης ,ο β΄αφορά μόνο τις συνεχείς συναρτήσεις

Καλά όχι ότι έχει και πολλή σημασία αλλά προκύπτει και από την ταυτότητα Lagrange αν αφαιρέσεις το δεύτερο τετράγωνο...

Πιο πολύ το γράφω για να σου θυμίσω κάτι Δημήτρη... :wink:

[quote="maniopas"]Απόδειξη με χρήση επαγωγής:
$ (\sqrt {(x_1^2 \plus{} x_2^2)(y_1^2 \plus{} y_2^2)} \plus{} x_3y_3 \plus{} ... \plus{} x_{k \plus{} 1}y_{k \plus{} 1})^2$
$ > \equal{} (\sqrt {(x_1y_1 \plus{} x_2y_2)^2} \plus{} x_3y_3 \plus{} ... \plus{} x_{k \plus{} 1}y_{k \plus{} 1})^2$
[/quote]
Αυτο το παραπανω γιατι ισχυει?
Εννοω πως απο την πρωτη συνεπαγεται η δευτερη?

Θετοντας
$ A\equal{}(x_1^2 \plus{} x_2^2)(y_1^2 \plus{} y_2^2)$ και
$ B\equal{}x_3y_3 \plus{} ... \plus{} x_{k \plus{} 1}y_{k \plus{} 1}$ και
$ C\equal{}(x_1y_1 \plus{} x_2y_2)^2$
τοτε λες και ξερωντας οτι $ A\geq C$ αρα και $ \sqrt A\geq \sqrt C$ (αφου $ C\geq 0$) οτι θα εχουμε:

και $ (\sqrt A\plus{}B)^2 \geq (\sqrt C\plus{}B)^2$

Αυτη η συνεπαγωγη δεν ειναι σωστη. Ας το παρουμε με αριθμους για να φανει το γιατι:
$ A\equal{}25 , C\equal{}16$ και $ B\equal{}\minus{}8$  Τοτε πραγματι:  $ \sqrt A\geq \sqrt C \Longleftrightarrow5\geq 4$ αλλα το:
$ (\sqrt A\minus{}8)^2 \geq (\sqrt C\minus{}8)^2 \Longleftrightarrow9\geq 16$  το οποιο δεν ισχυει.

Ισχύει από την B. C. S. για δύο αγνώστους που απέδειξα (όταν n = 2)

1)Ωραίες οι παρατηρήσεις και οι αποδείξεις που κάνατε,αλλά ξέχασα να πω ότι η απόδειξη της bcs ζητείται στο βιβλίο του απειροστικού ως άσκηση στα πρώτα μαθήματα,όπου δεν έχει αποδειχθεί η ύπαρξη των αρρήτων,άρα δεν μπορούμε να χρησιμοποιήσουμε τις έννοιες ρίζα και διακρίνουσα.Είμαι περίεργος να δω την απόδειξη υπο αυτές τις προϋποθέσεις(χωρίς τη χρήση της lagrandge,διότι αυτό ζητείται στην από κάτω άσκηση του βιβλίου).

2)Για τον miniopa:
Δεν καταλαβαίνω γιατί ισχύει η ανισότητα δυο γραμμές κάτω από εκεί που γράφεις:για $ n\equal{}k\plus{}1$

Υπαρχει ευκολη αποδειξη της BCS με την χρηση του εσωτερικου γινομενου. Αν θελεις μπορω να την γραψω.

Ζητάμε καθαρά αλγεβρικά

[quote="maniopas"]Είναι ισότητα (λάθος μου, αλλά δεν αλλάζει τίποτα)[/quote]
Δεν ειναι λαθος σου, αφου απλα ισχυει και η ισοτητα. :)
 Αλλα δεν καταλαβες αυτο που σου ειπα: Το παρακατω που χρησιμοποιησες που κανω quote ειναι λαθος και εξηγησα τους λογους.

[quote]$ (\sqrt {(x_1^2 \plus{} x_2^2)(y_1^2 \plus{} y_2^2)} \plus{} x_3y_3 \plus{} ... \plus{} x_{k \plus{} 1}y_{k \plus{} 1})^2$
$ > \equal{} (\sqrt {(x_1y_1 \plus{} x_2y_2)^2} \plus{} x_3y_3 \plus{} ... \plus{} x_{k \plus{} 1}y_{k \plus{} 1})^2$
[/quote]

Δεν ξερουμε αν ισχυει αυτο. Συνοπτικα (αν βαριεσαι να διαβαζεις αυτα που ειχα γραψει πανω):

Αν $ A\geq C \geq 0$ τοτε δεν ειναι απαραιτητο να ισχυει $ (A\plus{}B)^2\geq (C\plus{}B)^2$ για Β πραγματικο.

Έχεις πράγματι δίκιο. Η απόδειξή μου είναι λάθος. Να πως θα μπορούσε να διορθωθεί:

$ (x_1^2 \plus{} ... \plus{} x_{k\plus{}1}^2)(y_1^2 \plus{} ... \plus{} y_{k\plus{}1}^2) \equal{} (|x_1|^2 \plus{} ... \plus{} |x_{k\plus{}1}|^2)(|y_1|^2 \plus{} ... \plus{} |y_{k\plus{}1}|^2)$
όμοια με πριν (τώρα μπορώ να το εφαρμόσω, γιατί έχω θετικούς) $ >\equal{} (|x_1y_1| \plus{} ... \plus{} |x_{k\plus{}1}y_{k\plus{}1}|)^2$
$ >\equal{} (|x_1y_1 \plus{} ... \plus{} x_{k\plus{}1}y_{k\plus{}1}|)^2 \equal{} (x_1y_1 \plus{} ... \plus{} x_{k\plus{}1}y_{k\plus{}1})^2$

Εγώ πάλι δεν κατάλαβα γιατί ισχυει η  $ (x_1^2 \plus{} ... \plus{} x_{k\plus{}1}^2)(y_1^2 \plus{} ... \plus{} y_{k\plus{}1}^2)>\equal{} (\sqrt{x_1^2 \plus{} x_2^2}\sqrt{y_1^2 \plus{} y_2^2} \plus{} x_3y_3 \plus{} ... \plus{} x_{k\plus{}1}y_{k\plus{}1})^2$  στην απόδειξη του miniopa :wacko:.

[quote="DIMITRIS T"]Εγώ πάλι δεν κατάλαβα γιατί ισχυει η  $ (x_1^2 \plus{} ... \plus{} x_{k \plus{} 1}^2)(y_1^2 \plus{} ... \plus{} y_{k \plus{} 1}^2) > \equal{} (\sqrt {x_1^2 \plus{} x_2^2}\sqrt {y_1^2 \plus{} y_2^2} \plus{} x_3y_3 \plus{} ... \plus{} x_{k \plus{} 1}y_{k \plus{} 1})^2$  στην απόδειξη του miniopa :wacko:.[/quote]
Λογω υποθεσης της επαγωγης ισχυει:

Για καθε $ n \equal{} k$ και για καθε $ x_1,x_2,...,x_k,y_1,y_2,...,y_k$ πραγματικο αριθμο ισχυει:
$ (x_1^2 \plus{} ... \plus{} x_k^2)(y_1^2 \plus{} ... \plus{} y_k^2) > \equal{} (x_1y_1 \plus{} ... \plus{} x_ky_k)^2$

Τωρα στο:
$ (x_1^2\plus{}x_2^2\plus{} ... \plus{} x_{k \plus{} 1}^2)$ εχουμε κ+1 ορους, οποτε παιρνουμε το $ x_1^2\plus{}x_2^2$ σαν ενα ορο.
Ομοια και στο $ (y_1^2\plus{}y_2^2\plus{} ... \plus{} y_{k \plus{} 1}^2)$ παιρνουμε το $ y_1^2\plus{}y_2^2$ σαν ενα ορο.

 Οποτε λογω υποθεσης και αφου εχουμε μειωσει του κ+1 ορους κατα εναν, αρα εχουμε κ ορους αρα μπορουμε να χρησιμοποιησουμε την υποθεση με πρωτο ορο τα $ x_1^2\plus{}x_2^2$ και $ y_1^2\plus{}y_2^2$ θα εχουμε:

$ (x_1^2\plus{}x_2^2\plus{} ... \plus{} x_{k \plus{} 1}^2)(y_1^2\plus{}y_2^2\plus{} ... \plus{} y_{k \plus{} 1}^2)$
$ \equal{} ((x_1^2\plus{}x_2^2)\plus{} ... \plus{} x_{k \plus{} 1}^2)((y_1^2\plus{}y_2^2)\plus{} ... \plus{} y_{k \plus{} 1}^2)$ (ΚΑΙ ΛΟΓΩ ΥΠΟΘΕΣΗΣ):
$ > \equal{} (\sqrt {x_1^2 \plus{} x_2^2}\sqrt {y_1^2 \plus{} y_2^2} \plus{} x_3y_3 \plus{} ... \plus{} x_{k \plus{} 1}y_{k \plus{} 1})^2$

Ακριβώς αυτό είναι!!!  :D

Ευχαριστώ!!! :yup:


superior algebra
superior algebra unsolved

I know if $ n\equiv 3 \ (mod \ 4 )$& n is prime , then $ \mathbb Z_n[i]$ is a field .

but about to find all units and zero divisors in $ \mathbb Z_6[i]$ , $ \mathbb Z_{100}[i]$,&$ \mathbb Z_n[i]$, whenever $ \mathbb Z_n[i]$ is not a field .I think there is rules for that , what is it please ?, if it's exist

 :)

You could analyse how elements behave under the norm mapping $ N: \mathbb Z_n[i]\to\mathbb Z_n$ defined by $ N(a\plus{}bi)\equal{}a^2\plus{}b^2$. In particular have a look at the norm $ N(a\plus{}bi)$ being a unit or a zero divisor in $ \mathbb Z_n$, respectively.


Let t be the smallest positive real number for which cos t = cos t (degrees). W are taking the cosine of t radians on the left and of t degrees on the right. Compute the greatest integer less than t.

Note that $y=\cos (\text{x degrees})$ barely decreases over $[0,2 \pi ]$. The first intersection between $y=\cos (\text{x degrees})$ and $y=\cos x$ will occur just before the second peak of the $\cos x$ curve.

If $t$ is in radians, then we have:

$\cos \left( \frac{180}{\pi}t \right) = \cos t \implies \frac{180}{\pi}t = 360-t$

This rearranges to: $t = \frac{360 \pi}{180+\pi}$
$6 < t < 2 \pi < 7$, so the answer is $6$.


[quote="fjwxcsl"]Let  $ {x,y,z}\in{R},{t,u}\in{N},k\le{1}$ prove that
\[ \sum_{cyc}{(x \minus{} k y)^{2t \plus{} 1}(x \minus{} kz)^{2u \plus{} 1}}\ge{0}.\]
[/quote]

 We have specially

   $ \sum_{cyc}{(x\minus{}y)^3(x\minus{}z)^3}\ge{\frac{15}{4}(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2},$

 $ \frac{15}{4}$ here is the best one.

[quote="fjwxcsl"]

 We have specially

   $ \sum_{cyc}{(x \minus{} y)^3(x \minus{} z)^3}\ge{\frac {15}{4}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2},$

 $ \frac {15}{4}$ here is the best one.[/quote]
$ \sum_{cyc}{(x \minus{} y)^3(x \minus{} z)^3}\minus{}{\frac {15}{4}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}\equal{}$
$ \equal{} \frac{1}{4}\left(\sum_{cyc}(2x^3 \minus{} 3x^2y \minus{} 3x^2z \plus{} 4xyz)\right)^2.$

[quote="fjwxcsl"]  We have specially

   $ \sum_{cyc}{(x \minus{} y)^3(x \minus{} z)^3}\ge{\frac {15}{4}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2},$

 $ \frac {15}{4}$ here is the best one.[/quote]
Indeed, fjwxcsl and arqady  :lol: , we have

$ 4\sum(x \minus{} y)^3(x \minus{} z)^3 \equal{} 15\prod (x \minus{} y)^2 \plus{} \prod (2x \minus{} y \minus{} z)^2$.

We have also:

$ (a) \ \ \ \ \sum x^2(x \minus{} y)^2(x \minus{} z)^2\ge 2(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2$;

$ (b) \ \ \ \ \sum (x \minus{} 2y)(x \minus{} 2z)(x \minus{} y)^2(x \minus{} z)^2\ge 5(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2$;

$ (c) \ \ \ \ \sum (x \plus{} 2y)(x \plus{} 2z)(x \minus{} y)^2(x \minus{} z)^2 \plus{} 3(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\ge 0$.

a)
$ \sum{ x^2(x\minus{}y)^2(x\minus{}z)^2} \equal{} (\sum{x(x\minus{}y)(x\minus{}z)})^2 \minus{}2\sum{(x\minus{}y)^2(x\minus{}z)(z\minus{}y)xy} \equal{}(\sum{x(x\minus{}y)(x\minus{}z)})^2 \minus{}2\prod{(x\minus{}y)}\sum{xy(y\minus{}x)} \equal{} (\sum{x(x\minus{}y)(x\minus{}z)})^2\plus{}2(x\minus{}y)^2(y\minus{}z)^2(x\minus{}z)^2$

we get similar identities for others

I have obtained the following :

$ \prod(a\minus{}b)^2 \leq \frac{ 4(a^2\plus{}b^2\plus{}c^2\minus{}(ab\plus{}ac\plus{}bc))^3}{27}$ :)

[quote="anas"]I have obtained the following :

$ \prod(a \minus{} b)^2 \leq \frac { 4(a^2 \plus{} b^2 \plus{} c^2 \minus{} (ab \plus{} ac \plus{} bc))^3}{27}$ :)[/quote]
See here:

http://www.mathlinks.ro/viewtopic.php?t=297050

[quote="anas"]I have obtained the following :

$ \prod(a \minus{} b)^2 \leq \frac { 4(a^2 \plus{} b^2 \plus{} c^2 \minus{} (ab \plus{} ac \plus{} bc))^3}{27}$ :)[/quote]
It's equivalent to the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193555

arqady do u think that the above inequality works for this pb of nguoivn :

$ \sum{\frac{a^2}{b^2}} \geq \frac{9(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} ac \plus{} bc}\minus{} 6$ ? :wink:


calculus
integration
trigonometry
function
derivative
absolute value
calculus computations

I am currently solving the arc length for one curve of $r = 1-\sin\theta$

I have arrived to the equation:
$\int_{0}^{2\pi}\sqrt{2-2\sin\theta}d\theta$

Problem is I don't know how to integrate this one. Hope someone can help. Thanks.

Hint : $1-\sin\theta=(\sin\frac{\theta}{2}-\cos\frac{\theta}{2})^{2}$.  :wink:

I see sir but how can i integrate with the $\sqrt2$. I couldn't seem to use integration by substitution cause theres no cosine outside the squareroot. Another  hint would be nice.

The hint that N.T. Tuan gave is the best hint you're going to get - you don't seem to have used it yet. And $\sqrt{2}$ is a constant that can be factored out of the integral.

$\int^{2\pi}_{0}\sqrt{2-2\sin x}\,dx = \sqrt{2}\int^{2\pi}_{0}\sqrt{1-\sin x}\,dx$.

Now, knowing that $\sin 2x = 2\sin x \cos x$, and also $1 = \sin^{2}(\frac{x}{2})+\cos^{2}(\frac{x}{2})$, then

$1-\sin x = \sin^{2}(\frac{x}{2})+\cos^{2}(\frac{x}{2})-2\sin(\frac{x}{2})\cos(\frac{x}{2}) = \left[\sin(\frac{x}{2})-\cos(\frac{x}{2})\right]^{2}$.

Next, from the formula for the sinus of the difference of two angles,

$\sin(a-b) = \sin a \cos b-\cos a \sin b$,

make $a = \frac{x}{2}$ and $b = \frac{\pi}{4}$, and we get

$\sin(\frac{x}{2})-\cos(\frac{x}{2}) = \sqrt{2}\sin\left(\frac{x}{2}-\frac{\pi}{4}\right)$,

and so

$1-\sin x = 2\sin^{2}\left(\frac{x}{2}-\frac{\pi}{4}\right)$.

ow our original integral becomes

$\sqrt{2}\int^{2\pi}_{0}\sqrt{1-\sin x}\,dx = \sqrt{2}\int^{2\pi}_{0}\sqrt{2}\sin\left(\frac{x}{2}-\frac{\pi}{4}\right)\,dx = 2\int^{2\pi}_{0}\sin\left(\frac{x}{2}-\frac{\pi}{4}\right)\,dx$.

Carefully! $\sin (\frac{x}{2}-\frac{\pi}{4})\geq 0$? Are you sure?

One hint: in integrating a periodic function, any interval whose length is one period can be used. In this case we can do this:

$\int_{0}^{2\pi}\sqrt{2-2\sin\theta}\,d\theta= \int_{\frac{\pi}2}^{\frac{5\pi}2}\sqrt{2-2\sin\theta}\,d\theta$

$=$ (intermediate steps as in Carcul's post)

$=2\int_{\frac{\pi}2}^{\frac{5\pi}2}\sin\left(\frac{\theta}2-\frac{\pi}4\right)\,d\theta$

A second idea: from the very beginning, make the substitution $u=\theta-\frac{\pi}2$ and shift the interval of integration as above to get

$\int_{0}^{2\pi}\sqrt{2-2\cos u}\,du$

The advantage here is that the half-angle trig identity needed to proceed further is more familiar and easier to find.

Yes N.T.Tuan, you are right.
Professor Merryfield, thanks for the hint, it makes sense.

Sir isn't the variable u in the substitution supposed to be $u = \frac{\pi}{2}-\theta$ because $\sin(\frac{\pi}{2}-\theta) = \cos\theta$.

[quote="Carcul"]$2\int^{2\pi}_{0}\sin\left(\frac{x}{2}-\frac{\pi}{4}\right)\,dx$[/quote]
Hm....

$\int_{0}^{2 \pi}\sqrt{2-2 \sin x}~~dx=8$
$2\int^{2\pi}_{0}\sin\left(\frac{x}{2}-\frac{\pi}{4}\right)\,dx = 4 \sqrt{2}$

Note.. after solving it using the method Kent described I noticed all you're missing is an absolute value, pointed out by N.T.

My solution:
[hide]$\int_{0}^{2 \pi}\sqrt{2-2 \sin \theta}~~d \theta$

Let:
$u-\frac{\pi}{2}=\theta \iff u=\theta+\frac{\pi}{2}$
$du=d \theta$

$=\sqrt{2}\cdot \int_{\frac{\pi}{2}}^{\frac{5 \pi}{2}}\sqrt{1-\sin \left ( u-\frac{\pi}{2}\right)}~~du$
$=\frac{2}{\sqrt{2}}\cdot \int_{\frac{\pi}{2}}^{\frac{5 \pi}{2}}\sqrt{1+\cos u}~~du$
$=2 \cdot \int_{\frac{\pi}{2}}^{\frac{5 \pi}{2}}\sqrt{\frac{1+\cos u}{2}}~~du$

Note that $\cos \left( \frac{u}{2}\right) = \pm \sqrt{\frac{1+\cos u}{2}}\iff \cos^{2}\left( \frac{u}{2}\right) = \frac{1+\cos u}{2}$

$=\boxed{2 \cdot \int_{\frac{\pi}{2}}^{\frac{5 \pi}{2}}\left| \cos \frac{u}{2}\right| ~~du}$; You can evaluate on your own.. separate integrals from $\frac{\pi}{2}$ to $\pi$ and subtract the integral from $\pi$ to $\frac{5 \pi}{2}$.

$=\boxed{8}$[/hide]

Since $|\cos\frac{u}{2}|\le 1$, the integral is less than $4\pi$. It can't be 16.

[quote="apoapo"]Sir [b]skimnc[/b], Are you sure the right answer for integrating the integral $2\int^{2\pi}_{0}|\cos \frac{u}{2}| du$ is $8$ because when i evaluated it i got $16$ after using the substitution $v = \frac{u}{2}$. Hope you could reply soon.[/quote]

Yes, I'm sure: (Make sure if you're doing a substitution to change your bounds as well as substitute the derivative which is most likely where your error was (forgetting to put $\frac{1}{2}$ along with the $dv$.)

[hide]$2\int^{2\pi}_{0}\left| \cos \frac{u}{2}\right| du$

$\cos \frac{u}{2}\geq 0$ on $0 \leq u \leq \pi$
$\cos \frac{u}{2}\leq 0$ on $\pi \leq u \leq 2 \pi$

$=2\int^{\pi}_{0}\cos \frac{u}{2}du-2\int^{2\pi}_{\pi}\cos \frac{u}{2}du$
$=2 \cdot 2 \left[ \sin \frac{x}{2}\right]_{0}^{\pi}-2 \cdot 2 \left[ \sin \frac{x}{2}\right]_{\pi}^{2\pi}$
$=4 (1-0)-4 (0-1)$
$=4+4$
$=\boxed{8}$[/hide]

$\sqrt{1-\sin \theta}=\frac{|\cos \theta|}{\sqrt{1+\sin \theta}}$


geometry
rhombus
rectangle
perimeter
geometric transformation
reflection
trigonometry

Rhombus $PQRS$ is inscribed in rectangle $ABCD$ so that vertices $P$, $Q$, $R$, and $S$ are interior points on sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$, respectively. It is given that $PB=15$, $BQ=20$, $PR=30$, and $QS=40$. Let $m/n$, in lowest terms, denote the perimeter of $ABCD$. Find $m+n$.

[hide]
aftering drawing an acurate diagram and find all the lengths we can with Pythagorean, we find that $APRD$ and $CRPB$ is congruent by symmetry.  therefore $AP = CR = a$ and $DR = PB = b$.  the area of $APS$ and $CQR$ together is $ab$, while the area of the rectangle is $(15+a)(20+b) = 300 + 15b + 20a + ab$.  Subtracting the area of $APS$ and $CQR$ from the rectangle we get $300 + 15b + 20a= 900$ because the rest of the six right triangles are all congruent.  simplifying the equation we get $4a + 3b = 120$.  Since $a^2 + b^2 = 625$ by Pythagorean, we can use some painful algebra to find the 2 order pair solutions $(a, b)$ to the two system.  We get the solutions $(15, 20)$ and $(\frac{117}{5}, \frac{44}{5})$. obviously the first one is not valid because it will ruin the rectangle, so having $70 + 2a + 2b$ we get $\frac{672}{5}$ as the perimeter.  So the answer is $677$[/hide]

For your solution, I think you meant $AS = QB = b$ instead of $DR = PB = b$.  Otherwise, nice solution :)

[hide="Another solution"]Let $O \in PR \cap QS$ ($O$ is the center of the rhombus)
The perpendicular line from $O$ to $BC$, intersects $BC$ to its midpoint (because $O$ is the midpoint of $PR$).
So this line is the perpendicular bisector of $BC \Rightarrow BO=BC$

The triangles $QOB,OBC$ are isosceles, and similar (because they have $\angle QOB = \angle OBC$).

Hence $\frac{BQ}{OB} = \frac{OB}{BC} \Rightarrow OB^2 = BC \cdot BQ$

The length of $OB$ could be found easily from the area of $BPQ$:

$BP \cdot PQ = \frac{OB}{2} \cdot PQ \Rightarrow OB = \frac{2BP\cdot PQ}{OB} \Rightarrow OB = 24$

So $OB^2 = BC \cdot BQ \Rightarrow 24^2 = (20+CQ) \cdot 20 \Rightarrow CQ=\frac{44}{5}$

From the right triangle $CRQ$ we have $RC^2 = 25^2-(\frac{44}{5})^2\Rightarrow RC = \frac{117}{5}$

$($or we can define a similar formula : $OB^2=BP \cdot BA$ , and then we find $AP$
in other words the segment $OB$ is tangent to the circles with diameters $AO,CO)$

perimeter:
$2(PB+BQ+QC+CR) = 2(15+20+\frac{44+117}{5}) = \frac{672}{5}\Rightarrow$

$672+5=677$[/hide]

[quote="ehehheehee"][hide="a third way based only off of reflections"]You can just lable the points.  After drawing a brief picture, you can see 4 right triangles with sides of 15,20,25.

So the points of triangle RDS are (0,0) (0,20) and (15,0)
Since each right triangle can be split into two similar triangles, point (0,0) is 12 away from the hypotnuse.  By reflecting (0,0) over the hypotnuse, we can get 3rd point of the second right triangle (A.K.A the intersection of the diagonals of the Rhombus) which is (19.2,14.4)

By reflecting (15,0) over diagonal SQ we get P (23.4,28.8).  By adding 15 to the x value we get B(38.4,28.8)

So the perimeter is equal to $(38.4+28.8)*2$ which equals $\frac{672}{5}$[/hide][/quote]
what do u mean by "By reflecting (0,0) over the hypotnuse"?

the hypotnuse  is of right triangle RDS (I don't have the diagram in front of me, this is just from reading the solution)

you reflect the point (0,0) over the line that contains the hypotnuse if that makes more sense.

You can kind of figure out the solution by seeing the 4 congruent triangles...

Wow, this is some ancient thread....
After looking back at it again:
[hide="simple trig solution"]
for convenience, let $\angle PQS = \theta$, then since all triangles are congruent we have that $\angle BQR = 3\theta$, and therefore $\angle QRC = 3\theta - 90$.  let $RC = a$, then we have $\sin{(3\theta - 90)} = \frac{a}{25}$, or $-\cos{3\theta} = \frac{a}{25}$, one of the popular formula is $\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}$, and we know that $\cos{\theta} = \frac{4}{5}$, the rest is trivial.
[/hide]

btw, how do you figure out the coordinates of the image of the point under reflection??

Yeah, it probably is ancient, but I came here from the recourses section so I wouldn't know the difference...(wait they have the dates, duh)

I think I just found a nice solution.  :)

[hide=" Cyclic Quadrilaterals"]Let the intersection of $\overline{PR}$ and $\overline{SQ}$ be $I$ so that $PI=IR=15$ and $SI=IQ=20.$ Applying the Pythagorean Theorem on triangle $PBQ$ yields $PQ=25.$

Now, note that quadrilaterals $PBQI$ and $SIRD$ are congruent and both cyclic, since their opposite angles are both of measure $90^\circ.$ Hence, by applying Ptolemy's theorem on quadrilateral $PBQI,$
\[(PB)(QI)+(BQ)(IP)=(15)(20)+(20)(15)=(PQ)(BI)=(25)(BI)\implies BI=24. \]
It follows that $BD=2BI=48.$

Returning to $BPQI,$ note that $\angle PQI\cong \angle PBI \cong ABD.$ Hence, $\triangle ABD \sim \triangle IQP$ by $AA$ similarity. From here, it's clear that
\[\frac{AD}{BD}=\frac{IP}{PQ}\implies \frac{AD}{48}=\frac{15}{25}\implies AD=\frac{144}{5}. \]
Similarly,
\[\frac{AB}{BD}=\frac{IQ}{PQ}\implies \frac{AB}{48}=\frac{20}{25}\implies AB=\frac{192}{5}. \]
Therefore, the perimeter of rectangle $ABCD$ is $2(AB+AD)=2\left(\frac{192}{5}+\frac{144}{5}\right)=\frac{672}{5}.$[/hide]

Patience is a virtue.  :wink:

[quote="TheHeinrich"]if we have ab=300 and a^2+b^2=625

if we use  a^2+2ab+b^2=1225
                          srbs
                  a+b=35

did i do something wrong because the perimeter is 2(a+b)+70[/quote]

That is true, but what are $ a$ and $ b$? As far as I can tell you let $ a \equal{} 15, b \equal{} 20$, so that $ a \plus{} b \equal{} 35$ should be trivial. There's a diagram here [url=http://www.artofproblemsolving.com/Wiki/index.php/1991_AIME_Problems/Problem_12](wiki)[/url] and a couple more solutions there to illustrate also that the perimeter is not $ 2(a \plus{} b) \plus{} 70$. You probably drew your diagram wrong; remember you want the perimeter of the rectangle.

Edit: Never mind, I understand what you are saying. But where did you get $ ab \equal{} 300$ from?

Also you might want to use latex, as you can see spacing like that doesn't work very well. Just add $ \$$ signs around your equation, like [code]$a^2 + b^2$[/code]


limit
algebra
polynomial
calculus
trigonometry
binomial theorem
calculus computations

(I'm still a novice at calculus.   :oops: )

How do you solve these (easy) limits?  They're probably pretty straightforward, but here they are anyways.

I don't want just the answers, I want to know exactly how to solve them, not the answer.

$ \lim_{n \to \infty} \frac {\sqrt {n \plus{} 5} \minus{} \sqrt {n \plus{} 3}}{\sqrt {n \plus{} 1} \minus{} \sqrt {n}}$

$ \lim_{n \to \infty} \frac {\sqrt {2n \plus{} 1} \minus{} \sqrt {2n \minus{} 1}}{\sqrt {4n \plus{} 1} \minus{} 2\sqrt {n}}$

$ \lim_{n \to \infty} \frac {\sqrt {n} \minus{} \sqrt {n \plus{} 2}}{\sqrt {n^2 \plus{} 5} \minus{} n}$

Again, I want a method and the answer, not just the answer.  (Or just the method would be fine.)  I'm trying to learn the concept, not get the answer by asking people.

Thanks in advance!

the method in each of these cases is to rationalize the expression...

Obviously, but how?  I have little experience with this stuff; it's mostly new to me.

All of these would fall to the purely algebraic technique of rationalizing by multiplying by the conjugate. For instance, in the first one, you could multiply through by
\[ \frac {\sqrt {n \plus{} 5} \plus{} \sqrt {n \plus{} 3}}{\sqrt {n \plus{} 5} \plus{} \sqrt {n \plus{} 3}}\cdot\frac {\sqrt {n \plus{} 1} \plus{} \sqrt {n}}{\sqrt {n \plus{} 1} \plus{} \sqrt {n}}\]
My own instincts are to use the binomial series. To use the binomial series, first factor out the biggest item under the root, and then use the following:

For arbitrary $ \alpha$ and as $ y\to0,$ $ (1 \plus{} y)^{\alpha} \equal{} 1 \plus{} \alpha y \plus{} O(y^2).$

If we need more terms than that, we can have as many as we need:

$ (1 \plus{} y)^{\alpha} \equal{} 1 \plus{} \alpha y \plus{} \frac {\alpha(\alpha \minus{} 1)}2y^2 \plus{} \frac {\alpha(\alpha \minus{} 1)(\alpha \minus{} 2)}{3!}y^3 \plus{} \cdots.$

So for instance:

$ \sqrt {n \plus{} 5} \minus{} \sqrt {n \plus{} 3} \equal{} \sqrt {n}\left(\left(1 \plus{} \frac5n\right)^{\frac12} \minus{} \left(1 \plus{} \frac3n\right)^{\frac12}\right)$

$ \equal{} \sqrt {n}\left(\left(1 \plus{} \frac {5}{2n} \plus{} O(n^{ \minus{} 2})\right) \minus{} \left(1 \plus{} \frac {3}{2n} \plus{} O(n^{ \minus{} 2})\right)\right)$

$ \equal{} \sqrt {n}\left(\frac1n \plus{} O(n^{ \minus{} 2})\right) \equal{} \frac1{\sqrt {n}} \plus{} O(n^{ \minus{} 3/2})$

Similarly, $ \sqrt {n \plus{} 1} \minus{} \sqrt {n} \equal{} \frac1{2\sqrt {n}} \plus{} O(n^{ \minus{} 3/2}).$

We can now see in a straightforward way what happens to the original limit.

In retrospect, I should have divided numerator and denominator by $ \sqrt {n},$ getting

$ \lim_{n\to\infty}\frac {\sqrt {1 \plus{} \frac5n} \minus{} \sqrt {1 \plus{} \frac3n}}{\sqrt {1 \plus{} \frac1n} \minus{} 1}$

and then used the binomial expansion from there.

This can all be done much more rapidly than I can type this response; I'm used to this sort of argument and can find all three of those limits in my head. (Hint: one of them is not finite.)

[quote="AdithyaGanesh"](I'm still a novice at calculus.   :oops: )

How do you solve these (easy) limits?  They're probably pretty straightforward, but here they are anyways.

I don't want just the answers, I want to know exactly how to solve them, not the answer.

$ \lim_{n \to \infty} \frac {\sqrt {n + 5} - \sqrt {n + 3}}{\sqrt {n + 1} - \sqrt {n}}$

$ \lim_{n \to \infty} \frac {\sqrt {2n + 1} - \sqrt {2n - 1}}{\sqrt {4n + 1} - 2\sqrt {n}}$

$ \lim_{n \to \infty} \frac {\sqrt {n} - \sqrt {n + 2}}{\sqrt {n^2 + 5} - n}$

Again, I want a method and the answer, not just the answer.  (Or just the method would be fine.)  I'm trying to learn the concept, not get the answer by asking people.

Thanks in advance![/quote]

$ \lim_{n \to \infty} \frac {\sqrt {n + 5} - \sqrt {n + 3}}{\sqrt {n + 1} - \sqrt {n}}$

$ = \lim_{n \to \infty} \frac {\sqrt {n + 5} - \sqrt {n + 3}}{\sqrt {n + 1} - \sqrt {n}} \cdot \frac {\sqrt {n + 5} + \sqrt {n + 3}}{\sqrt {n + 5} + \sqrt {n + 3}}\cdot\frac {\sqrt {n + 1} + \sqrt {n}}{\sqrt {n + 1} + \sqrt {n}}$

$ = \lim_{n \to \infty} \frac {2 ( \sqrt {n + 1} + \sqrt {n} ) }{\sqrt {n + 5} + \sqrt {n + 3}}$

Dividing the whole expression by $ n$, we get

$ \lim_{n \to \infty} \frac {2(\sqrt {1 + \frac1n} + \sqrt1)}{\sqrt {1 + \frac5n} + \sqrt {1 + \frac3n}}$

${ \text{We note that as n} \to \infty, \frac {\alpha}{n} \to \infty}$  Where $ \alpha$ is an integer that does not equate to $ 0$.

Therefore, $ \lim_{n \to \infty} \frac {\sqrt {n + 5} - \sqrt {n + 3}}{\sqrt {n + 1} - \sqrt {n}} = \boxed{2}$

Okay, I think I understand it now.  (I presume the other two can also be solved by this method.)

The binomial expansion makes sense, but it seems simpler to simply rationalize the denominator.  Is binomial expansion more useful for more difficult limit problems?

Once you're used to it, the binomial expansion is both faster and less error-prone, but I admit that you do have to be used to it.

The big weakness of the "rationalize" idea is that its easy use is pretty much confined to square roots, while the binomial expansion works for any kind of root or power.

$ \lim_{n\to\infty}n^3\left(\left(n^5\plus{}1\right)^{\frac25}\minus{}n^2\right)\equal{}?$

That's really interesting.  I have only done binomial expansion with integer exponents, but what does it involve with fractional exponents?  By this I mean, say you were given an arbitrary binomial to a fractional exponent.  Wouldn't you need to evaluate the following?

$ \binom{\frac ab}{k}$

You would have to evaluate combinations with fractions, leading to fractional factorials.

How do you evaluate that?  (Could someone please direct me to a link or something?)

Follow this link :
[url]http://en.wikipedia.org/wiki/Binomial_series[/url]

Do not hope to understand everything at the first reading besides you need to know what Taylor series are

[quote="AdithyaGanesh"]Wouldn't you need to evaluate the following?

$ \binom{\frac ab}{k}$[/quote]
By convention - and the convention is established by this series -

$ \binom{\alpha}{k}\equal{}\frac{\alpha(\alpha\minus{}1)(\alpha\minus{}2)\cdots(\alpha\minus{}k\plus{}1)}{k!}.$

The series $ (1\plus{}x)^{\alpha}\equal{}\sum_{k\equal{}0}^{\infty}\binom{\alpha}{k}x^k$ converges for (at least) $ \minus{}1<x<1.$ For some $ \alpha,$ the series converges at one or both of the boundary points; when $ \alpha$ is a nonnegative integer, all but finitely many of the terms of the series is zero, making this a polynomial expansion (agreeing with the binomial theorem for nonnegative integer exponents that you may already know).

The development of this series is largely due to Isaac Newton. He had this series before he had the rest of his calculus, and he used it as one of his most important tools in developing calculus. In particular, he used it to derive a series for the arcsin, then used that to derive series for the sine and cosine, and then he knew the calculus of the trigonometric functions.


My state, fl, gives out awards to top ten teams and top twenty individuals. Please state your state and the number of people that get awards. Thank you.

Kansas: Top 6 teams, Top 6 individuals

EDIT:

Top 4 individuals: Future astronaut training camp at Kansas cosmosphere and space center.

[quote="goldendomer"]Theres A LOT of people who gets awards.[/quote]
How many is A LOT

I think it was the top 10 individuals and the top 10 teams.

CT: Top 4 Countdown (Unofficial) Top 4 Written (Official) Top 4 Teams

I got kinda confused in my state because I think they handed out like top 5 for written and top 4 for countdown, and something like top 5 teams. I got three trophies  :P
[hide="My state"]WA[/hide]

[quote="mathgeniuse^ln(x)"]Michigan:Top 4 Individuals, Top 4 Teams
I was the only person to get two trophies in my state. :lol:[/quote]

Shut up  :mad: !

I blame my team, yeah.... it was sad.

Well, I still have the spiffy 4th place trophy, Michigan.

The first place guy got some 1000 dollar scholarship ONLY if he went to this unheardof grand valley state university AND ONLY if he studied engineering or something. :rotfl:  :rotfl:

Top 10 individuals, top 10 teams.
For 4th, I got a ti-84 plus silver edition and a $2000 scholarship to UT austin engineering school.
Not bad, especially since UT is at least HEARD OF.

[quote="Klebian"]Top 10 individuals, top 10 teams.
For 4th, I got a ti-84 plus silver edition and a 2000 scholarship to UT austin engineering school. Not bad, especially since UT is at least HEARD OF.[/quote]2000?!?!?!Isn't that what you get at nationals or something???
And the TI84...don't you get that at nationals too??

This is Kentucky's awards:
Top Four get a 4-year full tuition scholarship to either UK or UofL.
Fifth to 10th get a 1000 dollar scholarship to either UK or UofL

I was so pissed.
I got 11th!!!
And one of my best friends got 10th.
I felt better when I got home.
Since last year, I didn't even make it to State.

Also Top 4 Countdown (Unofficial) winners get a trophy.
Top 10 Written (Official) get a trophy. Again, I WAS SO PISSED.
Top 4 (?) Teams get a trophy.
Top 6 Written get to have their coach get a small trophy.

[quote="Klebian"]Idk what you get at nationals, but it's probably more.
BTW, i recommend you edit your last post and turn BBCode off, its screwing up the whole page.[/quote]

Heh. Sorry.
I had to delete the dollar sign in your message because I didn't know how to undo the little BBcode thing or w/e it's called.

20000 is the absolute maximum one can possibly get at nationals. It's quite rare to achieve.

In Massachusetts, last year, the top 4, and the top girl, all got FULL scholarships, to the Wentworth institute of Technology, which isn't that bad. But FULL scholarships, whee :D

Haha, Sports Awards.

I guess we "Mathletes" practice Mathletics like Atheletes do Athletics.

Our chapter had better awards/organization (AND DONUTS/FOOD) than the state.  Our chapter had both Countdown and written awards while the state just gave trophies to the top 4.

[quote="Xantos C. Guin"]My chapter gives a 13" TV, 5" LCD player, an MP3 player, and a pair of walkie-talkies to the top 4. They choose in order of rank. They also have a decent buzzer system for the countdown round and a competent person resetting the buzzers after each contestant buzzes in. The room in which we take the test is above 50$\circ$F.

My state gives Shell T-Shirts, mouspads, and pens to everyone. They don't have a decent buzzer system for the countdown round and a competent person resetting the buzzers after each contestant buzzes in. The tent in which we take the test is below 50$\circ$F.

Odd. You'd think state competition would have better prizes than a chapter competition.[/quote]

My chapter had nothing but those trophies. You had to raise your hand last year, I forgot if we had buzzers this year or not.
My state had nothing but those trophies, no free hotel, no airplane tickets, no anything, transportation you're responsible for, it's 1 day, starting at 9, free lunch, that's it.

my chapter gave out trophies and hastings giftcards to the countdown round people. everyone got donuts and lunch and a shirt and pins.
they're getting smarter about the shirts.  two years ago the oklahoma state shirt was "Mathcounts: the Oklahoma brand of fun!" it had a picture of a cow on it. yippee. ;)
this year they looked sort of like shirts you buy from college stores that have the college's logo on it.

There's nothing wrong with the shirt with the cow that says 'Mathcounts; the Oklahoma brand of fun!' I think I have two shirts that say the same thing except instead of Oklahoma it says Texas (duh). They also had cows on them.

Last year they took a picture of a 5-pointed star with a grumpy face off the side of a liquor store in a city near me. maybe it was chickasha. i don't remember.  anyway, they gave him a sheriff's badge, a calculator, and a pencil and put the caption "solve or draw" by him.  it was odd.

You have not convinced me yet that your state shirts suck. See, we had that exact same shirt.

it's a conspiracy!  :o
if texas got their shirts from a liquor store in chickasha, oklahoma, they have no imagination.

of course, neither do we. :(

Texas has lots of other (better) prizes though.

[quote="binonunquineist"]Haha...what state are you from?[/quote]
Connecticut, I saw some schools at chapter with T-Shirts but they had their schools name on them, it said something like "St. Bernards, Strength in Numbers"

[quote="Captain Sugar"]calculators. i wish we got calculators worth $\$$100.
have you ever heard of Sugarton, Texas? it's supposed to have really great schools.  i met some people at a summer camp in North Carolina.[/quote]
Are you sure you're not talking about Sugarland?

[quote="Captain Sugar"]calculators. i wish we got calculators worth $\$$100.
have you ever heard of Sugarton, Texas? it's supposed to have really great schools.  i met some people at a summer camp in North Carolina.[/quote]

for making it to nats u get a 100$$$ calc i think...

So, you get $100 dollars for state and $100 dollars for countdown, and then $100 dollars for nats
that is different from $100 dollars for nats


function
ratio
geometry
geometric transformation
reflection
rotation
homothety

[b][u]Theorem 1.[/u][/b] $(O)$ is a circle and $A_{1},A_{2},...,A_{n}$ are collinear points in the plane. For every $X_{0}$ point on the circle set
\[X_{i+1}\in{X_{i}A_{i+1}}\cap{(O)}\] ($X_{i+1}\cap{X_{i}}=\emptyset,i=1,2,...,n-1$).
If there exist at least one $X_{0}$ point on the circle such that $X_{n}=X_{0}$, then the same is true for every point of the circle.

P.S. It is suggested above that $X_{0}$ does not collinear with $A_{i}$s and that $X_{i}A_{i+1}$s do not touch the circle.

Dear Mathlinkers,

I would like to hear comments and information about this nice and useful theorem. If you know any source where you have already saw this theorem please let me know. I suggested this former hypothesis about one year ago and only recently found enough time to prove it(though I can't say that the proof is difficult).

Also I would like to note that it is rather easy to prove some famous facts such as Pascal's or Butterfly's theorems(and not only) using the one above.

In the end here is an illustration for the theorem in the particular $n=8$ case.

Tiks , thats true and the proof is very neat, its absoloutly beautiful.

Generally if a function from a line( or a circle ) to itself holds cross ratio it has at most $2$ fixed point otherwise its identity.
I think grobber use to call it involution.

So you need to prove your function preserve the cross ratio(how :wink: )

Have a look here:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=circles+lots&t=5860

http://www.mathlinks.ro/Forum/viewtopic.php?t=113361

Dear Lomos, thank you for your reply! :)

Firstly,I would like to know if you proved that my function preserves the cross-ratio.

Secondly,how are you going to show that there are more than 2 fixed points on the circle in case when the line $A_{1}A_{n}$ do not intersects the circle or when $n$ is odd?

In fact my solution is mostly synthetical  :P :D,though it has some boring work with various cases to make sure that it works for all of them(it is always so for that kind of solutions  :oops: ) .

In the end I would like to know if you have heard about this kind of theorem before or its the first time you see it?

P.S. One more note: the theorem also holds when we replace circle with angle.(Like in the Pascal's and Papu's theorems)

[quote="Tiks"]
Firstly,I would like to know if you proved that my function preserves the cross-ratio.
[/quote]

Your function is projecting a circle on itself by couple of points.
Such functions  preserves the cross-ratio.

[quote]Secondly,how are you going to show that there are more than 2 fixed points on the circle in case when the line $A_{1}A_{n}$ do not intersects the circle or when $n$ is odd?[/quote]

Let $f(x_{0})=x_{n}$
then by your hyphothesis $f(x_{0})=x_{0}$ which leads to
$f(x_{i})=x_{i},1 \leq i \leq n$. which means $x_{i}$ are all fix points for $f$, and since there are at least $3$ of them then $f$ is identity which means for any point like $p$ on circle $f(p)=p$.

[quote] I would like to know if you have heard about this kind of theorem before or its the first time you see it?[/quote]

I have seen it in a couple of books. you can find it in

Section 5.
[i]Geometricheskie preobra zovanita III (Russian)[/i]
Or
[i]Geometric Transformations III (translated to English) [/i]
By I.M Yagolom

[quote]In fact my solution is mostly synthetical,though it has some boring work with various cases to make sure that it works for all of them(it is always so for that kind of solutions )[/quote]
synthetical ?Its gonna be really cool if you can proof it synthetical.

[quote="lomos_lupin"]
I have seen it in a couple of books. you can find it in

Section 5.
[i]Geometricheskie preobra zovanita III (Russian)[/i]
Or
[i]Geometric Transformations III (translated to English) [/i]
By I.M Yagolom

[/quote]

Dear Lomos,are you sure  that there is such kind of book you mentioned above?
If I am not mistaken there are only two toms of that book
Geometricheskie preobrazovania 1 i Geometricheskie preobrazovania 2.

Here is a link to the second book
[url]http://ilib.mccme.ru/djvu/yaglom/tom2.htm[/url]
maybe you ment that one?
If yes please tell me exactly(or approximatly) the location of the theorem in the book ,if you could.

Thank you!

Tiks, I am very interested in your syntetical solution, too!

Could you write it, please?

[quote="lomos_lupin"]The persian edition i have is in $4$ books.

The first one discusses Movement, Reflection,rotation,...
The second one discusses normal and spirual Homothecy
The third one discusses the projective geometry
the fourth one discusses Inversive geometry.

In the thrid one(one about projective geometry), Chapter one, section 5, projective transformaiton of line and circle, problem 77, discusses this idea.

But i am afriad, the book has been edited in this persian edition.

Let me know if you can find it there.[/quote]

Dear Lomos,

Thank you very much for this information!

 I eventually was able to find that book (the first edition, in Russian) in our national library, and actually I realized that this theorem is a rather simple one from the point of view of projective geometry. In fact, I really faild to learn that beautiful theory of my beloved Geometry ,but I hope  I will fix that mistake soon :) .

About my synthetical(more or less) solution, I will post it here soon, very soon dear $edriv$ :wink: .

I think your problem is true for all conic and circle is a special case.I am not sure about solution but I think this is a problem of projective geometry and projective coordinate is useful tool to solve this problem (with conic). I will try with this direct.
This problem make me to remember a quite similar problem with circle:

Given two circle $(O_{1}),(O_{2})$ and $(O_{1})$ contain $(O_{2})$ if exist $n$ point $A_{1},..,A_{n}$ on $O_{1}$ such that $A_{i}A_{i+1},\ (i=\overline{1,n},n+1\equiv 1)$ are tangent of $O_{2}$ then that property is true for all points of $(O_{1})$.

And really this property is true for all conic, circle is only special case.


abstract algebra
number theory proposed
number theory

$x,y,z$ E N such that $(x-y)(y-z)(z-x)=(x+z+y)$ . Prove that $x+z+y$ is divisible by $27$

First, suppose that two of the numbers $x$, $y$, $z$ have the same residue module $3$,
$x+y+z=(x-y)(y-z)(z-x)\cong 0 mod(3)$
then the three of then have the same residue module $3$, and
$x+y+z=(x-y)(y-z)(z-x)\cong 0 mod(27)$
Now, if they have distinct residues module $3$,
$x+y+z\cong 0+1+2 mod(3)\cong 0 mod(3)$
but
$(x-y)(y-z)(z-x)\ncong 0 mod(3)$


floor function
quadratics
inequalities
function
calculus
derivative
algebra

1-3. $\displaystyle\text { If there are infinite number of m and n satisfying the following equation, }$
$\displaystyle\text { then find a. }$

$\displaystyle \frac{m^2 - n^2 - 6n - 14}{m + n + 2} = a$

1-4. $\displaystyle\text { Find the closest integer to the following:}$

$\displaystyle\sqrt{13} + \sqrt{15} + \sqrt{17} + \sqrt{19} + \sqrt{21} + \sqrt{23}$

1-5 $\displaystyle\text { If the root of the following equation is a, }$
$\displaystyle\text { find } a^2$

$a\not= 1$
$\displaystyle \lfloor {x} \rfloor+ \frac{2004}{\lfloor {x} \rfloor} = x^2 + \frac{2004}{x^2}$

[color=blue][b](3)[/b][/color] Rewriting the equation in a way to leave no doubts:

$\text{floor}(x) + \frac{2004}{\text{floor}(x)} = x^2 + \frac{2004}{x^2}$

Let $k = \text{floor}(x)$ be an integer. Clearly, $k > 0$ and by the problem condition, also $k > 1$. Just from looking at the resulting quadratic equation for $x^2$, it always has 2 roots $x^2 = k$ and $x^2 = \frac{2004}{k}$. These 2 roots are kind of equivalent: either $x^2 = k$ and $\frac{2004}{x^2} = \frac{2004}{k}$, or $x^2 = \frac{2004}{k}$ and $\frac{2004}{x^2} = k$.  It is then evident that for $k > 1$, the original equation can hold iff

$\text{floor}(\sqrt k) = \frac{2004}{k}$

Since the left side is an integer, the right side must also be an integer and consequently, $k$ must be a divisor of 2004. Factoring this integer into primes,

$2004 = 2^2 \cdot 3 \cdot 167$

Even without trying the divisors one by one in the increasing order, it is clear that $k = 167$ is the only divisor of 2004 with this property:

$12^2 = 144 < 167 < 169 = 13^2,\ \ \text{floor}(\sqrt{167}) = 12$

and

$\frac{2004}{167} = 12$

As a result, $a = k = \sqrt{167}$ is a root of the original equation and the answer to the problem is $a^2 = 167$.

[quote="probability1.01"]I hope there is a quicker way for #2[/quote]
me too  :D
and i just realised my right boarder was below 25,5 already, so i found the complete answer :)

[color=blue][b](2)[/b][/color] The expression can be written as

$\sqrt{13} + \sqrt{15} + \sqrt{17} + \sqrt{19} + \sqrt{21} + \sqrt{23} =$

$= \sqrt{18 - 5} + \sqrt{18 - 3} + \sqrt{18 - 1} + \sqrt{18 + 1} + \sqrt{18 + 3} + \sqrt{18 + 5} =$

$= \left(\sqrt{18 - 1} + \sqrt{18 + 1}\right) + \left(\sqrt{18 - 3} + \sqrt{18 + 3}\right) + \left(\sqrt{18 - 5} + \sqrt{18 + 5}\right)$

The upper bound is easy to find: Let $n > k > 0$ be 2 positive integers.

$\sqrt{(n - k)(n + k)} = \sqrt{n^2 - k^2} < n$

$n - k + n + k + 2\sqrt{(n - k)(n + k)} < 4n$

$\left(\sqrt{n - k} + \sqrt{n + k}\right)^2 < (2 \sqrt n)^2$

Since both expressions under the square root are positive, we can take square roots of both sides of the inequality:

$\sqrt{n - k} + \sqrt{n + k} < 2 \sqrt n$

Using the result for the original expression,

$\left(\sqrt{18 - 1} + \sqrt{18 + 1}\right) + \left(\sqrt{18 - 3} + \sqrt{18 + 3}\right) + \left(\sqrt{18 - 5} + \sqrt{18 + 5}\right) <$

$<\ 6 \sqrt{18} = 18 \sqrt 2 \doteq 25.456$

The lower bound is more difficult and I am going to do it with calculus. The original expression is written as

$\sqrt{18}\left[\left(\sqrt{1 - \frac{1}{18}} + \sqrt{1 + \frac{1}{18}}\right) + \left(\sqrt{1 - \frac{3}{18}} + \sqrt{1 + \frac{3}{18}}\right) + \left(\sqrt{1 - \frac{5}{18}} + \sqrt{1 + \frac{5}{18}}\right)\right]$

Consider the function $f(x) = \sqrt{1 - x} + \sqrt{1 + x},\ x \in (-1, 1)$. If $x$ is small, the function $f(x)$ can be approximated by the McLaurin series (Taylor series at $x = 0$):

$f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f''''(0)}{4!} x^4 + ... $

$f(0) = 2$

$f'(x) = -\frac 1 2 \left(\frac{1}{\sqrt{1 - x}} - \frac{1}{\sqrt{1 + x}}\right),\ \ \ f'(0) = 0$

$f''(x) = -\frac 1 4 \left(\frac{1}{(\sqrt{1 - x})^3} + \frac{1}{(\sqrt{1 + x})^3}\right),\ \ \ f''(0) = -\frac 1 2$

$f'''(x) = -\frac 3 8 \left(\frac{1}{(\sqrt{1 - x})^5} - \frac{1}{(\sqrt{1 + x})^5}\right),\ \ \ f'''(0) = 0$

$f''''(x) = -\frac{15}{16} \left(\frac{1}{(\sqrt{1 - x})^7} + \frac{1}{(\sqrt{1 + x})^7}\right),\ \ \ f''''(0) = -\frac{15}{8}$

Hence,

$f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + R_4 = 2 - \frac{x^2}{4} + R_4$

The remainder $R_4$ is equal to

$R_4 = \frac{f''''(\alpha)}{4!} x^4$

where $\alpha \in (0, x)$ is unknown. Since $f'(0) = 0$ and $f''(0) < 0$, the function $f(x)$ has a maximum at $x = 0$. From the calculation of the derivatives, it is clear that all odd derivatives are equal to zero at $x = 0$ and all even derivatives are negative at $x = 0$. Hence, $f''''(x) < 0$ also has a maximum at $x = 0$. For $x < \frac 1 3$, the absolute value of the remainder $R_4$ is then bounded by

$f''''\left(\frac 1 3\right) = -\frac{15}{16} \left[ \left(\sqrt{\frac 3 2}\right)^7 + \left(\sqrt{\frac 3 4}\right)^7\right] = -\frac{15}{16} \cdot\frac{27 \cdot \sqrt 3 (1 + 8 \sqrt 2)}{128}$

$|R_4| < \frac{\left|f''''\left(\frac 1 3\right)\right|}{4!} x^4 = \frac{5 \cdot 27 \sqrt 3 (1 + 8 \sqrt 2)}{2^{14}} x^4 = q x^4$

As a result, for $x < \frac 1 3$,

$\left|\sqrt{1 - x} + \sqrt{1 + x} - 2 + \frac{x^2}{4}\right| < q x^4$

Applying this reult to the problem with $x = \frac{1}{18}$ or $\frac{3}{18}$ or $\frac{5}{18} < \frac 1 3$, we get

$\left|\sqrt{1 - \frac{1}{18}} + \sqrt{1 + \frac{1}{18}} - 2 + \frac{1}{4 \cdot 18^2}\right| < \frac{5q}{18^4}$

$\left|\sqrt{1 - \frac{3}{18}} + \sqrt{1 + \frac{3}{18}} - 2 + \frac{9}{4 \cdot 18^2}\right| < \frac{81q}{18^4}$

$\left|\sqrt{1 - \frac{5}{18}} + \sqrt{1 + \frac{5}{18}} - 2 + \frac{25}{4 \cdot 18^2}\right| < \frac{625q}{18^4}$

$\sqrt{18} \left|\sqrt{1 - \frac{1}{18}} + \sqrt{1 + \frac{1}{18}} + \sqrt{1 - \frac{3}{18}} + \sqrt{1 + \frac{3}{18}} + \sqrt{1 - \frac{5}{18}} + \sqrt{1 + \frac{5}{18}} - 6 + \frac{35}{4 \cdot 18^2}\right| <$

$<\ \frac{707q \sqrt{18}}{18^4}$

$\frac{707q \sqrt{18}}{18^4} = \frac{707 \cdot 5 \sqrt 3 (\sqrt 2 + 16)}{2^{16} \cdot 18^2} < \frac{3535 \sqrt 3}{2^{16} \cdot 18} < \frac{100 \sqrt 3}{2^{15}} \doteq 0.0053$

$\sqrt{18} \left(6 - \frac{35}{4 \cdot 18^2}\right) = \sqrt{2} \left(18 - \frac{35}{432}\right) = \sqrt 2 \left(18 - \frac{1}{12} + \frac{1}{432}\right) \doteq$

$\doteq \sqrt 2 \left(18 - 0.0833 + 0.0023\right) \doteq  17.9190 \sqrt 2 \doteq 25.3413$

$\left|\sqrt{13} + \sqrt{15} + \sqrt{17} + \sqrt{19} + \sqrt{21} + \sqrt{23} - 25.3413\right| < 0.0053$

$25.3360 < \sqrt{13} + \sqrt{15} + \sqrt{17} + \sqrt{19} + \sqrt{21} + \sqrt{23} < 25.3466$

Wow. That was some serious overkill :P

By my random guestimating I got about 25.3, which is closer than I thought :lol:


[quote="DIMITRIS T"]how can we proove that between 2 real numbers always there is a rational number :?:[/quote]

Actually, for most it is intuitive.

I don't know if this level proof actually belongs here.

[hide="Solution"]Let the real numbers be $ x$ and $ y$, and assume WLOG $ x < y$. Also, assume at least one of $ x$ and $ y$ is irrational; if both are rational, then $ \displaystyle\frac{x\plus{}y}{2}$ is valid.

Consider the decimal expansions for $ x$ and $ y$. Since they are not equal, there must exist at least one digit where they differ. Let the numerical digit of $ x$ be $ a$, and the numerical digit of $ y$ be $ b$. The rational number matches the digits of $ x$ and $ y$ up to this difference in $ a$ and $ b$. If $ y$ is irrational, then append $ b$ followed by infinitely many zeroes. If $ y$ is rational, then append the digit $ a$, followed by either the digits of $ x$ or the digits of $ x$ incremented by $ 1$ if and only if the digit is not $ 9$. If all the digits are $ 9$, then the number is rational, so $ y$ must be irrational; contradiction. Since $ b > a$, this constructed number is less than $ b$, yet greater than $ a$.[/hide]

thank you very much!!!
i looked for a proof without using archemedian property,that's why i posted this topic in this forum


Hi folks I'm new to this site - and it seems a lot of fun.

I would also like to tell you about another site like this one, created especially for former, current of future IMO contestants (International Mathematical Olympiad). Lots of coutries are represented there, we have about 700 topics and 2900 articles posted.

Try it out! Also a weekly MathContest - USAMO like - questions.

the site-s address is [url=http://www.mathlinks.ro]www.mathlinks.ro[/url]

With regards,
Valentin Vornicu

PS i know this is not a nice way to get someone's attention, but I hope that you won't mind. [/url]

:mrgreen: This is just so...de javu :mrgreen:
We were planning to do the same thing... :mrgreen:

Anyways, welcome to AoPS, I guess. :mrgreen:

(In the background)[size=75]Mr. Rusczyk! We got competition![/size][size=59]And their site rocks!...I mean What shall we do? :mrgreen:[/size]

P.S. I think this should be in the "Other Problem Solving Topics Forum" or the infamous "Etc. Forum" because this isn't exactly a math problem.

wow - didn't expect this - I was rather expecting my username / post deleted  :)

you can post about your site there :D

and well it's not quite competition, cuz we mostly deal with USAMO - IMO level questions (i mean at least that's what we want to do :D)

Mr. Ruskzyk is a kinf, compationate, forgiving, wide-hearted, overall great person who is respected through out this community, and I'm sure he won't do that.

...I'm sure he'd ask permission before advertising, too, but oh well.
...*sigh* I believe the first rule of posting is "If you think it's going to get deleted, don't post it" and I'm sure you heard of that because you're an adult.
With that said, *sigh* I know I'm sounding mean, but I didn't quite expect this either...

P.S. This site is the property of AoPS Incorporated, which is owned by our most respected Mr. Rusczyk. (Also Mr. Crawford, see [url=http://www.artofproblemsolving.com/AboutUs/AoPS_A_Company.php]here[/url] for details.)

once again sorry for "breaking in" like this - I have posted about your forum on mine (sound weird :D)

regards,
Valentin

Thank you. (course I sound weird too because I'm just a lowly comoderator responding to an Administrator leveled topic...)

By no means is your site competition, at least in my eyes. The web has a shortage of great math sites unfortunately, and I think we would all be very happy to see more. Thank you very much for sharing another excellent math site!


vector
superior algebra
superior algebra unsolved

Let A be an unitary ring.
a) If a,b are in A and ab in U(A) does it result that a,b are in U(A)?
b) If $ a$ in A has the property that there exists $ b$ in A s.t. $ ba\equal{}1$ and there is no $ c$ in A s.t. $ ac\equal{}0$ then $ a$ is in U(A).

1) not in general. 2) you have to assume $ c \neq 0$. $ ba\equal{}1$ implies $ (ab\minus{}1)a\equal{}0$, hence $ ab\equal{}1$.

a) If $ a\not\in U(R)$ has a right-inverse, then $ ab\equal{}1\in U(R)$ for some $ b\in R$. Contradiction!

For example take some vector space $ V$ over some field $ F$ with an infinite basis $ B$.
For $ b\in B$ take some bijection $ \sigma: B\to B\setminus \{b\}$ and also $ \tau: B\to V,b\mapsto 0,c\mapsto \sigma^{\minus{}1}(c)$ for $ c\in B\setminus \{b\}$.
Then $ \sigma,\tau$ lift uniquely to $ F$-endomorphisms $ f,g: V\to V$ with $ g\circ f\equal{}\text{id} _V$
and neither $ f,g$ are invertible. Here we take $ R: \equal{}\text{End}_F(V)$ the ring of all $ F$-endomorphisms of $ V$, i.e. all $ F$-linear maps.

b) If $ ba\equal{}1$, then $ (ac\equal{}0\Rightarrow c\equal{}bac\equal{}0\ \forall c)$.
I think you want $ ca\not\equal{}0$ for all $ c\not \equal{}0$.
Then $ ba\equal{}1$ gives $ (ab\minus{}1)a\equal{}0$, so $ ab\equal{}1$ and $ a$ has inverse $ b$.


You run a boarding house that has 90 rooms.  You have 100 guests registered, but on any given night only  90 of these guests actually stay in the boarding house.  Each evening a different random set of 90 guests will show up.  You don't know which 90 it will be, but they all arrive for dinner before you have to assign rooms for the night.  You want to give out keys to your guests so that for any set of 90 guests, you can assign each to a private room without any switching of keys.

What is the smallest number of keys necessary so that each guest will have a key to a private room?  Describe how you would distribute these keys and assign the rooms.  Prove that this is the smallest number of keys.

This is pretty hard....

Has anyone attempted this?  Does anyone have some cases or interesting hints to give to others?

[color=cyan]Oh, I just figured out what this question was asking on third reading.  So you give out multiple copies of every room key, and teh question is how many do you need to give out to be assured that no matter what 90 people show up, they each can use their keys to let themselves into a room that no one else will need to use.  That was a very inelegant way of phrasing it . . .

I haven't actually tried it yet, having just understood it, but I'll get back to you.  I've seen related questions, though, involving a bank safe or something.[/color]


[hide]The only number I can think of is 45.[/hide]

[quote="236factorial"]Let a%b be the least common multiple of a and b. What is the sum of all the natural number values of x such that 15%x=45?[/quote]

[hide="Answer"]  Finding the lowest common multiple involves cancelling out common prime factors the multiplying the other number with the primes leftover in each numebrs prime factorization.

$\frac{45}{15}=3$  Thus the factorization of the other number must have one more 3, in other words it must have two 3's in its prime factorization.  It can have a 5 or it can not, regardless there lowest common multiple is still 45.  The 3's have been taken care for, but to have any other prime factors in X's factorization would result in a lowest common multiple greater than 45.  So our only values for x are:

$ 3 \cdot 3 = 9$

$ 3 \cdot 3 \cdot 5=45$

  And there sum is: $45 + 9 =\boxed{54}$ [/hide]


abstract algebra
real analysis
real analysis solved

I know that the solutions for 6^x+3^x=4^x+5^x are x=0 and x=1. I don't view what conection is between Lagrange's theorem and this equation, becaue this equation is exercise for Lagrange's theorem. How can I prove that  x=0 and x=1 are the only solutions ?

Rewrite the equation as f(6)-f(5) = f(4)-f(3) , where f(t) = t^x . Now , by Lagrange's theorem :

[f(6)-f(5)]/(6-5) = f'(z1) , z1 in (5,6)
[f(4)-f(3)]/(4-3) = f'(z2) , z2 in (3,4)

f'(z1) = f'(z2) => x z1^(x-1) = x z2^(x-1) . If t =0 , it holds , so x=0 is a sol . The same if x=1 . Otherwise we get (z1/z2)^(x-1) = 1 , but they are from distinct intervals , that can't be true!

be f(y)=yx,x constant, we can write equation like 6x -5x =4x -3x, so f(6)-f(5)=f(4)-f(3) with Lagrange's theorem there are c1 \in (5,6) and c2 \in (5,6), and f'(c1)=f'(c2)  <=> x*c1x-1=x*c2x-1 => x=0 so x=1 are the only solutions.

What's your point  :?  ?! You repeated the same thing all over again ..


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Let $ F$ be a field and $ \alpha$ with the irreducible polynomial $ g$ over $ F$. If none of the remaining roots of $ g$ are in $ F(\alpha)$ does that mean that $ Gal F(\alpha)/F$ has order 1?

No, it means that the extension isn't Galois, and has no associated Galois group. The splitting field of $ g$ does have a Galois group, and its order is larger than the degree of $ g$.

This books calls it Galois Group but I see now that I actually meant $ Aut(F(\alpha)/F)$. So what is the order of this group?

It's pretty normal notation to refer to the automorphism group as the Galois group.

What's troubling you about the question, bike_me?  Something in the automorphism group must fix $ F$, that's a given.  Where could the automorphism send $ \alpha$?

It could send $ \alpha$ to any of the roots of $ g$ that are in $ F(\alpha)$ and since $ \alpha$ itself is the only one, the only element of the group will be the identity function. If I'm wrong please correct me.

[quote="ZetaX"][quote="LydianRain"]It's pretty normal notation to refer to the automorphism group as the Galois group.[/quote]
Doubt that  :wink:[/quote]
I think I've seen it in at least two texts...

I wouldn't call that a "pretty normal notation" then. Well, about one third of people seem to use this (here on the forum), but I don't see what that notations is supposed to help. Essentially, I don't care as long as people know they are dealing with multiple definitions (and hopefully meantion them when posting their problems).

Hmm, I've almost never actually heard anyone refer to it as anything other than the Galois group.

What about "automorphism group"¿  :wink:

Ive heard that theres a convention for referring to $ Aut(E/F)$ as the Galois group only when $ E$ is a galois extension of $ F$, in any other case it is called the automorphism group


:D let a,b,c$ \ge$ 0 such as: X=a+b+c. and a,b,c$ \le$ $ \frac{X}{2}$

compare A and B if:

A=($ \frac{X}{2}$ -a)($ \frac{X}{2}$ -b)($ \frac{X}{2}$ -c)

B=abc

but you must CAREFULLY  :D  :D  :)  :D

[quote="trannguyenquyND_VN"]:D let a,b,c$ \ge$ 0 such as: X=a+b+c. and a,b,c$ \le$ $ \frac {X}{2}$

compare A and B if:

A=($ \frac {X}{2}$ -a)($ \frac {X}{2}$ -b)($ \frac {X}{2}$ -c)

B=abc

but you must CAREFULLY  :D  :D  :)  :D[/quote]

$ B\ge\ 8A$

PS:ra chat nhe

$ A\equal{}\frac{1}{8}(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)$.
Now, $ (b\plus{}(c\minus{}a))(b\minus{}(c\minus{}a))\equal{}(b^2\minus{}(c\minus{}a)^2)\le b^2$.
Similarly getting the other three relations and then taking the product yields the result:
$ A\le\frac{1}{8}abc$
giving $ B\ge8A$. :)
PS: $ (b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)\le abc$ is just a form of the schur's inequality of $ r\equal{}1$... I forgot if it is BDT schur or not....... :rotfl:

[quote="Potla"]$ A \equal{} \frac {1}{8}(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)$.
Now, $ (b \plus{} (c \minus{} a))(b \minus{} (c \minus{} a)) \equal{} (b^2 \minus{} (c \minus{} a)^2)\le b^2$.
Similarly getting the other three relations and then taking the product yields the result:
$ A\le\frac {1}{8}abc$
giving $ B\ge8A$. :)
PS: $ (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)\le abc$ is just a form of the schur's inequality of $ r \equal{} 1$... I forgot if it is BDT schur or not....... :rotfl:[/quote]
Yes ,this inequality very easy :)


Refer to the[url=http://www.geocities.com/mario_the_sheep/geom.JPG][u]Given Figure [/u][/url]

W(a;b) is the centre of the circle.
TE is a tangent to the circle at E, which lies on the y-axis.
The circle touches the x-axis at D.
The equation of TE is $y = \frac{3}{4}x - 9$

Calculate the values of a and b.

[hide]First, we seek the coordinates of $E$, which are $(0,-9)$
Let's call the intersection of $TE$ and the x-axis $P$. coordinates $(12,0)$
Now we can say that $PE = PD = 15 \rightarrow$ coordinates of $D$ are $(-3,0)$
$\rightarrow$ $a=-3$
Now we construct the line through $E$, perpendicular (I don't know if this is the correct word, the symbol is $\bot$) to $TE$, and we call it $\psi$ (I just love that letter  :D )
We seek the intersection of $\psi$ and $x = -3$
$\psi = \frac{-4x}{3} - 9$
Intersection = $(-3,-5)$ so $a = -3$ and $b = -5$ [/hide]
Correct?

yep, nice,  although i did it slightly differently,
[hide]
Find E -> (0,-9)
Find line WE equation -> $y = \frac{-4}{3}x - 9$
Use distance formula for WE which gives us -> $WE^2 = a^2 + 81 + b^2 + 18b$

sub (a,b) into WE line equation -> $3b = -4a - 27$

Then the equation of the circle would be -> $(x-a)^2+(y-b)^2 = radius^2$
and since "The circle touches the x-axis at D" .'. $radius^2 = b^2 = WE^2$

.'. we have $3b = -4a - 27$
    and  $b^2 = a^2 + 81 + b^2 + 18b$

Then we solve simulataneously to reveal,
 $a = 27, b = -45$ OR $a = -3, b = -5$

And we know that (27,-45) is'nt a valid solution looking at the picture
.'. $a =  -3$
    $b = -5$

[/hide]

The point E is given by $(0, -9)$
Hence the circle that touches $3x-4y-36=0$ at E is given by  ${{x}^{2}}+{{(y+9)}^{2}}+2k(3x-4y-36)=0$.
It can be written as ${{x}^{2}}+{{y}^{2}}+6kx+(18-8k)y+81-72k=0$.
If this circle touches x-axis, we have $8{{k}^{2}}=81-72k\Rightarrow {{k}^{2}}+8k-9=0\Rightarrow k=1,\,-9$
Hence the equation to the required circle is ${{x}^{2}}+{{y}^{2}}+6x+10y+9=0$ or ${{x}^{2}}+{{y}^{2}}-54x+90y+\text{729}=0$
Hence the possible values for (a, b) are $(-3, -5)$ ot $(27, -45)$


algorithm
number theory
greatest common divisor
Euclidean algorithm

Hi,

I was wondering, what "things" you can do with the euclidean algorithm, that is, what rules are there and what can you not do. Just for example, can you do

$ GCD(a,b) \equal{}GCD(a, a \cdot b)$

Can you do $ GCD(a^2, a\plus{}b) \equal{} CGD(a^2, a(a\plus{}b)) \equal{} GCD(a^2,a^2 \plus{} b^2 \minus{}a^2)$?

So what limits are there with this kind of stuff. I would gratefully appreciate any help please =) Thankyou

The most general transformation the Euclidean algorithm permits you to do is $ \gcd(a, b) \equal{} \gcd(ax \plus{} by, az \plus{} bw)$ where $ xw \minus{} yz \equal{} \pm 1$.

Would it be possible to do the ones i mentioned?

None of those are valid, as you can check using small counterexamples.  I don't even know what you're doing in the third one.

But in a question i posted recently the following solution was given.

By the Euclidean Algorithm,

$ \gcd(n^2 + 4,n + 5) \& = \gcd\left(n^2 + 4 - (n^2 + 5n),n + 5\right) \\
=\gcd(-5n+4,5n+25)\\  =\gcd(29,5n+25)$

Could you please explain to me how this is possible?

Try the first IMO problem (not for this year, I mean the first ever...)  :D

[quote="t0rajir0u"]In the first manipulation an arbitrary multiple of one of the numbers was subtracted from the other.  This is perfectly fine and not what you did.  In the second step this was simplified and in the third step one of the numbers was added to the other.

Do you understand the proof of the Euclidean algorithm?  It would probably help.[/quote]

Okay, so correct me if im wrong.

The Euclidean algorithms proof is essentually because

If $ a|b$ and $ a|c$ then $ a|(b\minus{}c)$ and $ a|(b\plus{}c)$

Now the aforementioned statement can be proven as such

if $ a|b$, then $ b \equal{} a \cdot k$ for some integer $ k$

if $ a|c$, then $ c \equal{} a \cdot j$ for some integer $ j$.

$ b \minus{} c \equal{} a \cdot k \minus{} a \cdot j$

$ \equal{} a(k \minus{} j)$

Hence $ a| a(k\minus{}j) \implies a|(b\minus{}c)$

So we can take another example:

If we multiply $ b$ by any random integer $ n$, then

$ n \times b \equal{} n \times (a \cdot k)$

So as $ a| (n\cdot b)$ and $ a|c$, then

$ \boxed{a|(n\cdot b \minus{} c)}$

So is the boxed statement true?

So i can say,

$ GCD(b,c) \equal{} GCD( (n\cdot b)  \minus{} c, c)$  ?

[quote="kurt.math"][quote="t0rajir0u"]In the first manipulation an arbitrary multiple of one of the numbers was subtracted from the other.  This is perfectly fine and not what you did.  In the second step this was simplified and in the third step one of the numbers was added to the other.

Do you understand the proof of the Euclidean algorithm?  It would probably help.[/quote]

Okay, so correct me if im wrong.

The Euclidean algorithms proof is essentually because

If $ a|b$ and $ a|c$ then $ a|(b \minus{} c)$ and $ a|(b \plus{} c)$

Now the aforementioned statement can be proven as such

if $ a|b$, then $ b \equal{} a \cdot k$ for some integer $ k$

if $ a|c$, then $ c \equal{} a \cdot j$ for some integer $ j$.

$ b \minus{} c \equal{} a \cdot k \minus{} a \cdot j$

$ \equal{} a(k \minus{} j)$

Hence $ a| a(k \minus{} j) \implies a|(b \minus{} c)$

So we can take another example:

If we multiply $ b$ by any random integer $ n$, then

$ n \times b \equal{} n \times (a \cdot k)$

So as $ a| (n\cdot b)$ and $ a|c$, then

$ \boxed{a|(n\cdot b \minus{} c)}$

So is the boxed statement true?

So i can say,

$ GCD(b,c) \equal{} GCD( (n\cdot b) \minus{} c, c)$  ?[/quote]

No, you cannot say that $ GCD(b,c) \equal{} GCD( (n\cdot b) \minus{} c, c)$.

To see this more clearly, you can set $ b\equal{}xd$ and $ c\equal{}yd$ , where $ GCD(x,y)\equal{}1$ .we have that

$ nxd\minus{}yd\equal{}d(nx\minus{}y)$.But  $ nx\minus{}y$ can share a common divisor with $ y$.This happens because $ y$ can divide $ n$ ,for example.

More easy to understand , pick $ b\equal{}6, c\equal{}4, n\equal{}2$.

Okay, so could you please explain this to me:

Assume $ p|(8n \plus{}2)$ and $ p|(6n \plus{} 1)$. Then $ p|(8n \plus{} 2) \minus{} (6n \plus{} 1) \equal{} 2n \plus{} 1$ and therefore $ p|3(2n \plus{} 1) \minus{} (6n \plus{} 1) \equal{} 2$. Since 2 does not divide $ 6n \plus{} 1$, we obtain $ p\equal{}1$.

What's to explain?  Each step is correct.

The reason your other statement is false is that $ a | nb$ doesn't imply $ a | b$; in other words, to make a statement about $ \gcd$ all of your implications need to be invertible, whereas this is not necessary for the example in your post above.  I think you're confused partially because it's hard to think about all these variables; plug in some numbers and you'll quickly see that what you claimed is false.

The correct statement is that $ \gcd(b, c) \equal{} \gcd(b \minus{} nc, c)$.  See the difference?

[quote="kurt.math"]So we can take another example:

If we multiply $ b$ by any random integer $ n$, then

$ n \times b \equal{} n \times (a \cdot k)$

So as $ a| (n\cdot b)$ and $ a|c$, then

$ \boxed{a|(n\cdot b \minus{} c)}$

So is the boxed statement true?

So i can say,

$ GCD(b,c) \equal{} GCD( (n\cdot b) \minus{} c, c)$  ?[/quote]
That's correct if you didn't forget that you have [u]implicitly assumed[/u] that $ a|b$. Without this assumption, your conclusion is false as shown by others. As t0rajir0u explains, $ a|b\implies a|nb$ clearly holds, but the converse (ie $ a|nb\implies a|b$) does not generally follow.

What is learned here? No you cannot say $ \gcd(b,c)\equal{}\gcd((nb\minus{}c,c)$ in general for all integers $ b,c$ [u]unless[/u] you assume $ a|b$ and $ a|c$, which is pretty much obvious that $ a|nb\minus{}c$. Be clear about your assumptions.

The example you gave on your last post is correct. Notice that 2 is a prime and that's why $ p\equal{}1$.

Two lessons here: you need to keep track of your assumptions and you need to handle well with your variables (define what the variables (integers, primes, etc) are and stick with it).


Some naughty boys from [b]Mr. M[/b]'s neighbourhood play a joke with him. They hide [b]a hat of his under $ 1$ of $ n$ boxes placed in a circle[/b]. [b]Mr.M[/b] now has to check all $ n$ boxes to find out if his hat is there, but each time, he is just [b]allowed to open 4 boxes[/b]. And after each time, the hat will move to $ 1$ of its [b]"neighbour"[/b] boxes ($ 1$ of the boxes which are right next to its old box). Ask how many times at least then [b]Mr.M[/b] will be able to find out his hat? :D

[quote="vinhhop"]It is such an old problem, see [url]http://www.mathlinks.ro/viewtopic.php?t=274838[/url][/quote]
But no solution, can you recommend one :blush: ?

Ok. sorry because of my bad english! my meaning is one problem should be only post at one time...  :roll:


[quote="jhredsox"]What is the chance of it being the day you go into dst or out of dst if you don't know what year it is?
-jorian[/quote]
[hide]There are 97 leap years out of 400 and two days that you can go in or out of dst in. $2(\frac{97}{200*366})+\frac{303}{200*365})$
$\frac{97}{36600}+\frac{303}{36500}$ I don't feel like simplifying[/hide]

[quote="jhredsox"]its a little easier than that

hint:[hide]

just assume there's a 1/4 chance of a leap year

@bpms, where did you get 97/400?! :rotfl: [/hide]

-jorian[/quote]
I got $\frac{97}{400}$ from the fact that a year isn't a leap year when it ends in 00 even though it is divisible by 4 unless it is divisible by 400. Example, $1900 \neq$ leap year. $2000=$ leap year.

[quote="solafidefarms"]What is dst?[/quote]
Daylight savings time. Also, why did you say there will be no marathons for a while?

oh,sry bpms, i knew there was something different about what happened every 100 years, but i didn't know what it was

....so does anyone want to simplify from what bpms is doing?
-jorian

[quote="Ignite168"]The answer is 0.[/quote]
He's right, if you know the date then you're fine :) .

no, im assuming you're picking a random date from a hat (not really, just to clear it up)

.......so does anyone wanna solve?fully?
-jorian

[quote="solafidefarms"]Ah, duh, I'm not terribly quick :).

As for why no marathons? I have clarified in that locked thread.[/quote]
Is that only in mathcounts? I was thinking of starting a trig one in either intermediate topics or AMC...


trigonometry
function
calculus
calculus computations

[b]Show that $ \sin{(x} +  \pi x )$ cannot be periodic with rational period.[/b]

You do realize that's a strictly increasing function?

OK , Then it must be ${\sin(x+ \pi x)} $ !!!


This is probably an easy problem but I still can't figure out how to solve it and need help.

We have 50 integers.

27 of them are divisible by 3.
18 of them are divisible by 5.
16 of them are divisible by 6.
11 of them are divisible by 10.
7 of them are divisible by 15.
4 of them are divisible by 30.

4 of the integers are not divisible by 2, 3 or 5

How many of the integers are even?

Let $ A_2,A_3,A_5$ be set of integer divisible by $ 2,3,5$ in this $ 50$ integer.
We use formula:
$ |A_2\cup A_3\cup A_5|\equal{}|A_2|\plus{}|A_3|\plus{}|A_5|\minus{}|A_2\cap A_3|\minus{}|A_3\cap A_5|\minus{}|A_5\cap A_3|\plus{}|A_2\cap A_3\cap A_5|$
is OK!

PS: In this $ |A_2\cup A_3\cup A_5|\equal{}50\minus{}4\equal{}46$


modular arithmetic
search
number theory
greatest common divisor
number theory proposed

Let $ p$ be a prime such that $ p^2 \mid (2^{p\minus{}1}\minus{}1)$. Prove that for all positive integers $ n$, $ (p\minus{}1)(p!\plus{}2^n)$ has at least three distinct prime divisor.

[quote="cnyd"]$ p^{2}|2^{p \minus{} 1} \minus{} 1$

if $ q|p \minus{} 1$,$ q$ is prime and $ q > 2$

$ 2^{c}\parallel{}p!$  ,$ c > n$  $ \implies$

$ (p \minus{} 1)2^{n}[2^{c \minus{} n}\frac {p!}{2^{c}} \plus{} 1]$

$ (q,2^{c \minus{} n}\frac {p!}{2^{c}} \plus{} 1) \equal{} (q,2) \equal{} 1$  $ \implies$  $ q.\frac {p \minus{} 1}{q}2^{n}[2^{c \minus{} n}.\frac {p!}{2^{c}} \plus{} 1]$ has at least three distinct prime divisor

if $ c\leq n$  $ \implies$  $ q\frac {p \minus{} 1}{q}.2^{c}[\frac {p!}{2^{c}} \plus{} 2^{n \minus{} c}]$

$ (q,2) \equal{} (q,\frac {p!}{2^{c}} \plus{} 2^{n \minus{} c})$ has at least three distinct prime divisor

if $ p \minus{} 1 \equal{} 2^{k}$,$ p^{2}|2^{p \minus{} 1} \minus{} 1$  $ \implies$

$ (2,p) \equal{} 1$,$ ord_{p}2 \equal{} d$  $ \implies$  $ ord_{p^{2}}2 \equal{} pd$  or $ ord_{p^{2}}2 \equal{} d$

$ ord_{p^{2}}2|p \minus{} 1$  $ \implies$  $ ord_{p^{2}}2 \equal{} d$

$ \implies$  $ ord_{p^{2}}2 \equal{} ord_{p}2 \equal{} d|p \minus{} 1$

$ \implies$  $ 2^{2k}\equiv 1\pmod{p}$  $ \implies$  $ d|2k$

$ p \minus{} 1 \equal{} 2^{k}$,$ p^{2}|2^{2k} \minus{} 1$

$ \implies$  $ p^{2}|p^{2} \minus{} 2p \plus{} 1 \minus{} 1$  $ \implies$  $ p \equal{} 2$  Contradiction because $ 4|2^{1} \minus{} 1$[/quote]
I quite don't understand well your proof. Can you give it more detailed?

[i]Sorry , At present , I'm too busy . I will give some main ideas of the solution

    If nobody can finish it , I will post full solution as soon as possible

   The frist observation is $ gcd( p \minus{} 1 ; p! \plus{} 2^n)$ is of the form a power of $ 2$

   So we only need to prove each number $ p \minus{} 1 ; p! \plus{} 2^n$ have an odd prime factor .

     We will prove that $ p \minus{} 1\ \equal{} \ 2^{2^k} \ ; \ k \in \mathbb{N}$

    and the following lemma will be helpful :

 [b]  Lemma[/b] : Two arbitrary elements in the sequence : $ A_n \ \equal{} \ 2^{2^{n}} \ \plus{} \ 1$ are relative pime :

   ( Prove that lemma by using the identity  $ 2^{2^{n \plus{} 1} } \minus{} 1 \ \equal{} \ \prod_{i \equal{} 0}^{n} (2^{2^i} \plus{} 1)$ :wink: )[/i]

cnyd made  a mistake because 2^k = p-1. so p^2 divides 2^{2^k} -1. cnyd wrote p^2 divides 2^{2k }-1

[quote="cnyd"]
$ ord_{p}2 \equal{} d$  $ \implies$  $ ord_{p^{2}}2 \equal{} pd$  or $ ord_{p^{2}}2 \equal{} d$
[/quote]

Sorry if this is trivial ,but can you tell why  :oops:


geometry
trigonometry
cyclic quadrilateral
geometry unsolved

ABCD is a cyclic quadrilateral then prove that $ AC.BD \ge 2S$

it's used(without proof) here:
http://www.mathlinks.ro/Forum/viewtopic.php?p=512856&sid=5bc64e02c76ce6459d665ac10c94d640#512856

Fourrier wrote:
"If ABCD is cyclic, then AC.BD>=2S".

Proof:
Let E be the common point of AC and BD, then 2S(ABD)=BD.AE.sin(<AED) and 2S(BCD)=BD.CE.sin(<BEC), consequently 2S(ABCD)=AC.BD.sin(<AED). See sin<=1.

Here S(XYZ) means the area of XYZ

Best regards,
Sunken rock

A blind man sees better the details

[quote="darij grinberg"]You don't need $ ABCD$ to be cyclic for this. Generally, if $ ABCD$ is a quadrilateral, and $ \phi$ is the angle between its diagonals $ AC$ and $ BD$, then the area of the quadrilateral $ ABCD$ is $ \frac12\cdot AC\cdot BD\cdot\sin\phi$. Do you see how to prove this in one line?

  darij[/quote]

Yes I see it now

thanks


Here's another one I found a little bit later in the book.

[quote="Page 103"]...in other words, we could start and $ A$ and work our way to $ B$. [/quote]

I suppose this is what was meant to be written:
...in other words, we could start at $ A$ and work our way to $ B$.

[quote="The Screen"]The topic or post you requested does not exist[/quote]

[hide= ][/hide]

[quote="bluecarneal"][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170598[/url]

P.S how do you change the link into something else, like "try here". I know how to in hide tags, but this is my first url post[/quote]

[code][url=YOUR URL]YOUR TEXT[/url][/code]

[quote="myyellowducky82"][quote="r15s11z55y89w21"][quote="Introduction to Counting and Probability, 2nd edition, Page 94"]In this problem that people and desks are distinguishable, but books are not. [/quote]

I believe it was meant to read:
In this problem, people and desks are distinguishable, but books are not.

BTW is this the appropriate thread to post this in?[/quote]
If you're taking the C&P class then you can post it there.[/quote]

he probably hasn't...

i did that is why i put the link. i will put the errors there for r15


calculus
integration
calculus computations

Evaluate $ \int_{\minus{}1}^{1}\frac{x^{2}}{1\plus{}e^{\arctan x}}dx$.

You can apply the well-known relation (prove easily) $ \int_{\minus{}a}^{a}f(x)\ dx \equal{}\int_{0}^{a}\left[f(x)\plus{}f(\minus{}x)\right]\ dx$.

[url=http://www.mathlinks.ro/viewtopic.php?t=18738&search_id=309618549]Take a look here.[/url]

haha nice alternative solution kunny, I've never thought of that before.

thanks everyone


calculus
AMC
AIME
articles
geometry
function
analytic geometry

Hmm... I would first do a self assessment. Ask: Where am I in terms of mathematical ability when it pertains to problem solving skills?

Do some past AMC's and just test yourself and find out what your current potential is. Looking at your post, I would judge that you are beyond getting started. So, work on old AMC, AIME, MAML, etc. Whatever works for you. The more problems you encounter, the more experienced you become as a problem solver.

Other than that I highly suggest reading the AOPS volumes. You should read 1 first because there is some excellent material in there. Then do volume 2. I am reading that currently, even though I don't understand most of it. All you can do sometimes is try your best and just keep challenging yourself.

There is no sure way to make your score go up 20 points overnight, but I would also recommend reading the article on how to reduce stupid mistakes, I think it's by Mr. Rusczyk. Be an avid learner - it's the only way to improve.

Read this article, see if you are a "victim" of the "calculus trap"

[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php[/url]

Then, identify your weakness in problem solving and try to work through the basics.  Given your background, you should be able to catch up quickly.  I would start with the "Art of Problem Solving" books and take some of the online classes this site offerred.

The book you mention is probably more appropriate for someone more experienced in problem solving.  The book is probably meant for students looking to do well on the AIME and make the USAMO.  The AOPS books would probably be better if you lack experience in basic algebra and geometry problem solving.

Given what you describe of your background, I recommend working through the problems in the books by Israel M. Gelfand (my online avatar, no relation to me) and his associates.

http://www.amazon.com/exec/obidos/tg/detail/-/0817636773/

(Algebra $19.95)
http://www.amazon.com/exec/obidos/tg/detail/-/0486425649/
(Functions and Graphs $7.95)

http://www.amazon.com/exec/obidos/tg/detail/-/0486425657/

(Method of Coordinates $6.95)
http://www.amazon.com/exec/obidos/tg/detail/-/0817639144/
(Trigonometry $19.95)

The book republished by Dover called Sequences, Combinations, Limits and edited by S. I. Gelfand (who I think is I. M. Gelfand's son) appears to be part of the same series.

http://www.amazon.com/exec/obidos/tg/detail/-/0486425665/

(Sequences, Combinations, Limits $8.95)

Check these links, I'm pasting in from an old FAQ. These books are excellent. There's a new book in the same series, by other authors, called Lines and Curves.

I think I may've fallen into this trap.
I skipped fifth grade math because I got a perfect score on the pretest for the year.  So in 5th grade I took 6th grade math.  In 6th grade I took Algebra 1.  Then Algebra 2 in 7th.  Trig  in 8th and am now in MACH (Math Analysis Calculus Honors) as my school calls it as a freshman.  It is basically the precursor to AP Calculus which I will take next year.  So, should I go on through and take Calc 2 at a university? Or should I take a math elective like AP Stats at my school instead?

Thanks

DO NOT TAKE AP STATS!!!

huge waste of time

Well, you will probably have to take Stats in college.  So taking the AP exam might exempt you from that.  I agree that Stats is pretty boring.  But if you find an open period on your schedule that you I would take Stats.  When I say Stats is boring....I mean relative to other math classes.  Compared to english and history I love Stats :D .  You might also consider self studying stats.  That's pretty much what I'm doing.  If I knew more about stats before I chose to do this at the beginning of the year I would have given it a little more thought, though.  If you are already taking a big schedule and have a bunch of extracurriculars I would give independent studying stats some extra thought.

I am currently and simultaneously taking AP Stat and a graduate level statistics class. I agree that stat is the least interesting of my math classes, but it does beat my non-math classes. The graduate class is much faster; we're going at least three times as fast as in AP. I expect we will start using calculus in two or three classes, as opposed to the calculator's normalcdf and so on.

Hey Curl, welcome.
I think you should find a summer math program. Some of the programs' applications are available now, so you should look into them. After trying their problems, maybe you'll get an idea of what you should be looking into and what is important to learn for competitions.

[quote="curl"]Hey guys,

First of all, I would just like to say that this is a great forum and I wish I had found it earlier! I am a very active math student, both in school and out. I skipped high school algebra and most of precalculus so that I could study calculus independently at my high school (multivariable and real analysis included). However, on this past AMC-12, I answered 14 questions, got 4 wrong, mostly on stupid mistakes and missed qualifying for AIME. Needless to say, I was upset and dissapointed with my score. As a junior and an avid math club attendee, this year has been really lackluster in the math contest department: from MAML and AMC to NEML, my scores have not been good considering the relative easiness of the contests. I guess what I'm trying to say is, by my senior year, I need to be able to do well on these contests. Are there any ways that I could prepare and do well for these contests in the future?[/quote]Your situation is remarkably similar to mine. This year and last (I'm a junior now too) I've been going through the Calculus AB, BC series (probably not the best choice for me..), and next year I will probably be going to OSU to take an honors introductory calculus/analysis series (single and multivariable). I read some math outside of school and have had some experience with number theory and proof writing (last summer at Northwestern). Like you, I also don't do exceptionally well with competition math. I made it to the AIME last year somehow but this year I did worse, despite knowing about 3 times as much math and about math as I did back then. Despite all the reading and the number theory, my problem solving skills for competition math haven't improved much at all, and for me it's very frustrating. I asked for AoPS for my birthday but didn't get it for some reason, so I'll probably need to order it on my own pretty soon. Good luck with your problem-solving..

[quote="curl"]I don't believe I fell for the so called "calculus trap." I'm not actually in a class; I got an independent study in calc, so I basically just review material I'll need for the AP exam and push on through multivariable and real analysis. In my opinion, the "calculus trap" is baloney - there are plenty of intriguing problems in analysis, plenty of great proof to be done. Analysis starts at single variable calculus, after all. It's really really interesting stuff (especially fourier analysis I hear  ;) ). But yea, I guess I do lack understanding of the tricks and strategies of problems solving in geometry, algebra, number theory, and combinatorics.

Does anyone known if "The Art and Craft of Problem Solving" by Paul Zeitz is a good book worth buying?
Ill look into AoPS too of course.[/quote]

I have mixed feelings about the 'calculus trap'.  I consider myself a decent mathematician for my age (I'm 18 now), but I got started on contests pretty late (about the middle of Grade 10), and I did poorly on them at first, but in about a year's time, I moved from scoring ~85 on the AMC 10 (the first contest I ever did) to scoring ~132 on the AMC 12 and 7 or 8 (can't remember) on the AIME in my Grade 11 year, just a year later.

In the math courses I'm taking now in university (Analysis II, Complex Analysis, Discrete Math II, Vector Calculus), I find the problem solving skills I've built up to be invaluable.  It's easy to say that, say, an AMC counting-type problem has nothing to do with problems in analysis and stuff, but the actual type of math is secondary; it's the skills you've built up in solving the counting problem that allow you to see what is significant in the statement of the analysis problems and what kind of approach you might consider.

On the flip side, I think a lot of people are misunderstanding what the calculus trap actually is, when they post links to that article.  It's good to learn calculus, I've used it on contests before, and I've certainly used it to solve olympiad level problems before.  It's always good to learn more math.  But there's no point in learning more 'advanced' math when you're not able to apply the math you know sufficiently.

If you're taking Real Analysis but not able to qualify for the AIME, I suggest you stop with that for awhile and try to build up some basic problem solving skills - it's an investment that will pay off a lot in the future.  Personally, I still like to solve AIME type problems now, because I think that kind of creative problem solving skill has applications in other areas.  I've gotten to the point where I can score ~140 on the AMC 12 and ~10/11 on the AIME, were I to write it again - I'm sure that if I had been at that level a few years ago, I would be far better now, not only at 'problem-solving' type math, but also at things like complex analysis and algebra, because once that foundation is in place, it's that much easier to learn EVERYTHING.

Regarding blahblahblah's very helpful reply, yes, I think what Mr. Rusczyk is driving at in his "[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]The Calculus Trap[/url]" article is not saying that early study of calculus is bad (quite the contrary), but that sticking solely to the standard curriculum in mathematics is not enough. As he says, "the standard curriculum is not designed for the top students." To be a top student of math, you have to learn about very "elementary" principles from multiple angles until they become second nature. Another writing that makes the same point at greater length, a very interesting book, is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0805829091/learninfreed]Knowing and Teaching Elementary Mathematics: Teachers' Understanding of Fundamental Mathematics in China and the United States[/url] by Liping Ma. In that book you'll find some explanations about why so many Chinese and Chinese-American young people are so conspicuously good in math contests: better elementary instruction in math.

I desist - I prefer English Literature to Statistics actually - but you can more probably argue that the two subjects are like comparing apples and oranges, completely different fruits. In any case, I did somewhat fall into the trap and I did calculus for a year or two (or three), but I never actually finished Calculus C.... didn't take the final exam. That was when I was 14-ish. Then Mathscamp happened, and somehow I kept working at my mathematics, doing decently well in AMC (130++) but not AIME (at most 5-7). But one day I suddenly became quite good - IMO standard (= I guess its just to do with inundation of mathematics. Its not to do with the amount of number theory/inequalities/geometry you know, but actually the knowledge of how to apply them. That is why I love Combinatrics problems - you know everything already! Just whether you can use them creatively.... well not quite, you have to mess around with ideas like inductions, invariants, colouring etc. These are very interesting things to me though. The problem in exams are usually the time limit. The best way to improve is to not have the time limit at first, and work on them, and when you solve them, they will stick to your head. Sometimes they may even take a few days, but solving them is wonderful. Then, inexplicably, your standard will rise, just as mine did. Don't read model solutions. You may wake up one day, and do an exam and realise you can do all the problems quickly and efficiently !

If I buy the AoPS books, should I skip volume 1 and go straight for 2? I have participated in many a math contest (local not national) and I feel like I have covered most of the stuff in vol. 1... then again, maybe I haven't covered it enough? what should I do given my mathematical background?

edit: keeping in mind my goal is to score higher on the AMC, AIME, and other math olympiads

Most of the stuff in AMC I think can be solved with thoroughly understanding the topics in Volume I.  For AIME though I would go through Volume II.


$ p(k)$ is a polynomial of degree $ n$ such that $ p(k) \equal{} \frac {(n \plus{} 1 \minus{} k)!k!}{(n \plus{} 1)!}$ for $ k \equal{} 0, 1, ... , n$. Find $ p(n \plus{} 1)$.

The notations might be a little different (because I got this off of Kalva) but it should be the same question...and there was no solution on Kalva.

I tried setting this thing equal to $ Q(k)$ where $ Q(k)$ had roots at $ 0,1,...n$ but I cant find the constant term (leading coefficient) for $ Q(k)$...

[quote="MellowMelon"]Hint: http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381[/quote]
Amusing timing... intentional?

[quote="MellowMelon"]Blogs have date glitches. That post was made awhile back, actually.[/quote]Yeah, just noticed that... oops. Disregard that.

[hide="Thoughts?"]I don't see how Newton polynomials help. The function is equal to $ \frac {1}{\binom{n \plus{} 1}{k}}$ for $ 0\le k\le n$. Can't think of any identities that will turn that into a polynomial like that.[/hide]

[hide="please explain for me "]I'm must be crazy  :blush:  :blush:  :blush:
let $ g(k)\equal{}p(k)\minus{}\frac{(n\plus{}1\minus{}k)!k!}{(n\plus{}1)!}$
obviously , g(k) is also a polynomial of degree n , yet $ g(0)\equal{}g(1)\equal{}...\equal{}g(n)\equal{}0$ in other words g(k) has n+1 roots . g(k) is a polynomial of degree n having n+1 roots which means all the coefficients are equal to 0 ....  :?:  :?: [/hide]

[quote="HTA"]$ g(k) \equal{} p(k) \minus{} \frac {(n \plus{} 1 \minus{} k)!k!}{(n \plus{} 1)!}$
obviously , g(k) is also a polynomial of degree n [/quote]

$ (n \plus{} 1 \minus{} k)! k!$ is not a polynomial function in $ k$.  Be careful.  And I believe MellowMelon's hint is telling you to take the first differences :)

sorry can someone elaborate on how to construct this polynomial?
because this doesnt look like in the form of something that is in t0rajir0u's blog

Does anyone know when a! = 0 or is that not possible, even more negative a's?