AMC
USA(J)MO
USAMO
email
summer program
MathPath
geometry

[i]I've done one round of editing here, and I encourage Mathfanatic to delete my comments to earlier suggestions for changes to the first draft. Further comments on the current draft are very welcome.[/i]

Here's a Frequently Asked Questions (FAQ) file on the American Mathematics Competitions (AMC) program. I recommend this program to all parents of children who are curious about math. Read appropriate sections of this FAQ for more details.

1) Who can take the AMC tests?

The AMC

http://www.unl.edu/amc/index.html

tests are available worldwide, and are a big part of shared experience of young math-interested people around the world. Participation can be entirely by postal mail, and is growing in several countries of east and southeast Asia and eastern Europe. The USAMO test is restricted to citizens and permanent residents of the United States, but the other tests are open to anyone, making the AMC tests an especially good program for people living overseas.

2) How young can a person start in the AMC program?

Anyone anywhere can sign up for the entry-level competition, the [url=http://www.unl.edu/amc/e-exams/e4-amc08/amc8.html]AMC 8[/url]. The AMC 8 can be taken by anyone up to eighth grade, or by any homeschooler up to an equivalent age. Children as young as age eight take the AMC 8. You or your school pay money to get a test packet for ten test-takers. (A homeschooling parent can share the cost with other homeschooling parents.) Then you take the test on the designated day (always in November, the week before the Thanksgiving holiday in the United States) under specified proctored conditions. The proctor mails back the answer sheets and a signed statement about the proctoring conditions, and after processing at AMC headquarters you get back the scores, which can come first by email, very conveniently. The AMC 8 is a 25-question, multiple-choice, forty-minute test of STRICTLY prealgebra mathematics, but that doesn't mean it is easy. Perfect scores on the AMC 8 appear to be rarer than perfect scores on the SAT I math section (even by persons of the same young age). Test-takers from Taiwan, including test-takers from the local junior high in my former neighborhood in Panchiao, Taiwan, make up about one-third of all perfect scorers on the AMC 8 each year. Whole states of the United States, including Minnesota, go years in a row without any perfect scores.

Kids who score high on the AMC 8 can be invited to apply to cool summer programs, such as [url=http://www.mathpath.org/]MathPath[/url], which young people from several parts of the United States say is an interesting program. My son and one other guy in our local testing group got this year's beautiful MathPath brochure in the mail as a result of their AMC 8 scores. Other than that, the main benefit of AMC 8 is simply that it increases motivation to think deeply about math problems that go beyond the standard school curriculum. I like the AMC 8 a lot and will probably have all my children take it repeatedly as they grow up.

3) What are the tests that lead to the United States of American Mathematics Olympiad?

The next level of the AMC competitions, taken in February each year, is the [url=http://www.unl.edu/amc/e-exams/e5-amc10/amc10.html]AMC 10[/url] and the [url=http://www.unl.edu/amc/e-exams/e6-amc12/amc12.html]AMC 12[/url], with presumed grade levels corresponding to the numbers in the names of the tests. In other words, the AMC 10 assumes completion of beginning high school algebra and geometry, while the AMC 12 assumes completion of all the usual precalculus math courses taught in United States high schools. Each test is a 25-question, multiple-choice, seventy-five minute test, with some questions in common between the AMC 10 and the AMC 12 each date the tests are given. But, again, that doesn't mean these are easy tests. [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05perfect1012AB.html]Perfect scores on the AMC 12[/url] are a LOT rarer than 800s on the SAT I math section (and even much rarer than 800s on the SAT II Math IIC subject test), and are taken very seriously by college admission officers, for example those at M.I.T.

Young people who score sufficiently high on those tests (top 5 percent of all test-takers on the AMC 12, or top 1 percent of AMC 10, or a roughly equivalent cut-off raw score level on either test) are invited to take the [url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]American Invitational Mathematics Examination[/url] (AIME), a 15-question, integer-answer, three-hour test with no penalty for guessing but almost no hope of obtaining a correct answer by guessing, which is legendarily difficult. Even though the AIME is based strictly on precalculus mathematics, it is based on a THOROUGH understanding of that level of mathematics, and thus perfect scores usually number only one or two among all the more than 10,000 invited test-takers. The average (mean) score is just above 3 out of the 15 possible, and the modal score is usually 2. My oldest son took his first AIME on 23 March 2004.

Young people who score sufficiently high on the AIME are selected for the [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html]United States of America Mathematics Olympiad[/url] (USAMO) with rank-order for eligibility for the USAMO determined by a combination of AIME and AMC 10/12 scores. This is the one phase during which there is some advantage to being a younger rather than older test-taker: more of the below-twelfth-grade test-takers will go on to the USAMO at a given combined score level than will twelfth graders. The USAMO is a 6-question, essay-writing mathematical proof, two-day test designed to be as much as possible like the International Mathematical Olympiad (IMO) tests and is used to choose the United States team sent to the IMO. If your child makes it to the USAMO, then you will have to make careful local arrangements, because the USAMO goes online at the AMC Web site JUST before the official testing hours (which are matched across time zones) and you download the test to administer it locally. I don't know this by personal knowledge--my son's not at that level this year--but by reading the AMC Web site. Test security rules get more punctilious as you go up the AMC ladder.

The AMC program is now getting into talent development of the younger budding mathematicians by setting up opportunities for ninth graders to go to the [url=http://www.unl.edu/amc/a-activities/a6-mosp/mosp.html]Mathematical Olympiad Summer Program[/url] (which I call MOP for historical reasons). It is anticipated that as many as thirty ninth graders may go to MOP this summer of 2004 for a special summer program--free of charge--that will prepare them for the main MOP program in following years. You can read about the culture of MOPpers on a Web site

http://moppers.kaseorg.com/

kept by a homeschooled young man who has been to MOP several years in a row. (He was also one of the [url=http://www.ditd.org/public/article.aspx?cid=41&mid=121&tp=2]Davidson Fellows[/url] in math last year, as was a fellow MOPper.)

On the whole, I like the AMC competitions a lot because they go well beyond the standard school curriculum, which marches relentlessly to calculus without smelling the mathematical roses along the way, and because they have a high credibility with summer program directors, university admissions officers, and others who may be able to help my son develop his mathematical abilities beyond what I can do for him. Yes, the out-of-pocket expense for the tests (before the AIME, which is free) is a little high for a homeschooling family that can't find other test-takers to share the cost, but it is worth it.

4) What if my school doesn't offer the AMC tests?

If your school doesn't offer the AMC tests to you, then you can offer the tests to it. You could make your local school an offer it would be silly to refuse by signing up for the AMC test of your choice at your own expense, and then letting the local school know you have an opportunity for other students at the school (as described above) that is already paid for. I basically used this approach to find local public schools willing to proctor the tests for my homeschooling group: I said I had test booklets already ordered, and just needed a space for taking the test and a neutral proctor. One year no students from the school that hosted my group joined the test, but the next year, at a different school, some students from that school used some of the left-over test booklets to take the test at the same time. Urge your local principal, math teacher, gifted education coordinator, or anyone else you can think of to help you out and schedule the test session. (I had my best luck by contacting ASSISTANT principals at local middle schools: the principals are more likely to ignore such requests, in my observation, but see what you find out locally.)

I hope other participants here will join in with their descriptions of math programs for young people, or their been-there-done-thats regarding the AMC tests.

[End of draft FAQ]

That's it. Edit ruthlessly. Let me know what you really think.

[quote="BHorseMath"]okay...I had never heard of the AMC before I joined AoPS.  I would be interested in taking it but my school doesn't do it.  How would I be able to work this?  Is it possible?[/quote]

You could talk to your school or math teacher about it.  If your school isn't interested in proctering the exam, you could contact the AMC and take it at another local school.

my suggestion: Make the whole thing a LOT shorter. Also, I would discuss the high school exams before the AMC 8, since a lot more people take them. I think some of your readers would turn aside before they got thru all the AMC 8 stuff and even realized that there was something for high schoolers. Maybe your general discussion of how beneficial math competitions are should be kept separate from a description of the AMC exams. I usually describe this to people something like this:

American Mathematics Competitions (AMC) is an organization which offers a series of exams leading to selection of the US team to the int'l mathematics olympiad. 300,000 students all over the US and several foreign countries take the first exam, a 70 minute multiple choice test given at students' schools in the spring.  Approx. the top 5% are then invited to take a higher level 3 hour exam. From that group, 300 are selected to take the USA Mathematical Olympiad exam, a 9 hour proof-based exam.  On the basis of that exam, approx 30 students are chosen to attend a specialized training camp to select the team for the int'l competition and for younger students to prepare for future competitions.

AMC also offers a separate exam for grades 8 and below called the AMC-8. This is a one-hour multiple choice test given in November. Middle school students who do well on that exam freguently choose to take the preliminary high school exam in the spring.

[quote="texas137"]my suggestion: Make the whole thing a LOT shorter.[/quote]

That's very standard advice for all online writing, so I'll keep that in mind. I think, however, that the computer industry, which studies this issue, probably studies people who are at the nadir of reading ability among educated adults, and I'm usually writing for people who can get through longer passages, especially when I'm writing FAQs. The much shorter piece you wrote as an example is food for thought about how much is enough to get people interested. I just hate to type the same thing twice, which is why I'm taking an email I wrote recently and trying to turn it into a FAQ. Everyone can read faster than type.

I thank you all for doing some of our "public relations" work here, especially TokenAdult for actively promoting the AMC contests, we appreciate it!

As long as you are at it, I would like to ask two questions:
(1) This summer I will be working on an update/revision/addition of the Frequently Asked Questions page that appears on our webiste.  What Frequently Asked Questions need to be addressed?  How can we make the FAQ more informative?

(2) In our adveritising/information/promotion brochures that go to teachers, schools, previously registered groups and the like, what needs to be included?  Omitted?  Changed?

Steve Dunbar
AMC Director

Steve,
It's hard to be part of AMC for homeschoolers or any kids who do not have the support of their schools. I offered to pay for everything to local schools (and their students) in exchange for a room and someone other than myself to menitor the test. I was rejected each every time. Other home school families have since asked me about AMC. I wonder if your FAQ could include a section for homeschoolers or those students whose schools have no interest in AMC.

For example, if a home schooler can not take AMC at any local school, then what are other options? What will be the rule on proctors (non-parents?)? What about the location (has to be public? libraries, most of the time, don't let us reserve the rooms for more than 15-30 min. And we can only reserve or check the availability of the room a day or two ahead of the time.)? Where can homeschoolers get study help (a link to AoPS would be great!), etc.

I knew several kids who are in regular schools; but the schools have no interests at all. I went with some of the parents to the schools and talked to math teachers and so on. They often say "no funding" (although I offered to pay), "no time", or "not enough interests", even something like "make other kids feel bad", "why are you asians always wanting to compete?", "learning is more important than winning."

These kids in a way face the same problem as home schoolers do. Therefore, FAQ for homeschoolers will also help them.
Thanks!

I have a few questions for Mr. Dunbar as well.

When I attended middle school, I was able to the the AMC - 8 all three years. The first year, my parents paid for the test to be taken. The other two, I attended a different school.

This new school has some strange rules. They only allow students in their most advanced math class (Geometry 8) to take the AMC8. That was ok for me, since I had finished geometry by the time I got there.

On the other hand, my sister took Algebra in 7th grade. She naturally inquired about taking the contest as a 7th grader. She was told that she couldn't and had to wait till the next year. So my parents came to speak to the person in charge of the math competition, and later the principal. Each time, they were hit with the same argument : "we decide who takes the test. only this set of people is eligible." My parents offered to pay any extra amount of money that was necessary, and arrangements for proctoring and rooms had already been arranged - there was plenty of space for any amount of extra students. Nonetheless, my sister was not allowed to take this competition, even though she was completely eligible by the standards of the AMC, and my parents were willing to go the distance to make arrangments for her. Ultimately, the school claimed that my sister would miss valuable school time by taking the AMC 8 (one hour of one day, which is laughable considering that for various sports tournaments, other students missed many days on end.)

A few months later, my sister registered to take the AMC-10 at my high school. She took it succesfuly and received a prize.

Perhaps the AMC's commision could make some sort of rule that would help out kids in the position of my sister, with stubborn schools that have no interest for their students?

Though not in my case, it seems many people have trouble getting schools to participate.  I think math competitions (beginning with MathCounts) have been the main reason why I've been interested in math.  AMC has played an important role (and most likely still will if not more so for the next 4 years) in my math education.

It would be beneficial to a lot of other people if AMC could arrange perhaps regional contests proctored by official guys, even if it costs a little more for us.

[quote="istudio"]It's hard to be part of AMC for homeschoolers or any kids who do not have the support of their schools. I offered to pay for everything to local schools (and their students) in exchange for a room and someone other than myself to menitor the test. I was rejected each every time.[/quote]

Well, if you live in the Twin Cities metropolitan area of Minnesota you can join my homeschool math team. My local public school district provides us with space and proctoring for the AMC 8 each year, no problemo.

[quote="istudio"]Other home school families have since asked me about AMC. I wonder if your FAQ could include a section for homeschoolers or those students whose schools have no interest in AMC.[/quote]

I'll agree here with the suggestion that the AMC Web site ought to address both these questions: how do homeschoolers participate (which I think IS addressed in the Web pages for most of the individual tests) and how do students in clueless schools participate?

[snip of list of problems and questions homeschoolers have]

The manual for each testing program is reasonably clear, or at least has seemed reasonably clear to us in Minnesota, on how homeschoolers can participate. I have found space and proctoring in my local public school district. Another homeschool math team tests at a chance, and hires a retired teacher as a proctor.

Definitely there need to be more channels for taking the tests for kids who are stuck in overly bureaucratic schools and can't get the tests in school.

I am not sure if this is a frequently asked question, but still: what about administering AMC-10 to middle schoolers? I am working towards organizing a math team in my kid's middle school. The kids are all 7 and 8 grade, but many of them have completed Algebra I and take 9th grade geometry courses, honors courses etc. Besides, some of them have taken AMC- 8 in 6th grade and did pretty well. So it feels like they are ready at least to try AMC-10. Are there are any formal or other obstacles for that?

Also, what happens to student's scores? Are they recorded and stored forever? Are there any restrictions about how many times you can take AMC-10 altogether, and how you average your scores from different years?

Thanks,

Irina

I don't think giving the test in middle school's is a problem at all.  I took the AMC 10, and AIME last year with my junior high (grades 7 and 8), and not with the High School, so it should be ok.  All you need to do is register.

There shouldn't be any restricitons on how many times you can take the AMC 10, or any of the tests run by them.  Just that you can't take the AMC 10 when you're in 11th or 12th grade.  And the score you get that year is the score that counts for that year.  None of the scores from previous years should matter for the tests.  The may matter though, for scholarships.

Hope this helps.

From my experience, if you spend time on this site solving all of the AMC and AIME problems posted or reading the solutions of others for the problems you don't understand, you should do pretty well on the AMC/AIME. It's really helped me so far; I got a 3 on the AIME last year, but now I feel very confident of qualifying for USAMO (as a sophomore) barring some big mistakes on my part.

[quote="trisana"]The AMC is a big deal at my school (I'm a freshman, and the AMC 8 wasn't offered at my middle school). A BIG deal. We are recieving old AMCs to work through over the holiday, and from the day we get back to school  until the day the test is administered, we will be doing an AMC a day. I'm just in geometry, (taking Algebra in 7th or earlier wasn't offered at my school, even though I was ready--all of 7th grade was a review), and I want to know whether I should get someone who has already completed geometry to sum up the second semester curriculum, or should I just study my fact sheets and AoPS book? I really want to do well on this test. My main objective is to make it to the AIME, because probably only one geometry kid in the class (not me!) stands a chance without some serious study time.

Thanks,
Caroline[/quote]

Wow, what school do you go to? Sounds like a really good school to me. :)

On my school website it doesn't say anything about getting higher than 100 or 120. It just says getting in the top 1% and 5%. Is both qualifications good for each test EVERYWHERE? Or is it more like a someone in charge of your school gets to decide thing?

It is either 120 or top 1% to qualify by the AMC 10 and either 100 or top 5% to qualify by the AMC 12.
Getting a 120/100 guarentees that you qualify for the AIME. The top 1%/5% is there incase the AMC 10/12 is too difficult and less than 1%/5% of people get a 120/100.

When is the cutoff for the USAMO released?

The USAMO teachers' manual says that the qualifiers will be selected no later than Tuesday (April 3), and invitations will be mailed out sometime next week.  It also says not to call before Friday (the 6th).  I'm guessing qualifiers will be out late Tuesday afternoon or Wednesday.  Cutoffs will be released at the same time as the qualifier list.

Teachers' manual is here:
[url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/07-USAMO-TM/TM-USAMO,2007.pdf[/url]

um does anyone know the major changes for AMC 2008? I heard a rumor that you cannot use a calculator anymore right? How about the problems' level of difficulty (comparing to AMC 2007) , scoring policy, the score cut-off for AIME? Do they change or just remain the same as the one in 2007?

Thanks

Read Frequently Asked Questions
Q: AMC 10/12 #5, #6, #7 at the
newly revised 2007-2008 Frequently Asked Questions page of the AMC website.
http://www.unl.edu/amc/f-miscellaneous/faq.shtml#amc1012

(After some editing, revision, and testing this fall the newly revised Frequently Asked Questions page will be more prominently displayed.)

See also the 2007-2008 Information Brochure.
See also the AMC 12 Chair's letter to teacher's and participants in the 2007 Summary of Results and Awards.  All are available at our website.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote]Perfect scores on the AMC 12 are a LOT rarer than 800s on the SAT I math section (and even much rarer than 800s on the SAT II Math IIC subject test), and are taken very seriously by college admission officers, for example those at M.I.T.[/quote]what an understatement..

Can you take both the AMC 10 and 12 if you take them on the two dates offered?  I haven't found the answer anywhere! :?:

> Can you take both the AMC 10 and 12 if you take them on the two dates offered? I haven't found the answer anywhere!

This is answered explicitly and completely in Frequently Asked Question AMC 10/12 #12 at
http://www.unl.edu/amc/f-miscellaneous/faq.shtml
See also FAQ AMC 10/12 #10 before it.

Steve Dunbar
MAA Director, American Mathematics Competitions

a question for the amcdirector-
if someone skipped grades and was, in say,, 11th grade but ""agewise" they were "supposed" to be in 10th grade, would they be allowed to take the amc10?
wouldnt it be more fair to make eligibity based on age as opposed to grade?

How so? Math ability is dictated more by the material you have learned, not your biological age.
And I believe the answer is that if you're in 11th grade, you're in 11th, and have to take the AMC 12.

EDIT:
[quote]Any student that is officially enrolled in grades 10 and below, taking at least one course."[/quote] is indeed the wording on the website.

Now that we've reached the end of the 2007-2008 contest year, we're in the process of looking at and updating our Frequently Asked Questions.  The current FAQ can be found [url=http://www.unl.edu/amc/f-miscellaneous/faq.shtml]here[/url].

1) Are there any questions not addressed on the current FAQ that you would like to see?
2) Are there any answers to current questions that are unclear?

There is no guarantee that your question will be included in the update as it may be too specific for a general answer.

Thank you in advance for your input.  We do read everything here on AoPS and enjoy the excitement and interest in the AMC.

Elizabeth Claassen
AMC Senior Accounting Clerk
(and resident number cruncher)

Wait is it possible to take the AMC 10 AND the the AMC 12 since some of the questions are the same?

[quote="chess64"]$ 6 \cdot \# \text { correct } \plus{} 2.5 \cdot \# \text { skipped }$

Click [url=http://www.unl.edu/amc/]here[/url] to find out more.[/quote]
Actually, I think the number skipped is 1.5 points.


I think this should be in Intermediate, I thought only basic trig was allowed here?
[hide]Use formula $A=\frac{1}{2}ab\sin{C}$.  So the answer is $\frac{1}{2}\cdot 2\cdot 6 \sin{56^{\circ}}=6\sin{56^{\circ}}$[/hide]

I'll let this one go since area formula using sin is covered in Volume 1 of AoPS books.

[quote="buzzer11"]in a triangle, the sides are 2, 6, and x. The angle included between the two sides 2 and 6 is 56 degrees. To the nearest hundreth, what is the area of the triangle?

Hint: use trigonometry and a calculator.[/quote]

Area = $\displaystyle \frac{1}{2}ab \sin{C}$. Plug in the numbers:

$\frac{1}{2}\cdot 2 \cdot 6 \cdot \sin{56}$

$6\cdot \sin{56}$

Since he tells you to use a calculator:

$\approx 0.86\cdot 6 \approx 4.974$


Fire is one of the leading causes of accidental deaths. It is important for everyone to take every preventative measure and precaution possible to be ready to deal with a fire emergency.
More than half of all fatal fires occur between 10 p.m. and 6 a.m. when everyone in the home is usually asleep. Smoke alarms are necessary to alert you to fires when you sleep. Will smoke alarms allow enough time to evacuate safely?
Build a mathematical model to determine the number and locations of smoke alarms to provide the maximum time for evacuation. Also include a model to determine the number and location of at-home fire extinguishers to have available. Build a mathematical model for evacuation of a family from both one and two story homes.

I am probably missing something, but a lot of these mathematical modeling questions seem just dumb.  Is a mathematical modeler really going to be able to figure out how many smoke alarms to get, and where to put them, better than an ordinary person with common sense?  Are mathematical equations somehow going to help with this?

I would like to see what's considered to be a good solution to this problem.


A circle with radius 3 cm intersects a circle with radius 4 cm. At points of intersection, radii are perpendicular. What is the difference in the areas of the non-overlapping parts of the circles?

[hide]
Set area of overlap = $x$
Non-overlapping area of circle with radius 4: $4^2\pi-x$
Non-overlapping area of circle with radius 3: $3^2\pi-x$
Difference: $(4^2\pi-x)-(3^2\pi-x)=\boxed{7\pi}$ :D [/hide]


From the 1979 APICS Competition, problem 2

Show that if n is a positive integer greater than 1, then n^4 + 4 is not prime.

This is called the Sophie Germain (sp?) theorem.

Germain is correct.  The Germain Idendity.

[quote]By the way, fiertytycoon, if this problem was worth 7 marks, that would probably be worth 5 or 6..[/quote]

Yah, I know, but it isn't worth 7 marks :D Anyways, everyone gets the idea...

Well if it was out of 1 mark then you'd get 0 :D but yeah.


Some people at my school left today. What do you guys do?

No idea  :wacko:

But it's fun, I've heard. It's actually "Beta Convention," not a beta convention, because the entire thing belongs to Beta Club.

Oh, Beta Club.  Hmm, I'm not in any honors societies, so I guess I wouldn't be going, then.

I was in beta sophomore year, but then i realized that all it was was forced volunteer work (at least at my school). i mean, if i'm going to volunteer somewhere, i don't want it forced on me.

most of the people at my school that do it are in it mainly for college aps.

[quote="Karth"][quote="drunner2007"]who won the math competition at the convention?[/quote]

1st: Me  :D (Northview High School)

2nd: Cyrus Vadafari (Mark's School, aka Columbus High School)

3rd: Sally Park (Joe's School, aka North Gwinnett High School)

w00t All of us were GHP 2006  :P[/quote]

cool. how hard was it? what type of format?

Haha, wow, I completely missed that there was a math competition involved here :roll: .  Anyhow, good job, Karthik.  Were there other competitions also, or does Beta Club just have a fondness for math?

Well, actually, there are all sorts of competitions. Some of them include acryllic painting, science, talent, public speaking, english, and creative writing. Actually, Vibin Kundukulam, who is also from Northview, placed first in state at science. Our quiz bowl team is also going to nationals with 2nd place.  :)  So yea, it's not only math. It's everything. And I'm happy that I'm not alone to go to nationals, cuz that'd sorta be boring  :)


Please help :? :
Consider a prime p, we know that there are infinitely many primes among numbers of the form kp+1.Prove that among these primes there is a prime q=tp+1 such that p^t -1 is not divisible by q.

I thik you were solving Problem 6 of IMO 2003.
I give you a hint we know that all divisors of $a^{p}-1$ are multiple of $p$ or in the pattern $kp+1$ ....


function articles algebra polynomial probability counting distinguishability

Could someone please give me a basic explanation of what generating functions are and how they work? Please don't quote Wikipedia or Wolfram MathWorld. Their articles are overly technical and use lots of confusing symbols.

Thanks.

A generating function is basically a power series that yields information about a sequence you're looking at.  Or, if $ f_{A}(x)$ is a function, then the generating function of $ f_{A}(x)$ is the infinite polynomial $ a_{0}\plus{}a_{1}x\plus{}a_{2}x^{2}\plus{}a_{3}x^{3}\plus{}\cdots$, which carries all of the information about the sequence.

As a simple example, the generating function for the exponential function, $ e^{x}$, is just $ 1\plus{}x\plus{}x^{2}/2!\plus{}x^{3}/3!\plus{}x^{4}/4!\plus{}\cdots$.  A more complicated example, which I will not prove is the following: let $ a_{n}$ denote the number of ways of associating multiplicands with parentheses in an ordered product of $ n$ numbers so that the resulting expression correctly defines a multiplication of the numbers.  Then $ a_{n}\equal{}\binom{2n\minus{}2}{n\minus{}1}/n$.

So $ e^{x} \equal{} 1 \plus{} x \plus{} x^{2}/2! \plus{} x^{3}/3! \plus{} x^{4}/4! \plus{} \cdots$? How do generating functions relate to probability?

[hide="How to find the coefficient using purely generating functions"]
Besides just multiplying everything out, there are tactics. For example, on that problem. It can be written as \[ \left(\frac{x(1-x^6)}{1-x}\right)^3=x^3(1-x^6)^3(1+x+x^2+x^3+\dots)^3.\] The "infinite polynomial" isn't as intimidating as it looks. It can be written as \[ (1+x+x^2+x^3+\dots)^3=1+3x+6x^2+10x^3+\dots.\] Now I think you can expand since they're only cubed. The first two terms become \[ x^3(1-3x^6+3x^{12}-x^{18})=x^3-3x^{9}+3x^{15}-x^{21}.\]The "infinite polynomial" becomes \[ 1+3x+6x^2+10x^3+15x^4+21x^5+28x^6+36x^7+\dots.\] We only look up to the seventh power because we want the coefficient of $ x^{10}$ and the lowest power in the other terms is $ x^3.$ Only considering the terms that have exponents summing to $ 10,$ we get the 2 terms \[ x^3\cdot36x^7\newline-3x^{9}\cdot3x.\] If we add these together, we get \[ 36x^{10}-9x^{10}=27x^{10}.\] So the number of ways to roll a $ 10$ is $ 27.$ Put this over the total number of outcomes: $ 6^3=216,$ to get \[ \frac{27}{216}=\boxed{\frac{1}{18}}.\]

This method looked like it took a lot longer, but if you are very familiar with generating functions and how to quickly find the coefficient, this is no slower than listing out the possibilities.

(If you didn't understand this, just take all this on faith for now, you'll understand when you get the book).[/hide]

Of course, a list would be just as good. However, when the sum and number of dice increases, a list like this would be near impossible, so don't completely rely on it.
[hide="List"] \[ \begin{tabular}{c|c|c\parallel{}c}
Throw \#1&Throw \#2&Throw \#3&Permutations\\\hline
6&3&1&6\\
6&2&2&3\\
5&4&1&6\\
5&3&2&6\\
4&4&2&3\\
4&3&3&3\\\end{tabular}.\] Add all the numbers in the fourth column and you get the number of ways to roll a $ 10$ with $ 3$ dice, which is $ 27$. Put it over $ 216$ to get \[ \frac{27}{216}=\boxed{\frac{1}{18}},\] same as before.[/hide]

[hide="Why does this work?"] Think about the polynomial \[ (x+x^2+x^3+x^4+x^5+x^6)^3.\] How many ways can you make an $ x^{10}$ term? The possible ways are \[ x^6\cdot x^3\cdot x\newline x^6\cdot x^2\cdot x^2\newline x^5\cdot x^4\cdot x\newline x^5\cdot x^3\cdot x^2\newline x^4\cdot x^4\cdot x^2\newline x^4\cdot x^3\cdot x^3.\] Of course, we have to consider the permutations of each. Hmm, the powers of the terms looks familiar. Let's refer back to our list.
\[ \begin{tabular}{c|c|c\parallel{}c\parallel{}c}
Throw \#1&Throw \#2&Throw \#3&Monomial&Permutations\\\hline
6&3&1&x^6\cdot x^3\cdot x&6\\
6&2&2&x^6\cdot x^2\cdot x^2&3\\
5&4&1&x^5\cdot x^4\cdot x&6\\
5&3&2&x^5\cdot x^3\cdot x^2&6\\
4&4&2&x^4\cdot x^4\cdot x^2&3\\
4&3&3&x^4\cdot x^3\cdot x^3&3\\\end{tabular}.\] The coefficients match the throws! That's also what our generating function was in the beginning. The $ x^1$ represented a throw that resulted in a $ 1,$ $ x^2$ represented a throw that resulted in a $ 2,$ and so on. Here, the $ x^6$ represented a throw that resulted in a $ 6,$ the $ x^3$ represented a throw that resulted in a $ 3,$ and so on.[/hide]

Generating functions is a [b]tactic[/b] not a theorem. I don't see any way to "derive" generating functions because there's nothing to prove.

Some of these problems [b]can[/b] be solved by listing, but we seek a generating functions solution. Others, good luck with listing...
Some of these problems will ask you to prove some of the statements I made in the first hide tag.

[hide="Problems"]
Difficulty is denoted by the $ \star$ out of $ 5$ stars.

[hide="1)"]
$ (\star)$
Show that the "infinite polynomial" (I will start calling it a power series now since that is its proper name) \[ 1+x+x^2+x^3+\dots\] can be written as \[ \frac{1}{1-x}.\][/hide]

[hide="2)"]
$ (\star)$
How many ways can you roll a $ 23$ with $ 5$ dice?[/hide]

[hide="3)"]
$ (\star)$
I have $ 3$ urns: one with $ 3$ red marbles, one with $ 3$ green marbles, and one with $ 4$ blue marbles. Assuming the marbles are indistinguishable, how many ways can I choose $ 4$ balls?[/hide]

[hide="4)"]
$ (\star\star)$
Expand the power series \[ (1+x+x^2+x^3+\dots)^2.\] Give the first $ 10$ terms of it.[/hide]

[hide="5)"]
$ (\star\star\star)$
We have $ 25$ people at my party. Everyone agrees to pitch in $ \$1$ or $ \$2$ except for one generous person, who will pay $ \$3, \$5$ or $ \$9.$ How many ways can we come up with $ \$39?$[/hide]

[hide="6)"]
$ (\star\star\star)$
Six boys and  eight girls go on a nature hike. Each boy will pick either $ 0$ or $ 2$ flowers while each girl will pick either $ 0$ or $ 3$ flowers. How many ways can the group end up with $ 20$ flowers?[/hide]

[hide="7)"]
$ (\star\star\star\star)$
Find an expression for the coefficient of $ x^k$ in the power series \[ (1+x+x^2+x^3+\dots)^n.\] Prove your answer.[/hide]

[hide="8)"]
$ (\star\star\star\star)$
I'm giving out $ 100$ identical pieces of candy to $ 5$ kids (of different ages). The two youngest kids want at most $ 1$ piece. The middle kid will take any number of pieces. The two oldest kids each want an odd number of pieces. How many ways can I distribute the candy?[/hide]

[/hide]

Credit goes to RoFlLoLcOpT

I wrote a handout for WOOT about generating functions which is now available [url=http://web.mit.edu/~qchu/Public/TopicsInGF.pdf]here[/url]; hopefully the introduction will be useful at least, although parts of it get involved.  Some other resources and comments are available [url=http://qchu.wordpress.com/2009/06/19/gila-iv-the-unreasonable-effectiveness-of-generating-functions-in-the-combinatorial-sciences/]here[/url].


superior algebra
superior algebra unsolved

Denote by $ s_m$ by the number of elements of order $ m$ in a group $ G$. Let $ |G|\equal{}p^aq^b$, where $ p,q$ both odd prime, and $ exp(G)\equal{}pq$. Suppose that $ s_p\neq s_q$ , prove  $ s_{pq}\neq s_p$ and  $ s_{pq}\neq s_q$.

I donot kown how to deal the "Hint for the proof: Use Brauer's characterization of characters"

[quote="rulin"]Denote by $ s_m$ by the number of elements of order $ m$ in a group $ G$. Let $ |G| \equal{} p^aq^b$, where $ p,q$ both odd prime, and $ exp(G) \equal{} pq$. Suppose that $ s_p\neq s_q$ , prove  $ s_{pq}\neq s_p$ and  $ s_{pq}\neq s_q$.

I donot kown how to deal the "Hint for the proof: Use Brauer's characterization of characters"[/quote]

Can you give more context?  Is this from a book?

Have you tried using the characters from Frobenius's n'th roots result: θ_n(g) is the number of n'th roots of g in G.  Then θ_n is a generalized character and its degree is related to your s_n.

thanks JackSchmidt. Not in a book, this is in my test question.

So you have solved it? Can you post the idea of the solution?


algorithm
combinatorics unsolved
combinatorics

Given a table $m\times n$, in each cell of which a number $+1$ or $-1$ is written. It is known that initially exactly one $-1$ is in the table, all the other numbers being $+1$. During a move, it is allowed to chose any cell containing $-1$, replace this $-1$ by $0$, and simultaneously multiply all the numbers in the neighbouring cells by $-1$ (we say that two cells are neighbouring if they have a common side). Find all $(m,n)$ for which using such moves one can obtain the table containing zeros only, regardless of the cell in which the initial $-1$ stands.

I used chessboard coloring + calculations modulo 4 to show that we cannot elliminate all numbers in case $2\,|\,n$, $2\,|\,m$.

We can do this if and only if $2|(m-1)(n-1)$
It needs only a simple double counting.
and for $m$ odd or $n$ odd the it's easy to give a solution

baraye omide hatami.
salam.

man ham ghabool daram bayad (m-1)(n-1) bar 2 bakhshpazeer bashad.

ebteda sabet khaham kard agar (m-1)(n-1) bar 2 bakhshpazeer bashad meetavan een kar ra anjam dad.

be ezaye paye badihi ast. hal farz mikonam baraye kamtar az m, n ke dar sharte bala sedgh mikonand
tavanesteh bashim een kar ra bokonim hala baraye m va n mikhaham een kar ra bokonam.

esme jadval haiee ke m-1*n-1 anha bar 2 bakhshpazeerand ra tablo migozareem.

farz konim m tedad satrha va n tedad radifhast bayad hadeaghal yeki az m ya n fard bashad
agar m fard bood tebghe farze ghavi dar esteghra(masahat) baraye m*t ham sabet shodeh
be torike ke t koochektar az n va m*t mostatile tablo bashad.
hala yek radif m*1 ezafeh kardeh az anja ke m fard ast ba yek algorithm sotooni ra ke dar an
adade -1 gharar darad ra entekhab va har bar yek khaneh az an ra ke -1 dar an neveshteh shodeh
ra entekhab va amale morede nazar dar masale ra anjam midahim ba kami deghat mifahmim tamam
khanehaye een radif 0 shodeh va dar samte rast va chab (dar soorate vojood) do jadval
ba  yek adade -1 va tablo bevojood amadeh ast. pas kar tamam ast . chon dar hengame amale masale faghat adade dakhel haman
mostatilha taghieer mikonand .
agar n fard bood esteslali moshabeh vali een bar yek sadre 18n ra hazf mikonim.

ama baraye anke sabet konim een shart kafeest  berahati mitavan did ke har bar ke rooye khanehaee ke dar satrhaye bala va paeen
va sotoonhaye chab va rast een amal ra anjam dahim -1 sabet ast. va har bar ke dar khanehaye
mostatile m-18n-1 anjam dahim tedad -1 haye jadval tagheer mikonad.

az tarafi faghat dar soorati mishavad tamame khanehaye jadval 0 shavand ke har khaneh yek bar entekhab shavad.
-1 dar ebteda be andaze fardta darim baraye anke kole jadval khali az -1 shavad bayad khanehaye
mostatil tooyeeye zoj bashad.

[quote="pbornsztein"]Hmmm...I guess he said the same thing too :D

Pierre.[/quote]
 :lol:  :lol:  :lol:  :lol:

Well please don't post solutions in farsi
Also I told the problem is simple.


integration
limit
function
calculus
real analysis
real analysis unsolved

Define the number $ e$ as the number for which $ \int_1^e \frac {1}{x}\ dx \equal{} 1$.
Prove that $ \lim_{n\to \infty} \left(1 \plus{} \frac {1}{n}\right)^n \equal{} e$.

[hide="one approach"]Define the function $ f: \mathbb{R}^ \plus{} \to \mathbb{R}$, $ f(x) \equal{} \int_1^x \frac {1}{\xi}d\xi$.  Prove using the substitution $ \xi \equal{}s^k$ that $ f(x^k) \equal{} kf(x)$ for all $ x,k\in \mathbb{R}^ \plus{}$.  Use L'Hospital's rule and the fundamental theorem of calculus to show $ \lim_{n\to \infty}nf\left(1 \plus{} \frac {1}{n}\right) \equal{} 1$, and note that $ f$ is a continuous strictly increasing function (hence injective) to arrive at the conclusion.[/hide]


modular arithmetic
quadratics
number theory unsolved
number theory

Hi.

I have been struggling with both directions of the following problem:

A natural number $ a$ is a primitive root $ (mod 83) \Leftrightarrow a^{41}\equiv \minus{} 1 \not\equiv a (mod 83)$

Thanks!

Firstly, 83 is prime, so there exists a primitive root.

By definition, $ a$ is a primitive root if and only if the least exponent $ k>0$ such that
$ a^k \equiv 1 \pmod{83}$ is $ 82$.

In particular, $ 83|a^{82}\minus{}1\equal{}(a^{41}\minus{}1)(a^{41}\plus{}1)$. But if $ 83|a^{41}\minus{}1$, $ a$ will not be a primitive root.
because the exponent 41 is less than 82.

So, the first direction is OK. For the return, I have made little progress :(

For the second part, the group of units $ mod 83$ has order $ 82\equal{}2*41$, so the possible orders of $ a$ are $ 1,2,41$, and $ 81$.

Sorry CaptainCanada, I didnt understand your direction or proof.

Can you elaborate?

The multiplicative order $ d$ of $ a$ mod 83 divides 83-1=82. This means d = 2, 41, 82. $ a\neq \minus{}1$ means $ d\neq 2$ and $ a^{41}\equal{}\minus{}1(mod 83)$ means $ d\neq 41$.
So d=82 and $ a$ is a primitive root.

A generalization - for a prime $ p$ of the form $ 2q \plus{} 1$ where $ q$ is also a prime (i.e. $ q$ is a Sophie Germain prime), then $ a$ is a primitive root mod $ p$ iff $ a^q \neq 1 \pmod p$ (or equivalently: $ a$ is not a quadratic residue mod $ p$) and $ a \neq \minus{} 1 \pmod p$.

Basically the same proof suggested by CaptainCanada and Baranonv - we need to rule out the possibilities $ a^1 \equiv 1 \pmod p$, $ a^2 \equiv 1 \pmod p$ and $ a^q \equiv 1 \pmod p$.


[quote="JRav"]I'm not going to do the math but it seems like this could be easily tackled using Heron's formula.[/quote]

Note that the height of the triangle is equal to the height of the square and that the base of the triangle is $ \frac 34$ of the side length of the square. Call the side length $ s$. Then $ h \equal{} s$ and $ b \equal{} \frac 34 s$, so the area of the triangle is $ \frac 12\left(s \cdot \frac 34 s\right) \equal{} \frac 38 s^2$. That divided by $ s^2$ yields a ratio of $ \boxed{\frac 38}$.

Donkeykingkong, if you have a lot of problems, I think you should put them together in 2 or 3 posts instead of 20 different posts.


What's currently on your playlist?
[b]Lynyrd Skynyrd- Free Bird[/b]

Third Eye Blind - Blinded...I just discovered the song today, it's awesome.

The White Stripes - The Air near my fingers

this topic was locked in the music forum in lieu of haviung more productive discussion or smething.

Wasting the dawn-69 eyes

[quote="shadysaysurspammed"]this topic was locked in the music forum in lieu of haviung more productive discussion or smething.[/quote]Yeah but this is G&FF, where it's imossible to have productive discussion  :lol:

Right now: U2 - I Still Haven't Found What I'm Looking For


[quote="Jean-Louis"]Can anyone show that the radius of the circle inscribed in this Pythagorean triangle is (x + y – z)/2 ?[/quote]

You mean right triangle?  :huh:

I think so. The problem is stated as such in my textbook.

This is untrue - consider a 3-4-5 triangle. The area is $ 3\cdot4 \equal{} 12$ and the semiperimeter is $ (3\plus{}4\plus{}5)/2 \equal{} 6$, so the inradius is $ A/S \equal{} 12/6 \equal{} 2$. However with your formula we have $ r \equal{} (3\plus{}4\minus{}5)/2 \equal{} 1$.

Edit: woops nvm..

[quote="Anaxerzia"]This is untrue - consider a 3-4-5 triangle. The area is $ 3\cdot4 = 12$[/quote]

The area is $ \frac{3\cdot 4}{2}= 6$.  ;)

[hide="Proof"] Consider the right triangle $ ABC$, where $ \angle C = 90^{\circ}$.  Let $ I$ be the incenter.  Drop perpendiculars to all three sides; let $ D$ be the foot of the perpendicular on $ BC$, let $ E$ be the foot of the perpendicular on $ AC$, and let $ F$ be the foot of the perpendicular on $ AB$.

$ IDCE$ is a square with side length $ r$; it is now easy to show that

$ BC = BD+r$
$ AC = AE+r$
$ AB = AF+FB = BD+AE$ (similar triangles)

Hence

$ r =\frac{BC+AC-AB}{2}$ [/hide]

and in general for any triangle $ r\equal{}\frac{\Delta}{s}$


An inversion on a circle with center in A carries the figure in the triangle C'D'B'. Then applying Menelaus theorem points M',N' and R' are collinear (and even A', which is the point to infinity), so their counter image, AMNR, is a circle  :)

[quote="marko avila"]Let C_1 and C_2 be two  circles  that  intersect  in A  and B. Draw a line through A that intersects C_1  and C_2  in C and D respectively. M is the midpoint of CD, and N and R are the midpoints of the arc CB and BD  not containing A  of C_1 and C_2 respectively. Prove that M,A,N and R are on a circle.[/quote]

This is equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=5433 (if you don't directly see the equivalence, read my post in that topic; there, I prove that M, A, N, R lie on one circle as an auxiliary result).

[b]EDIT:[/b] mitchan88, didn't see your post. It's basically the same as my proof...

  Darij


linear algebra
matrix
algebra
polynomial
function

Hello,

My exercise is the following :
Suppose that the square matrix $A$ has a real or complex entries and that $A^k=I$ for some natural number $k$. Prove that each eigenvalue of $A$ must be a $k$ th root of 1.

I can manage to prove that if an eigenvalue of $A$ is a $k$ th root of 1 then it works but i can't manage to prove it doesn't work otherwise.

I think we should use the propertie which says that if $\lambda$ is an eigenvalue of $A$ then $f(\lambda)$ is an eigenvalue of $f(A)$ with $f(x)=a_k x^k+a_{k-1} x^{k-1}+ \cdots + a_1x+a_0$.

Thanks a lot for your help!

Mark.

Suppose $Ax=\lambda x$ for $x\ne0.$

Then $A^2x=A(\lambda x)=\lambda^2x$ and so on.

$\lambda^kx=A^kx=x$ since $A^k=I.$ Thus $(\lambda^k-1)x=0.$ Since $x\ne 0,$ this means that $\lambda^k=1.$

More generally - and it's exactly the same argument, just follow what happens to an eigenvector - we have that if $f(A)=0$ for some polynomial function $f,$ then every eigenvalue of $A$ is a root of $f(\lambda)=0.$

An example of this in operation: let $J$ be the $n\times n$ matrix all of whose entries are 1. By simple explicit calculation, we see that $J^2=nJ.$ From the above argument, all of the eigenvalues must be solutions of the polynomial equation $\lambda^2=n\lambda.$ So the only possible eigenvalues are $0$ and $n.$ Since the trace (the sum of the eigenvalues) is $n,$ we have $n$ being a simple eigenvalue and $0$ being of algebraic multiplicity $n-1.$


A 15-millihenry coil with a resistance of 160 ohms is connected to a 24 volt battery. When the current is 80 milliamperes, at what rate is it increasing

It's quite simple:

$ U-IR=L\frac{\delta I}{\delta t}$

$ \frac{\delta I}{\delta t}=\frac{U-IR}{L}$

you're absolutely right, it is indeed quite simple.

extension: What happens when the current turns on in a real coil?

(Think of a real coil as a series combination of inductance and resistance.)


Let $ n>1$ and for $ 1 \leq k \leq n$ let $ p_k \equal{} p_k(a_1, a_2, . . . , a_n)$ be the sum of the products of all possible combinations of k of the numbers $ a_1,a_2,...,a_n$. Furthermore let $ P \equal{} P(a_1, a_2, . . . , a_n)$ be the sum of all $ p_k$ with odd values of $ k$ less than or equal to $ n$.

How many different values are taken by $ a_j$ if all the numbers $ a_j (1 \leq j \leq n)$ and $ P$ are prime?

If I understand the question correctly, the case $ n \equal{} 3$ is $ P(a_1,a_2,a_3) \equal{} a_1a_2a_3 \plus{} a_1 \plus{} a_2 \plus{} a_3$
Since $ P$ is a prime bigger than $ 2$ it must be odd. This requires that two pf $ a_i$ are even primes, i.e. $ 2$.
So take $ a_1 \equal{} a_2 \equal{} 2$ then $ P \equal{} 5a_3 \plus{} 4$, so for this case the question is asking how many primes are of this form with a_3 a prime.
This is a hard unsolved problem. Other cases are at least as hard, so the general problem seems insoluable.
Where did I go wrong?


From a group of four boys and two girls, two children are
selected at random. What is the probability that the second
child selected is a girl? Express your answer as a common
fraction.

if boy was selected first:

$ \frac{4}{6}*\frac{2}{5}\equal{}\frac{8}{30}$

if girl was selecte first :

$ \frac{2}{6}*\frac{1}{5}\equal{}\frac{2}{30}$

$ \frac{8}{3}\plus{}\frac{2}{30}\equal{}\frac{1}{3}$

answer : $ \frac{1}{3}$

Suppose the set {a,b} denotes our choice of children. We can do a bijective transformation from {a,b} to {b,a}. Therefore it is equivalent to calculating the probability of getting a girl in the first pick. There are 2 girls, and 6 children, so answer is $ {{2}\over{6}}\equal{}{{1}\over{3}}$

aw,how come 1/3 and not 1/2:after all there's more girls than boys,so why?!

oh,more buys,not more girls,isee thnks:)


superior algebra
superior algebra solved

Prove that in every finite group G there exists an element x with x^2=e,
x is not e.

Don't you mean finite group with an even number of elements? And isn't this fiercely trivial? :D

Just consider this: assume there isn't any blabla.. Then you have e, and you can make pairs from all the other elements, pairs of the form x,x^(-1). You then get an odd number of elements in G, false, so there exists an x=/=e s.t.x=x^(-1).


Quadratic function $f(x)$ satisfies:
(1) $f(-1)=0$
(2) For all $x$ there is always $x \leq f(x) \leq \frac{1+x^2}{2}$.
Find $f(x)$.

tricky question ;)

[hide]
we want to prove in the following, that there is no such function.

suppose there is such function, let $f(x) = ax^2+bx+c$

then we know \[ f(-1)=a-b+c=0 \iff a+c=b \]
and \[ x\le ax^2+bx+c \le \frac{1+x^2}2 \]
therefore the discriminant of $ax^2+(b-1)x+c$ is $\le 0$
so \[ (b-1)^2-4ac\le 0 \iff 0\le (b-1)^2\le4ac \]
because if $ax^2+(b-1)x+c\ge 0$ for all $x$, $a>0$ so also $c\ge0$
thus we get \[ 2\sqrt{ac}\le a+c=b\le 2\sqrt{ac}-1 \]

this is obviously is wrong, hence there can't be such function.
[/hide]

:blush:  :oops:  i guess i my transformations in the end arent all \emph{that} correct ... i will post a correct solution, as soon as i found one ;-)

alright, here is the correct solution ;) :

[hide]first \[ 0=f(-1)=a-b+c \iff a+c=b \]

second for $x=1$ \[ 1\le f(1) \le \frac{1^2+1}2 = 1\iff f(1)=1 \]
\[ \iff 1=a+b+c \stackrel{a+c=b}{\iff} b=\frac12 \]

now we consider the discriminants of $f(x)\ge x \iff ax^2-\frac12x+c \ge 0$:
\[ \frac14-4ac\le 0 \iff \frac1{16} \le ac \le \left( \frac{a+c}2\right)^2 = \frac1{16} \implies a=c=\frac14 \]

therefore the only quadratic function suffices the conditions is $f(x)=\frac14x^2+\frac12x+\frac14$, because $f(-1)=0$ and $f(x)=x \iff \frac14x^2-\frac12x+\frac14 = \left( \frac12x-\frac12\right)^2 \ge 0$ and $f(x)\le \frac{x^2+1}2 \iff \frac14x^2-\frac12x+\frac14 = \left( \frac12x-\frac12\right)^2 \ge 0$.

so the solution is indeed \[ \boxed{f(x)=\frac14x^2+\frac12x+\frac14} \][/hide]

[quote="Ezbakhe Yassin"]Why $\frac14-4ac\le 0$?[/quote]
you have that:
$ax^2 + bx + c \ge 0$ so it has one root or has none. So the dicriminant must be $0$ or negative.

By condition two, we gat two quadratic inequalities:
\[ ax^2-\frac{1}{2}x+c \geq0 \]
And:
\[ (2a-1)x^2+x+(2c-1) \leq 0 \]
So if $a \in (- \infty, 0)$, inequality one is fauls; if $a \in [ \frac{1}{2}, \infty)$, then inequality two is fauls, so we must have $a \in (0, \frac{1}{2})$. Thus we can use $\Delta$ without worrying anything.

[hide]by inspection, $y=x$, and $y=x^2/2+1/2$ are tangent at $(1,1)$, the desired curve is between those for all x, the new curve must be tangent to $y=x$ at $(1,1)$
say $f(x)=ax^2+bx+c$ and $f'(x)=2ax+b$
$f'(1)=1$, $f(1)=1$, $f(-1)=0$ are our conditions giving:

$2a+b=1$, $a+b+c=1$, $a-b+c=0$, solve the system for $(a,b,c)=\left(\frac{1}{4} , \frac{1}{2}, \frac{1}{4}\right)$[/hide]


I need a good game to play in my free time.  I don't have any gaming systems though; so it would need to be an online game.  Does anybody have any suggestions?

I prefer the tactical side of things, Shattered Galaxies or Ballerium are good choices.

Anything you are picky about?

Most things are good as long as you don't have to download or buy them.

I'm really looking for a free internet game.

theres shockwave.com and flashplayer.com

for free downloads, there's acid-play.com

too lazy to do hyperlinks :P

runescape..its free. like a cheap version of WoW or Everquest.

Also you can try Unreal Tournament 2004

Ragnarok Online.  There exist private servers which are free.
One which i'm on is discoveryRO...and there recently was a wipe, so you can start fresh on the server.

Do you have URL's to go with these sites?

You can check it out at http://www.discovery-ro.com if you want..

[quote="samath"]Does this game need to be downloaded though?[/quote]

Yeah.

notpron is fun, and challenging.  find lots of tips in the fun and games factory

oh heavens no, notpron is terrrrrrrrible


Putnam
induction
algebra unsolved
algebra

Prove that the sequence defined by $ a_0\equal{}a$, $ a_1\equal{}b$, $ a_{n\plus{}1}\equal{}\frac{{a_n}^x\plus{}{a_{n\minus{}1}}^x}{{a_n}^{x\minus{}1}\plus{}{a_{n\minus{}1}}^{x\minus{}1}}$, n\in N*, where $ a>0$, $ b>0$, $ x>1$, is convergent

[quote="Rust"]Therefore exist $ 0 < c < 1$[/quote]
Well.. That's the hard part of the problem to prove that.... Can you do it?

[quote="conan_naruto236"]Prove that the sequence defined by $ a_0 = a$, $ a_1 = b$, $ a_{n + 1} = \frac {{a_n}^x + {a_{n - 1}}^x}{{a_n}^{x - 1} + {a_{n - 1}}^{x - 1}}$, $ n\in N^{*Ư$, where $ a > 0$, $ b > 0$, $ x > 1$, is convergent[/quote]
I think this problem isn't hard.
This is my solution.
[b]Solution[/b] :
We have : $ a_{n+2}=\frac {{a_n}^x + {a_{n - 1}}^x}{{a_n}^{x - 1} + {a_{n - 1}}^{x - 1}}\geq \frac{a_{n+1}+a_n}{2}$
or $ 2a_{n+2} \geq a_{n+1}+a_n$
Let $ L=max${$ a_0,a_1$}. By induction we prove that $ 0 \leq a_n \leq L$.
or $ (a_n)$ is bounded.
Then exist $ A=sup (a_n)$ and let $ A_n=min${$ a_n,a_{n+1}$}. Clearly $ (A_n)$ is bounded.
But
$ A_n=min${$ a_n,a_{n+1}$}=$ min${$ a_n,a_{n+1},\frac{a_n+a_{n+1}}{2}$} $ \leq$ $ min${$ a_n,a_{n+1},a_{n+2}$} $ \leq$ $ min${$ a_{n+1},a_{n+2}$}$ =A_{n+1}$.
There fore by Weierstrass 's Theorem we have : $ lim A_n=B$, $ B \in R$
I will prove that $ lim a_n =B$
Indeed we have :
$ \forall \epsilon >0, \exists N\in N, \forall n\in N, n\geq N, B-\frac{\epsilon}{3}<A_n<B+\frac{\epsilon}{3}$
Any $ m>N$ we have :
$ B-\frac{\epsilon}{3}<A_{m-1}<B+\frac{\epsilon}{3}$
$ \Rightarrow B-\frac{\epsilon}{3}<A_{m-1}\leq a_{m-1}$
If $ a_m<B+\frac{\epsilon}{3} \Rightarrow B-\frac{\epsilon}{3}<A_{m-1}\leq a_m<B+\frac{\epsilon}{3}$ $ (1)$
If $ a_m \geq B+\frac{\epsilon}{3}$
$ \Rightarrow A_m=a_{m+1} <B+\frac{\epsilon}{3}$
$ \Rightarrow a_m \leq 2a_{m+1}-a_{m-1}<2(B+\frac{\epsilon}{3})-(B-\frac{\epsilon}{3})=B-\epsilon$
Therefore:
$ B-\epsilon<B-\frac{\epsilon}{3}<a_m<B+\epsilon$ $ (2)$
By $ (1)$, $ (2)$ we have :
$ \forall m\in N, m>N, B-\epsilon<a_m<B+\epsilon$
$ \Rightarrow lim a_n=B$
We are done.  :lol:


number theory
prime numbers
number theory proposed

Let $ p$ and $ q$ be two prime numbers greater than $ 3$. Prove that if their difference is $ 2^{n}$, then for any two integers $ m$ and $ n$, the number $ S\equal{}p^{2m\plus{}1}\plus{}q^{2m\plus{}1}$ is divisible by $ 3$.

It is obviosly, because $ p\equal{}\pm 1\mod 3,\ q\equal{}\pm 1\mod 3$ and $ p\not \equal{}q\mod 3$, therefore $ 3|(p\plus{}q)$.


Hi, I would like to solve this problem:

Let have a special set of keyboards. Each keyboard consists of just one row of buttons.
Every button has a frequency of typing usage.
Price of button can be expressed as  (frequency TIMES its position in the row).

Imagine that this set have M keyboards and N buttons.
On each keyboard we can place  arbitrary number of buttons equal or lower than N.

However, there is a restriction:
Buttons are numbered as a sequence of nonnegative integers from 0 to M-1  (i.e. if M==10, buttons are 0 1 2 3 4 5 6 7 8 9) and they must be placed on keyboards in order.

So, let have 4 keyboards and 10 buttons.
Then acceptable placement might be:

1. keyboard: |0|1|2|3|4|
2. keyboard: |5|6|
3. keyboard: |7|
4. keyboard  |8|9|

And frequencies of buttons might be
0  100
1   50
2   350
3   12
4   600
5   1000
6   100
7   5
8   800
9   35

So the prices of buttons are
0  100*1=100
1   50*2=100
2   350*3=1050
3   12*4=48
4    600*5=3000
5   1000*1=1000
6   100*2=200
7   5*1=5
8   800*1=800
9   35*2=70

Ideal placement of buttons on keyboards is such placement, that the sum of the prices of buttons is a lowest. (placement in the example might be not ideal, in fact, visually is clearly not ideal)

Task is - for any given number of keyboard, buttons and frequencies of buttons find an ideal placement of buttons on keyboards.
If there are more ideal placements, choose one with more buttons on tha last keyboard, if there are more ideal placements with the same number of buttons on the last keyboard, choose one
with more buttons on the penultimate keyboard and so on ...

How could be this solved using a dynamic programming technique?
We can suppose that M<=100 AND N<=100 AND frequency of each button <= 50 000

Your state should be (number of keyboards used, number of letters placed). Then the update step will be $ dp(i, j) \equal{} \max_{1 \le k \le \min(j, n)} p(j\minus{}k\plus{}1, j) \plus{} dp(i\minus{}1, j\minus{}k)$ where $ p(i, j)$ is the price of putting letters $ i$ through $ j$ on a single keyboard.

[quote="paladin8"]Your state should be (number of keyboards used, number of letters placed). Then the update step will be $ dp(i, j) \equal{} \max_{1 \le k \le \min(j, n)} p(j \minus{} k \plus{} 1, j) \plus{} dp(i \minus{} 1, j \minus{} k)$ where $ p(i, j)$ is the price of putting letters $ i$ through $ j$ on a single keyboard.[/quote]

Thank you, but could you make it a bit more clear? I'm very new to the concept of dynamic programming, so I am not really sure if I inderstood what you had written.

1. Are you sure that 2-dimensional state space is enough?
2. What is k?
3. How to get these states? Am I right that we should at first fill in that table of states? How to do this?

Yeah, sure. First off, I use dp(i, j) to represent "the minimum cost of putting the first j keys on the first i keyboards." The idea is to construct the dp table recursively starting with i = 0 as the base case.

1. I think 2 dimensions is enough. Each "step" in the DP is basically putting another set of keys on a keyboard, so you only need to know how many keys and keyboards you've used so far to determine what possible steps there are.

2. k is the variable for the number of keys you put on the next keybaord. You can put anywhere from 1 to min(j, n) in any step, so you should consider all such possibilities and take the best one.

3. So the initial state you know dp(0, 0) = 0 since placing 0 keys on the 0 keyboards costs 0. And you know dp(0, x) for x > 0 is impossible so you should assign it a cost of infinity. Then you can use the update step to compute the rest of the table (e.g. you can compute dp(1, x) for all x after that, then dp(2, x), etc.).

The update step basically answers the following question: "if I want to minimize the cost of putting the first j keys on the first i keyboards, how many keys should I put on the i-th keyboard?"


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algebra
polynomial
induction
number theory unsolved
number theory

Give $ n $ positive integers satisfying $ (x_1,x_2,...,x_n)=1 $.
Take $ S_k = x_1^k+...+x_n^k $.
Prove that:
$ [1,2,...,n] $ is divided by $ (S_1,S_2,...,S_n) $.

Why hard? By Newton's formula (see, for instance,
http://mathworld.wolfram.com/Newton-GirardFormulas.html,
coefficients $a_1,a_2,\dots,a_{n}$ of polynomial
$f(x)=(x-x_1)(x-x_2)\dots (x-x_n)=x^n+a_1x^{n-1}+\dots+a_n$
satisfy relations like $ia_i=P(a_1,a_2,\dots,a_{i-1},S_1,S_2,\dots,S_n)$
($P$ is a polynomial with integer coefficients)..
From this, if all $S_i$ are divisible by $p^k>n$ for some prime $p$, then by
induction all $a_i$ are divisible by $p$. So, $f(x)=x^n$ modulo p.
But hence $x_i^n=f(x_i)=0$ modulo $p$ and all $x_i$ are divisible by $p$, which contradicts to $(x_1,x_2,\dots,x_n)=1$.


function
topology
real analysis
real analysis unsolved

Can anybody help me with the following proofs?

1. If K is a compact subset of $R^{d}$ and
U1 ⊂ U2 ⊂ · · · ⊂ Uk ⊂ · · ·
is a nested upward sequence of open sets with K ⊂ ∪kUk, then prove that K is contained in one of the sets Uk.

2. Prove that if K is a compact subset of $R^{d}$, then K contains points of minimal norm and points of maximal norm. That is, there are points x0, x1 ∈ K such that

$\parallel{\it x0}\parallel \leq  \parallel{\it x}\parallel \leq \parallel{\it x1} \parallel$ for all x ∈ K.

3. Prove that if K is a compact subset of $R^{d}$ and y is a point of $R^{d}$, which is not in K, then there is a closest point to y in K. That is, there is an x0 ∈ K such that

$\parallel x0-y \parallel \leq  \parallel x-y \parallel$ for all x ∈ K.

4. Prove that a compact metric space is complete (that is, every Cauchy sequence converges).

Sorry but all problems are seem basic:
1) Just using Heine- Borel Lemma.
2)+3) the dist. fucntion is continous
4) Because every Cauchy sequence in compact space is bounded then exist a subsequence which is convergent.


Very nice problem!
My solution:
By Cauchy result:
1)  $ (\sum_{cyc}\frac{\sqrt{b+c}}{a}}})(\sum_{cyc}{\sqrt{b+c})\ge (\sum_{cyc}{\sqrt{\frac{b+c}{a}})^2 $
But,is well-known inequality,
2)  $ \sum_{cyc}{\sqrt{\frac{b+c}{a}}\ge\sqrt{\frac{6(a+b+c)^2}{bc+ca+ab}} }$
which also belongs to Ductrung!
From 1) and 2) result:
3)   $ \sum_{cyc}{\frac{\sqrt{b+c}}{a}}\ge\frac{6(a+b+c)^2}{(bc+ca+ab)(\sum_{cyc}{\sqrt{b+c}})} $
By Tchebyshev result:
4)   $ \sum_{cyc}{\sqrt{b+c}}=\sum_{cyc}{a\frac{\sqrt{b+c}}{a}}\le\frac{1}{3}(\sum_{cyc}{\frac{\sqrt{b+c}}{a}})(a+b+c) $
After some calculations from 3) and 4) results QED!!
____
Marin Sandu

If $ a,b,c>0 $  then:
[1]    $ \sum_{cyc}{\frac{\sqrt{b+c}}{a}}\ge\sqrt{\frac{3}{a+b+c}}\sum_{cyc}\sqrt{\frac{b+c}{a}} $
[2]    ${{ \sum_{cyc}{\frac{\sqrt{b+c}}{a^2}}\ge\frac{3}{a+b+c}}\sum_{cyc}\frac{\sqrt{b+c}}{a}} $
____
Marin Sandu

[quote="ductrung"]For positive real numbers $a,b$ and $c$, prove that
\[ \frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}+\frac{\sqrt{a+b}}{c} \ \ge\ 3\sqrt{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}  \][/quote]
By Holder $(\sum_{cyc} \frac{\sqrt{b+c}}{a})^2=\sum_{cyc}\frac{b+c}{a^2}+\frac{2}{abc}\sum_{cyc}a\sqrt{(a+b)(a+c)}=$
$=\sum_{cyc}\frac{b+c}{a^2}+\frac{2}{abc}\sqrt{(\sum_{cyc}a\sqrt{(a+b)(a+c)})^2\sum_{cyc}\frac{a}{(a+b)(a+c)}\cdot\frac{1}{\sum\limits_{cyc}\frac{a}{(a+b)(a+c)}}}\geq$
$\geq\sum_{cyc}\frac{b+c}{a^2}+\frac{2}{abc}\sqrt{\frac{(a+b+c)^3}{\sum\limits_{cyc}\frac{a}{(a+b)(a+c)}}}=\sum_{cyc}\frac{b+c}{a^2}+\frac{2}{abc}\sqrt{\frac{(a+b+c)^3(a+b)(a+c)(b+c)}{2(ab+ac+bc)}}\geq$
$\geq\sum_{cyc}\frac{b+c}{a^2}+\frac{2}{abc}\sqrt{\frac{(a+b+c)^3\cdot\frac{8}{9}(a+b+c)(ab+ac+bc)}{2(ab+ac+bc)}}=\sum_{cyc}\frac{b+c}{a^2}+\frac{4(a+b+c)^2}{3abc}$.
Thus, it remains to prove that $\sum_{cyc}\frac{b+c}{a^2}+\frac{4(a+b+c)^2}{3abc}\geq\frac{18(a+b+c)}{ab+ac+bc}$.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, it's obvious that the last inequality is equivalent to $f(w^3)\leq0$, where $f$ is a convex function.
Id est, by $uvw$ it remains to prove the last inequality for $b=c=1$, which gives $(a-1)^2(10a^2+20a+3)\geq0$. Done!


number theory
greatest common divisor
algorithm
search

Well this isn't much of a method but simply divide $4k^2-5$ by $2k+3$ and it is leaving a remainder($4k^2-5$ > $2k+3$ for $k\geq3$)(taking $k$ integer)

[quote="dondigo"]Prove, that:
$gcd(2k+3, 4k^2-5)=1$[/quote]

use the euclidian algorithm!!

$\gcd(4k^2-5,2k+3)=\gcd(2k+3,-6k-5)=\gcd(2k+3,4)=\gcd(2k-1,4)$ which is obviously $1$ since the left is odd, the right is even

search before asking what it is

[quote="mna851"]The GCD of an odd number and an even number can be different from one.  For example take $\ 6$, and $\ 9$.  Although it does work in this case because in order for their GCD to be different from one $\ 2k-1$ must contain a factor of $\ 2$.  It clearly doesn't because it is odd.[/quote]

sorry, i was not clear, obviously, for them to have a common factor with 4, it has to be divisible by $2$ (or $2\cdot 2$), but since the other term is odd, that cannot be


which do you think is more cheating. being underaged and lieing about it or testing positive for doping.

the womens chinese gymnastic team has been accused a lot of having underaged gymnasts. BUt there passports say they're 16. The IOC totally agrees and beleivesthem.the gymnasts obvioulsy do not look like 16 except a few.

or is this worse

Marion Jones(i think thats her name) admitted to doping and got stripped of her medals in 2000. Her teammates got stripped of there medals even though they didn't do anything wrong. Now doping is occuring more so athletes that arereally good like michael phelps have to do lots of drug tests and stuff just to prove they're innocent.

doping is taking banned substances, like steroids or human growth hormones

in that case i think doping is worse. i cant really see whats all that bad about being underaged...

i got to say doping. underage means you have extraordinary talent that surpasses many people older than you. i think thats a plus for hte gymnasts themselves. and its not cheating. however, lying about age is a bit iffy. i dont think that should be allowed considering taht the IOC has a rule on that. I think that would fall under the same category as underage drinking, only less severe. But looks can be decieving charmander ;]

Doping is worse cause its not even your own talent anymore. its stunted, and thats cheating. underage means you have extraordinary talent. doping means you dont have talent but wanna show up others so you cheat.

[quote="dreambold22"]what "doping"?

and whats so BAD about being underaged, seriously?[/quote]

well its againist the rules, people feel cheated if they know they lost because someone cheated. The reason that rule exists is to protect the athlete's if your wondering. The rule was instated so that young athletes can't cause permanent muscle damage or something. The athletes have to wait till their bodies are fully developed which is usually when they're 16.

I think that both forms of cheating are equally wrong.  A rule is a rule, and just because you think it's dumb or pointless doesn't meant you don't have to follow it.  If it is somehow proven that the Chinese gymnasts (or any other Olympic athletes) were underaged, they should face the same consequences as Marion Jones and the other athletes convicted of doping.

[quote="dreambold22"]and whats so BAD about being underaged, seriously?[/quote]

I will actually answer this question.

When you are younger, you are more flexible. Thus you can pull off bigger and better moves.

[quote]
and whats so BAD about being underaged, seriously?
[/quote]
Also, when you're younger, you're shorter, so you have a lower center of gravity, which enables you to pull more difficult moves.

[quote="jjx1"]first of all marion jones teamates were not convicteed of doping. only marion jones. and actually officially the chinese gymnasts are 16. it just doesn't look like it[/quote]

O please...those documents could have EASILY been changed by the government so they could compete. Also the chinese government doesnt exactly have a clean record in this sort of thing. Remember the chinese girl in a previous olympics who admitted she was underage? Since they've already cheated once we have no reason what so ever to trust them. The only reason the IOC and FGI (I think those are the initals) wont do an investigation is because their money greedy idiots who don't care about what the olympics stand for anymore. That felt good to get out.

well duh they're probably not 16. but no one will ever be able to covict them. but the ioc won't investigate because they know china is important

[quote="dreambold22"]

and whats so BAD about being underaged, seriously?[/quote]

the younger you are, the better you are able to perform, that itself is nto a bad thing, but if some country decided to ignore the rule, *cough* china *cough* and decided that gold medals worth more than the future of a young girl, they would have access to a wider pool of talents, giving them an unfair advantage.

it is cheating, the younger the person is, the more flexible they are, and they are also shorter, which cause them to have a lower center of gravity.

[quote]BEIJING - [b]Just nine months before the Beijing Olympics, the Chinese government's news agency, Xinhua, reported that gymnast He Kexin was 13[/b], which would have made her ineligible to be on the team that won a gold medal this week.

....

[b]The Associated Press found the Xinhua report on the site Thursday morning and saved a copy of the page. Later that afternoon, the Web site was still working but the page was no longer accessible. Sports editors at the state-run news agency would not comment for publication.[/b]

Gymnasts have to be 16 during the Olympic year to be eligible for the games. [i][b]He's birthday is listed as Jan. 1, 1992.[/b][/i][/quote]

jan. 1st...wow, wats the chance of that? :wink:

plus couple other of her teammates are also exactly 16..

nahh...nothing fishy about that...

[quote="crazychinesemaniac"][quote="serialk11r"]If you check the heights on most of these gymnasts, those "underaged" (supposedly) girls are not gonna grow anymore. Their heights are all like, 5'1", 5'3", 4'9", even the ones who are 20+ years old. Who says when you're younger you can pull off moves better? Got evidence to back that up?

I wouldn't call being underaged cheating. It would be lying.
Doping is cheating though.[/quote]

99.9999999999% of people 16-20 are taller than 5".  :wink:[/quote]

lol i'll b lucky to reach 5' (i think u meant five ft) i'm 14 and only 4' 10" =]

an and ding. some chinese ppls have b-days on new year cuz like new year one year older of something like that... but yes why not just wait a couple more years if they are in fact younger. they're doing gymnastics everyday they really aren't going 2 b much less flexible


geometry
ratio
AMC
AIME
trigonometry
parallelogram
area of a triangle

In the triangle $\triangle ABC$, the segments have the lengths shown and $x + y = 20$. Find its area.

[img]http://www.kalva.demon.co.uk/aime/soln/diag8915.jpg[/img]

you can do a lot with area ratios, but it gets kind of messy, then finish off with ceva's theorem, but i don't have an answer now...

using the area ratios, you find that x=15, and y=5, and
APE=APF=1/2EPC=1/2FPB=1/3PDB=1/3PDC
but i cannot solve from there...

The lines aren't medians, even though they look like it, right?

They look like it, but they aren't. It doesn't say that anywhere in the problem. And if they were, then $6/6=2/1$, but that's not true. ;) (The median split each other in a 2:1 ratio.)

Yeah, I was pretty sure that that was the case.

I'd think we might be able to use an approach having to do with the sums of the areas of each triangle by Heron's Formula being equal to two times the area of the triangle itself, also by Heron's Formula, and it might be possible to factor out the sides of the larger triangle itself from there.

This problem lends itself to masspoints very well it seems, especially since 6+6=9+3.

well, actually one of them turns out to be a median if you work out some of the problem

Here's my solution, but it's extremely long :roll:
[hide]
Using a different form of ceva's theorem, we have $\frac{y}{x+y}+\frac{6}{6+6}+\frac{3}{3+9}=1\Longleftrightarrow\frac{y}{x+y}=\frac{1}{4}$

Solving $4y=x+y$ and $x+y=20$, we obtain $x=15$ and $y=5$.

Then using mass points, we see that $AF: FB=1: 2$, $BD=DC$, and $AE: EC=1: 2$.

From here, we need another theorem (I forgot what it was called):
[hide="theorem"]
In general, if you have $\triangle XYZ$ and $M$ is the midpoint of $YZ$, then we always have $XY^2+XZ^2=2(XM^2+YM^2)$.

(This theorem can be proven in many ways.)[/hide]

Let $BD=p$, $AB=a$, and $AC=b$.

Applying the theorem to $\triangle PBC$,
$15^2+9^2=2(6^2+p^2)$ ($PD$ is the median)
$225+81=2p^2+72$
$p^2=117$

Applying the theorem to $\triangle DAB$,
$a^2+p^2=2(15^2+6^2)$ ($PB$ is the median)
$a^2+117=450+72$
$a^2=405$

Applying the theorem to $\triangle CAD$,
$b^2+p^2=2(9^2+6^2)$ ($PC$ is the median)
$b^2+11=162+72$
$b^2=117$

So $AB=9\sqrt{5}$, $AC=\sqrt{117}$, and $BC=2\sqrt{117}$.

By Hero's formula,
$\triangle ABC=\sqrt{\left(\frac{3\sqrt{117}+9\sqrt{5}}{2}\right)\left(\frac{9\sqrt{5}-\sqrt{117}}{2}\right)\left(\frac{\sqrt{117}+9\sqrt{5}}{2}\right)\left(\frac{3\sqrt{117}-9\sqrt{5}}{2}\right)}$

$=\frac{\sqrt{(9\cdot 117-81\cdot 5)(81\cdot 5-117)}}{4}$

$=\frac{\sqrt{648\cdot 288}}{4}$

$=\frac{\sqrt{9\cdot 72\cdot 4\cdot 72}}{4}$

$=\boxed{108}$

There must be a better way...[/hide]

Congratulations! You just solved a #15 on an AIME! :)

Does anyone have an elegant solution? :D

Ooh, pick me! :D I have to go to school now, but I shall post when I'm back.

Here's a summary of my trig solution...[hide]
Let the area of $\triangle{APE}=r$.  By various area ratios, one obtains $\triangle{APF}=r, \triangle{APE}=2r, \triangle{CPD}=3r, \triangle{BPD}=3r, \triangle{BPF}=2r$.  Let $m\angle{APE}=\theta$.  Then we have $\angle{APF}=\theta$, $\sin{\angle{CPE}}=\frac{4}{3}\sin{\theta}$.  Now let $\sin{\theta}=n$.  We have $2\theta+\sin^{-1}{\frac{4}{3}\sin{x}}=180^{\circ}$.  Take the sine of both sides to get \[ 0=\sin{(2\theta+\sin^{-1}{(\frac{4}{3}n)})}=\sin{2\theta}\sqrt{1-\frac{16}{9}n^2}+\cos{2x}\frac{4}{3}n \]
\[{ \implies 0=2n\sqrt{1-n^2}\sqrt{1-\frac{16}{9}n^2}+\frac{4n}{3}}(1-2n^2) \implies \frac{2}{3}-\frac{4}{3}n^2+\sqrt{1-\frac{25}{9}n^2+\frac{16}{9}n^4}=0 \] \[ \frac{16}{9}n^4-\frac{25}{9}n^2+1=\frac{16}{9}n^4-\frac{16}{9}n^2+\frac{4}{9} \] \[ \implies n^2=\frac{5}{9} \implies \sin{\theta}=\frac{\sqrt{5}}{3} \]
\[ \implies r=9n=3\sqrt{5} \] because the area of a triangle is $\frac{1}{2}ab\sin{C}$.  Hence, the area of the triangle is $12r=\boxed{36\sqrt{5}}$[/hide]

I probably messed something up within all of that plugging and chugging through trig...I recall once doing it  in a much easier way...

Well, using mass points you can get x=15, y=5 pretty easily.  You just have to remember each cevian has mass points in terms of a scale factor which isn't necessarily the same for each.  However, doing some arithmetic with the equations you can find x and y in terms of the scale factors, and find x = 3y.

Not sure how to go about getting the area from here though... I feel it shouldn't be that hard knowing all the cevians, I think there's some simple theorem I'm missing.

Probably stuff with $ab\sin{C}$ and $\sin{2\theta}=2\sin{\theta}\cos{\theta}$. I think I see a solution using that at the end.

I found another geometric solution. (I believe this one is more "elegant" than my previous one :) )
[hide]
We can find $BP=15$ and $EP=5$ by using mass points, Ceva's Theorem, etc. (I used Ceva's, as shown in my previous solution.)

Let $Q$ be the point on $AB$ such that $FC\parallel QD$.
Since $AP=PD$ and $FP\parallel QD$, $QD=2FP=6$. (Midline Theorem)

Also, since $FC\parallel QD$ and $\QD=\frac12FC$, we see that $FQ=QB$, $BD=DC$, etc. (Midline Theorem)
Similarly, we have $PR=RB$ ($=\frac12PB=7.5$) and thus $RD=\frac12PC=4.5$.

$PDR$ is a 3-4-5 right triangle, so $\angle PDR$ ($\angle ADQ$) is $90^\circ$.
Therefore, the area of $\triangle ADQ=\frac12\cdot 12\cdot 6=36$.
Using area ratio, $\triangle ABC=\triangle ADB\times 2 =(\triangle ADQ\times \frac32)\times 2=36\cdot 3=\boxed{108}$
[/hide]

[quote="frt"][quote="chess64"]What's the midline thm?[/quote]
Given $\triangle ABC$, point $D$ on $AB$ and point $E$ on $AC$.

Then, $AD=DB$ and $AE=EC$ $\Longleftrightarrow$ $DE\parallel BC$ and $DE=\frac12BC$.
Alternatively, if $AD=DB$ and $DE\parallel BC$, then $AE=EC$ and $DE=\frac12BC$.

There might be another name for this theorem though.[/quote]

It's also called the Midsegment Theorem I believe..essentially the same thing.
From what I remember, it states
In a triangle, if a line is drawn through the midpoint of one side parallel to the second, then it bisects the thrid side.  In fact, any two of those three statements imply the third.  The segment formed is called the midsegment and has length equal to half of the length of the side it's parallel to.

Also just a specific case of the theorem which states that parallel lines cut lines into proportional segments: namely, that when three or more parallel lines cut another line, the propoertions between the segments formed are always the same.

I remember doing this problem at a math club meeting before, though I only found the solution after I was given a hint.

[hide="The Hint"]Examine the angle between the cevians given  ;) [/hide]
[hide="Geometric Solution"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=596&sid=6468ca49c7f5848367ea9cbe964daad8[/img]

Because the length of cevian $BE$ is unknown, we can examine what happens when we extend it or decrease its length and see that it simply changes the angles between the cevians. Wouldn't it be great if it the length of $BE$ was such that $\angle APC=90^\circ$?  Let's first assume it's a right angle and hope that everything works out. :P

Extend $AD$ to $Q$ so that $PD=DQ=6$. The result is that $BQ=9$, $PQ=12$, and $BP=15$ because $\triangle CDP\cong \triangle BDQ$. Now we see that if we are able to show that $BE=20$, that is $PE=5$, then our right angle assumption will be true.

Apply the Pythagorean Theorem on $\triangle APC$ to get $AC=3\sqrt{13}$, so $AE=\sqrt{13}$ and $CE=2\sqrt{13}$. Now, we apply the magical Law of Cosines on triangles $CEP$ and $AEP$.  :)

Let $PE=x$. Notice that $\angle CEB=180^\circ -\angle AEB$ and $\cos CEB=-\cos AEB$, so we get two nice equations.

$81 = 52 + y^2 - 2y \sqrt{13}\cos CEF$
$36 = 13 + y^2 + y \sqrt{13} \cos CEF$

Solving, $y=5$ and we're happy!

Now, the area is easy to find. $[ABC]=[AQB]+[APC]=\frac12(9)(18)+\frac12(6)(9)=\boxed{108}$[/hide]

Lol, if I took the Aime and just guessed the angle measure, this problem would've taken 15 seconds.  :D

we are overcomplicating the problem...
[hide="solution"]using mass points, $w_C=1$, $w_D=m=w_A$ and $w_B=n$, $w_F=3n$, using mass point principles,
$1+m=3n$ and $1+n=m$, solving, $m=2$ and $n=1$, using this,
$\frac{x}{y}=3$ and $CD=BD$  (making $PD$ a median), solving, $x=15$ and $y=5$
due to the ratio of $PD$ to $AD$, $2[CPB]=[ABC]$, now focus on $\triangle CBP$
using the median formula, $BC^2+12^2=2(15^2+9^2)\implies BC=6\sqrt{13}$
Using heron's where the sides are $6\sqrt{13}$, $9$, $15$, $[CBP]=54 \implies [ABC]=108$[/hide]

Hm, I had the mass points part, but didn't notice the median there.

What formula is that you used?  I don't recognize it and mathworld has nothing under that name.

do you mean the median one or heron's?

heron's: http://mathworld.wolfram.com/HeronsFormula.html

the median one is just a specialized form of stewart's

the median thing is in aops, it is essentially if you have a parallelogram, the sum of the squares of the sides is equal to the sum of the squares of the diagonals, and you get the parallelogram by rotating the triangle 180 degrees about the midpoint of the side, and krusty is right, it also can be shown (pretty easily in fact) through stewart's theorem
notation: $\triangle ABC$ with $D$ mp of $BC$, $CD=n$, $BD=m$, other regular $a$, $b$, $c$ assignments
$cnc+bmb=dad+man$, $2m=2n=a$, so it simplifies to $2c^2+2b^2=4d^2+a^2$

[quote="Altheman"]we are overcomplicating the problem...[/quote]Heh, I liked my solution.  :P

I don't know how to use mass points. I might take the time to learn them later though.

[quote="Altheman"]4everwise, it is pretty similar to what i did except i did not bother to draw the altitude to one of the triangles, if you have a triangle with sides: $9$, $15$, and $6\sqrt{13}$, that screams to me draw an altitude as $2^2+3^2=13$, and i think it works out very similar to your solution since drawing an altitide to the 9 side should have a length of 12, your solution is interesting, but i think that is not a reliable way to do problems since you cannnot easily replicate those guesses.[/quote]After I learned mass points, I was able to solve the problem in the same way you did. (without referring to solution)

Mass points are awesome!  :D

4everwise, how (where) did you learn mass poitns quickly?


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Can some one explain to me this step by step? Like for a 3years old girl:)

If the atomic weight of platinum is 195.08g/m, what is the mass of a single platinum atom????

Calculate the number of moles of Phosphorus atoms in 100g of P4S10?

Thanks!
I'm new  to chemistry :(

Again, I'm not giving you the answer. Just follow my steps.
Since the gram molecular mass of $Pt$ is given, this is the wight of 1 mol of $Pt$ atoms or $ 6.023 \times 10^{23} $ atoms of $Pt$. Thus, if you divide the weight of $6.023 \times 10^{23} $ particles by the no. of particles = $ 6.023 \times 10^{23} $, we get the weight of one atom!!


AMC USA(J)MO USAMO AIME Putnam inequalities algebra

http://www.unl.edu/amc or you can just read this forum to find out...

People don't necessarily do alot of math. Like the people who have already experienced IMO...

actually, you probably do have to do a lot of math to get to IMO. To get to USAMO, luck could partially get you there.

You are mistaken, people who have already made IMO do even more math than those who haven't.

I would agree. Usually the better someone is at something, the more he or she does it (note the word USUALLY). Unless he or she hates it or something like that.

Even problems on the USAMO can be somewhat predictable.

Well, it's my lay opinion that USAMO-level problems are not predictable enough for practice alone to be sufficient for getting on the IMO team (USA IMO team that is...and there probably are some years where it's quite a bit easier than others).

[quote="MithsApprentice"]Well, for us non-supergeniuses, doing A LOT of math (but you also need quite a bit of natural talent) is the best way to make IMO. If you're intellectually above practicing, I salute you.

As for USAMO, I think the AMC and AIME are formulaic enough so that someone who is not all that talented but extremely determined could qualify.[/quote]

I think even the USAMO is formulaic to a certain degree, especially with those inequalities.  I'm not even sure it's much harder than AIME...perhaps it's much harder to say get 30+ on USAMO than 12+ or so on AIME, but last year I got 8 on AIME which I studied a decent amount for and a 16 on USAMO with about a month or so of sporadic studying after I found I got in (to my disbelief).  (Actually, I basically perused the kalva answers section lol).  I think the 3rd problems on each day are much harder than 12-15 on AIME.  But I don't really think the other 4 problems are much harder than the AIME.

I think #1 and #4 aren't usually too difficult once you get used to USAMO style problems.  #2 and #5 are considerably harder, and you might run into time trouble.  Also, sometimes rigor and clarity are hard to achieve in your proof.  I don't even think I could get a 3 on any of the #3's and #6's on recent USAMOs.

[quote] I think the 3rd problems on each day are much harder than 12-15 on AIME. But I don't really think the other 4 problems are much harder than the AIME.[/quote]

That's ridiculous. The only reason the last few AIME problems are "hard" is because of the time limit. Make the AIME a 4 1/2 hour exam and there would be a massive increase in the number of 15's. However, make the USAMO a 6 hr exam (each day) and the scores would not change dramatically for people who can't do the third and sixth problems. You get 4 1/2 hours per day, which is 2:15 for each of the first two problems if you don't do the third one. Get a 28 and then say that 1, 2, 4, 5 are not much harder than the AIME.

I agree with darktreb here. Saying 1,2,4, and 5 are "not much harder" than the last AIME problems is quite a ridiculous charge. The limiting factor on many AIME scores is the time limit, not the difficulty level. It's vice versa on the USAMO. I'd say the only people who have a right to complain about not having enough time or enough difficulty on the USAMO are those who earn at least 28 points. You'll find that the direct relationship between USAMO scores and USAMO indices is fairly limited because they are entirely different fields of problem solving. In my opinion, there is just enough correlation so that a claim saying someone who does well on USAMO has the "math skills" necessary to do well on the two preliminary tests would be mostly accurate.

I really don't find the AIME that easy or the USAMO that difficult.  (especially the harder AIME's in the 90's.  The 2002, 2003, 2004 ones are of course easier.)  I don't have a chance to prove myself with a 28 since I only took the USAMO that one time (i can only take putnam now), but I more or less got a 16 with no studying.  I don't think it's much of a stretch to think I could figure out how to do problems 2 and 4 with a year of study.  I guess I didn't make something clear.  I didn't mean to say that say problem 13 on the AIME is about the level of problem 2 on the USAMO.  I made that comment keeping in mind the different time limits for the two exams.  In other words, solving 13 in 30 minutes is just as hard as solving 2 in 90 minutes.

[quote]In other words, solving 13 in 30 minutes is just as hard as solving 2 in 90 minutes.[/quote]

First off, I think replace "13" with "14" or "15" and you are closer to being accurate. However, I think it comes down to this.

There very often are problems 2 or 5 that are quite difficult, and even spending 3, 4 hours may get you nowhere (time becomes less and less effectively spent as you spend more of it on a specific problem). Consider #5 in 2003, which aside from the strange normalization (which was somewhat rare), was a tough problem. #5 in 2004 was dependent on a substitution that, while arguably obvious, was pretty much the core of the problem and if hypothetically you just wanted to play with the fifth powers and use Schur's (for all I know there is a Schur solution but I don't really care) it would be very hard.

On the other hand, I would be hard pressed to find a #13 not doable in 30 minutes, if you are at the skill level to be able to make significant progress on 2, 5.

Finally, getting a 16 on the USAMO is very admirable, but the increase in difficulty between getting 16 and 28 is massive. Problems 1 and 4 are accessible to anyone with the determination to tackle them (look at #4 in 2004, which required no mathematical knowledge whatsoever aside from logic). You generally don't need to know too many special things to do 1 or 4, and often one good trick does the job almost instantaneously (specifically consider 2002, 2003, 2004 #1's).

It's not quite a "I have worked for a long time now I can always do the problems" kind of thing. Just look at Anders Kaseorg not getting higher than 24 (which is less than four problems) last year; this after going to two IMOs.

What is a typical 12th place score on USAMO?  Preferably, what have they been in past years?

[quote="darktreb"]hypothetically you just wanted to play with the fifth powers and use Schur's (for all I know there is a Schur solution but I don't really care) it would be very hard.[/quote]

Unlikely; Schur is most generally a statement about homogeneous polynomials; the polynomial you would get out of expanding that problem would be anything but homogenous.  Probably the best you would be able to hope for is a hideous sum of squares and other positive terms.

I agree as well :) .  I'm hoping to get problem #1, maybe #4, and that's about it.

By the way, have any of you actually sat down and done a 4.5 hour practice session in one go?  Cuz maybe that makes a difference.  I know I never have one 4.5 hour block of time.

Also, for those saying the USAMO is predictable, do you mind telling me what's going to be on it?

I don't think a 4.5 hour session would help.  I don't remember getting any practice out of taking the USAMO last year :D.

As for what will be on the USAMO, they are almost always from the following topics:

-- Inequalities
-- Geometry, especially with circles inscribed in quadrilaterals or triangles
-- Functional equations
-- Game theory type problems
-- Stuff having to do with sets of integers (so essentially number theory)

-- Combinatorics shows up once in a while, too, at least on other NMOs.

For geometry, there are plane geometry/solid geometry

Correct me if I'm wrong, but I don't think there's solid geometry on the USAMO anymore. (maybe because it was removed from IMO?)

I've looked at quite a few relatively recent USAMO problems and haven't seen one solid geometry problem yet.

There's never any solid geometry and as a general rule of thumb you probably wouldn't expect three consecutive years of "classical"  Inequalities (notice after the last two years, how there wasn't one this year).

3's and 6's generally alternate between Geometry problems and somewhat interesting (as in different) problems, such as 2002 #6, 2003 #3, #6, 2004 #3, and 2005 #6.

I would be very surprised to see a classical inequality or functional equation as #6 on the USAMO.


Let $ a,b,c\geq 0$ s.t. $ a^2\plus{}b^2\plus{}c^2\equal{}1$.Prove that:
$ \frac{bc}{a^2\plus{}1}\plus{}\frac{ca}{b^2\plus{}1}\plus{}\frac{ab}{c^2\plus{}1}\leq \frac{3}{4}$.

[quote="Inequalities Master"]Let $ a,b,c\geq 0$ s.t. $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.Prove that:
$ \frac {bc}{a^2 \plus{} 1} \plus{} \frac {ca}{b^2 \plus{} 1} \plus{} \frac {ab}{c^2 \plus{} 1}\leq \frac {3}{4}$.[/quote]
This problem will be solve by method pqr . Put a+b+c=p=1;ab+bc+ca=q;abc=r

[quote="onlylove_math"][quote="Inequalities Master"]Let $ a,b,c\geq 0$ s.t. $ a^2 + b^2 + c^2 = 1$.Prove that:
$ \frac {bc}{a^2 + 1} + \frac {ca}{b^2 + 1} + \frac {ab}{c^2 + 1}\leq \frac {3}{4}$.[/quote]
This problem will be solve by method pqr . Put a+b+c=p=1;ab+bc+ca=q;abc=r[/quote]

this ineq is really easy so we dont need to use  pqr
Here is my proof
$ \frac {bc}{a^2 + 1} + \frac {ca}{b^2 + 1} + \frac {ab}{c^2 + 1}\leq\ \sum\ \frac {(b + c)^2}{4(a^2 + c^2 + a^2 + b^2)}$
$ LHS \leq \ \sum\( \frac {b^2}{a^2 + b^2} + \frac {c^2}{a^2 + c^2})$ (since $ T_2 Lemma$
then the last is quality


real analysis
integration
function
real analysis unsolved

I've been studying for a midterm, and I came to this problem in Rudin that I've gotten rather hung up on.

$ \mu$ a $ \sigma$-finite measure, $ 1 \leq p \leq \infty$, $ g$ measurable such that $ f \in L_p$ implies $ fg \in L_1$.  Prove that $ g \in L_q$.

I see where this question is coming from, it's asking me to prove $ f \rightarrow \int_X \left|fg\right|\ d\mu$ is bounded, but this doesn't seem to get me very far.  Breaking up $ X$ into a countable collection $ \{E_i\}$ with $ \mu(E_i) < \infty$, I can of course prove $ g \in L_q(\mu, E_i)$, but I don't see any reason why the sum (to get $ g$) needs to converge.  I've also tried using simple functions that converge to various things, but again, brick wall.

Any hints anyone can give me?  Please, no full solutions.

Certainly a similar question, but I don't see how that link answers my problem.

:what?: The question in Kent's link and your question are [b]exactly[/b] the same up to the notation (your $ p$ is his $ q$ and vice versa) and switching to the contrapositive form of the assertion (well, I hope it is called "contrapositive" in English when you replace $ A\Rightarrow B$ with $ \neg B\Rightarrow \neg A$ :?).

Ah, right.  I didn't consider switching $ f$ and $ g$.


Sea $ABC$ un triangulo acutangulo con el incirculo $(I)$. $(K_{A})$ es el circulo tal que $A \in (K_{A})$ y $AK_{A}\perp BC$ y este es tangente internamente a $(I)$ en $A_{1}$, similarmente tenemos $B_{1}, C_{1}$.\\

a) Probar que $AA_{1}, BB_{1}, CC_{1}$ son concurrentes, llamemos a este punto $P$.\\
b) Asumir que los circulos $(J_{A}), (J_{B}), (J_{C})$ son simetricos para los excirculos $(I_{A}), (I_{B}), (I_{C})$ atravez de los puntos medios de $BC, CA, AB$, respectivamente.\\
Probar que $P_{P/(J_{A})}= P_{P/(J_{B})}= P_{P/(J_{C})}$.\\

Nota: Si $(O;R)$ es un circulo y $M$ es un punto entonces: $P_{M/(O)}= OM^{2}-R^{2}.$

[quote="carlosbr"]Sea $ ABC$ un triangulo acutangulo con el incirculo $ (I)$. $ (K_{A})$ es el circulo tal que $ A \in (K_{A})$ y $ AK_{A}\perp BC$ y este es tangente internamente a $ (I)$ en $ A_{1}$, similarmente tenemos $ B_{1}, C_{1}$.\\

a) Probar que $ AA_{1}, BB_{1}, CC_{1}$ son concurrentes, llamemos a este punto $ P$.\\
b) Asumir que los circulos $ (J_{A}), (J_{B}), (J_{C})$ son simetricos para los excirculos $ (I_{A}), (I_{B}), (I_{C})$ atravez de los puntos medios de $ BC, CA, AB$, respectivamente.\\
Probar que $ P_{P/(J_{A})} \equal{} P_{P/(J_{B})} \equal{} P_{P/(J_{C})}$.\\

Nota: Si $ (O;R)$ es un circulo y $ M$ es un punto entonces: $ P_{M/(O)} \equal{} OM^{2} \minus{} R^{2}.$[/quote]

a) Sea $ D$ el punto de tangencia de $ (I)$ en $ BC$ y $ D'$ el punto diametralmente.
Sean $ X,Y,Z$ los puntos de tangecia de los excíruclos $ (I_{A}), (I_{B}), (I_{C})$ en los lados $ BC,CA,AB$ respect.
Como los círuclos $ (K_a),(I)$ son homotéticos con centro en $ A_1$ los puntos $ A$ y $ D'$ son correspondientes en esta homotecia, es decir $ A,D',A_1$ son colineales. Además (por una homotecia en $ A$ del incírculo y el excírculo) $ A,D',X$ son colineales..
Así las rectas $ AA_{1}, BB_{1}, CC_{1}$ son las rectas $ AX, BY, CZ$ que concurren por ceva.

b)$ AX$ es el eje radical de los círculos $ (J_{B}), (J_{C})$:
$ A$ pertenece al eje radical porque las tangentes a cada círculo miden lo mismo (pues son tangentes del incírculo).
$ X$ pertenece al eje radical porque teine la misma potencia de punto con respecto a cada círculo:
Si $ M$ es punto medio de $ AC$, Calculando todo:
Por la formula de medianas en $ J_BI_BX$
$ XM^2\equal{}\frac{1}{2} (XJ_B^2\plus{}XI_B^2)\minus{}MI^2$
Por pitágoras:
$ XI^\equal{}c^2\plus{}r_B^2$
Por pitágoras:
$ MI^2\equal{}r_B^2\plus{}\frac{1}{4}(a\minus{}c)^2$
Por la formaula de medianas:
$ XM^2\equal{}\frac{1}{2}(XA^2\plus{}(s\minus{}b)^2)\minus{}\frac{1}{4}(b^4)$

Entonces $ XJ_B^2\equal{}[DA^2\plus{}(s\minus{}b)^2\minus{}\frac{1}{2}(b^2)]\plus{}[2r_B^2\plus{}\frac{1}{2}(a\minus{}c)^2]\minus{}[c^2\plus{}r_B^2]$
$ P_{X/(J_{B})} \equal{} DA^2\plus{}(s\minus{}b)^2\minus{}\frac{1}{2}(b^2)\plus{}\frac{1}{2}(a\minus{}c)^2\minus{}c^2$
Analogamente:
$ P_{X/(J_{C})} \equal{} DA^2\plus{}(s\minus{}c)^2\minus{}\frac{1}{2}(c^2)\plus{}\frac{1}{2}(a\minus{}b)^2\minus{}b^2$
Expandiendo cada uno se obtiene
$ P_{X/(J_{C})} \equal{}P_{X/(J_{B})}$

Así  $ AX$ es el eje radical de los círculos $ (J_{B}), (J_{C})$ y analogamente $ BY$ y $ CZ$ son los ejes de $ (J_{C}), (J_{A})$ y $ (J_{A}), (J_{C})$. Entonces su punto de intersección, $ P$, tiene la misma potencia de punto con respecto a los tres círculos.

Sean $ D,E,F$ los puntos de tangencia de $ (I)$ con $ BC,CA,AB$ y $ D',E',F'$ los puntos de tangencia de $ (I_A),(I_B),(I_C)$ con $ BC,CA,AB.$ La inversión con centro $ A$ y potencia $ AE^2 \equal{} AF^2$ transforma a $ (I)$ en si misma y a $ (K_A)$ en la recta paralela a $ BC$ tangente a $ (I)$ en el inverso $ T_1$ de $ A_1.$ Pero $ T_1$ pasa a ser el antipodal de $ D$ en $ (I)$ $ \Longrightarrow$ $ AT_1 \equiv AA_1$ es el A-rayo de Nagel de $ \triangle ABC$ $ \Longrightarrow$ $ AA_1,BB_1,CC_1$ concurren en el punto de Nagel $ N_a$ de $ \triangle ABC.$

Los círculos $ (J_C)$ y $ (J_B)$ son tangentes a $ AB,AC$ en $ F,E$ (simétricos de F',E' respecto a los puntos medios de  AB,AC) y sus centros forman los paralelogramos $ I_BAJ_BC$ y $ I_CAJ_CB$ $ \Longrightarrow$ $ BJ_C \perp IA$ y $ AJ_C \perp BI,$ en otras palabras $ J_C$ es el ortocentro de $ \triangle AIB.$ Similarmente, $ J_B$ es el ortocentro de $ \triangle AIC$ y por tanto $ J_B,J_C$ son los polos de las paralelas medias a $ AC,AB$ con respecto a $ (I)$ $ \Longrightarrow$ $ J_BJ_C$ es la polar del punto medio $ M$ de $ BC$ con respecto a $ (I).$ Pero si $ R_1 \equiv AD' \cap (I),$ diferente de $ T_1,$ la circunferencia $ (M)$ de diámetro $ DD'$ que pasa por $ R_1$ es ortogonal a $ (I).$ Entonces la tangente a $ (I)$ por $ R_1$ pasa por $ M$ $ \Longrightarrow$ $ J_BJ_C \equiv DR_1,$ por consiguiente $ AD' \perp J_BJ_C$ y como $ A$ tiene igual potencia $ AF^2 \equal{} AE^2$ con respecto a los círculos $ (J_B),(J_C),$ se sigue que $ AD'$ es el eje radical de $ (J_B),(J_C)$ $ \Longrightarrow$ $ N_a$ es centro radical de $ (J_A),(J_B),(J_C).$