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AIME
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AMC 12
AMC 10
AIME I

This year's brochure says that colleges and universities are encouraged to administer AMC 10 and 12 and the list of participating colleges will be on the AMC website. Is it available yet?

There are no colleges on that list in CT. Can a student from CT take a test in PA, NY or MA?

[quote="bpms"]Is there anywhere I can get a master list of all highschools offering the 10 and 12?[/quote]

I doubt it. The registration period is still going on, at least I think. Just call all the high schools in your area and ask.

[quote="tokenadult"]Browsing the AMC Web site is your friend.

[url=http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml]2007 Additional Locations for the AMC 10B/12B at Higer Education Sites[/url]

(There is a prominent link to this page on the home page of the AMC site.)

It is possible that other colleges will be added to this list before the testing date.[/quote]

Hope you wouldn't mind me asking but do those colleges offer the alternate AIME as well as the AMC B test?

[quote="13375P34K43V312"]Well you can't take both the AIME and the alternate AIME...[/quote]

I know that, but what if you can't make it to the first AIME testing day and your school doesn't offer the second one?

[quote="dkushnir"]I contacted a couple of colleges on the AMC list and the answer is Yes, out of state students can take AMC 10/12 there.[/quote]

But is there any colleges administrating the alternate AIME?

The colleges are generally only responsible for administering the AIME to students who qualify by taking the AMC 10/12 at the college.  It's up to the college whether to offer the regular date or the alternate date -- my understanding is that the AMC wants as many schools as possible to use the regular date, and only use the alternate date if there's a good reason to do so.

In other words, if you take the AMC at your regular school, and you qualify for the AIME, then your school is responsible for administering the AIME.  If you can't make the regular date (because your school is closed or you have an unavoidable conflict), then your school should offer the AIME on the alternate day -- talk to your school and make sure that they understand this.  If you can't make the AIME I and your school is closed on the AIME II, well, that's a problem.  You can try to contact a local college in that case (who gave the AMC) and see if they'll take you, but they're certainly not obligated to do so, as it's a bit of a hassle for them (see next paragraph).  You can also try a different high school in your area.  There are also provisions for giving the AIME/USAMO if the school is closed -- see the AIME Teacher's Manual on the AMC web site for details.

Note that taking the AIME anywhere other than your regular school requires permission from the AMC -- in particular, the AMC only mails out enough copies of the AIME to cover that school's qualifiers, so to take the AIME anywhere else requires the AMC to know well ahead of time so that the proper arrangements can be made.

I thought I'd tell everyone about my group's visit to the AMC college testing site at Bethel University in Minnesota. I'm very grateful to the AMC headquarters for their encouragement and assistance to colleges to offer testing sites.

Bethel University is in a suburb of St. Paul and generally offers good freeway access to most parts of Minnesota's major populated areas. Professor Eric Gossett organized colleagues and students to direct visitors to the campus and to proctor the test. When I heard he had that many proctors, I realized I could take a test unofficially for practice as a coach, which is what I did.

Bethel ordered only AMC 12 test kits. I knew about this far enough in advance that I was able to order an AMC 10 test kit for my testing group, having them sent to Bethel care of Professor Gossett. As it turned out, there was very little demand for the AMC 10 test kit I bought, but the AMC 12 test kit was fully subscribed by Monday before the test. Five students from my testing group and another five from two other high schools took the 12. I took the 10 unofficially, and one student in my testing group took it officially, for his first experience on the test. Bethel had a computer in the room with a spreadsheet open so that we could all log in our names, addresses, and other information before the test began. Two student proctors observed the test and kept time with a computerized countdown projection display and the verbal announcements of time remaining specified in the AMC teacher manual.

Bethel was extra gracious because they offered school T-shirts to all students, pens to all parents, and admission folders to any parent who was interested. During the test a Bethel student accompanied the parents to the campus coffee shop. After the test, they offered on-the-spot unofficial scoring and offered prizes (book store gift certificates) to the top scorers on the 12 in the room. AIME qualifiers by unofficial scoring received school paperweights as a prize. I was very deeply impressed by Bethel's hospitality and only wish that the U of MN follows its rival school's example next year and hosts an AMC testing date. All of the students were grateful for the opportunity to take an AMC 12 on the B date. I think six of the students yesterday had taken the A date AMC 12 but had been dissatisfied with their scores (even if they AIME-qualified on the A date).

Bethel took care of mailing in all the official answer sheets. Once the contest manager had all of those, student proctors distributed solutions to students, several of whom groaned at dumb mistakes they had made. I think this year's tests were hard, compared to historical standards--but that is a good thing, just something that takes some getting used to. This morning I got a telephone call to confirm some contact details and school affiliation for two students in my testing group.

I commend AMC for arranging the college testing sites and thank Bethel for its hospitality. One parent came from two hours away yesterday so that her son would have a chance to test. I recommend that parents next year look early and often for college testing sites if their children desire to take the AMCs on the B date, and I hope the college testing site program expands to many more locations.

P.S. I didn't qualify for AIME on my unofficial AMC 10B--no matter what the cut-off score is this year. That's why my math team needs some new coaching.

Oh, yes, and about the issue of AIME testing availability, my host at Bethel was not sure if they would be running an AIME test location. They definitely had qualifiers yesterday, but they haven't reserved a space for either AIME date. I have a plan B location (a private testing and tutoring center where my testing group took the A date tests) for AIME qualifiers in Minnesota who can't find an AIME test location.

tokenadult: At what time of day did Bethel administer the test? Was it during regular school hours or was it in the late afternoon? And if it was during regular school hours, did the participants' schools excuse their absence?

So far, I haven't attempted to set up a site at my college, in part because I'd really like to encourage more high schools to do it for themselves; I'm also not sure if it did whether during school hours or in the late afternoon would be better (and in any case, we'd have to work around whatever my teaching hours are, which vary widely from semester to semester.)

Congratulations to you and hats off to Bethel for hosting the contest so well.
This sounds exactly like what we were hoping for in all respects.  Donita (who answers the amcinfo requests and managed the questions and answers about college and university adminstration) read the post over my shoulder, and she is thrilled to hear that it worked do well.

We are definitely going to continue this program and hope to greatly expand the participation next year.  It workd out quite well for a first try.

Thanks for the report!

Steve Dunbar
Director, American Mathematics Competitions

[quote="Kent Merryfield"]tokenadult: At what time of day did Bethel administer the test? Was it during regular school hours or was it in the late afternoon? And if it was during regular school hours, did the participants' schools excuse their absence?[/quote]

My son also took the 12B at a college location.  It was offered at 11 a.m. and then all of the students and parents were invited to lunch in their cafeteria.  Since most of the schools in this area are on break this week, the timing didn't matter.  My son is in 8th grade, so it was a great way for him to have the opportunity to try an AMC test before he got to high school.

I would think that if the test date is a school day in your area, it would be better to offer a late afternoon hour, if possible.  It's wonderful that you're considering doing this. Making these tests available to more kids is a great thing to do.

[quote="Kent Merryfield"]tokenadult: At what time of day did Bethel administer the test? Was it during regular school hours or was it in the late afternoon?[/quote]

It was in the late afternoon, at 3:30pm Central. That was a bummer, because that conflicted with my son's linear algebra class, which usually only meets once a week. (I think the class shouldn't meet on Wednesdays, but they don't ask my opinion.) That was the only time the hosts could find a room, I was told. We had a bit of a search to find parking on campus at that hour.

That's why AIME at Bethel could be very problematic. The tutoring center where we did our A date testing costs me more money, but it has a bunch of empty rooms during the school hours when AMC prefers people to take the AIME. Most of its business begins in the afternoon and extends into the evening.

I think the AMC testing at college sites is a great idea.  Twenty-five middle and high school students from 10 different schools took the AMC10/12 at the University of Colorado at Boulder yesterday.  For some, this was their only chance to take the AMC this year.  For others, this was an opportunity to improve on their A test score, or to try the AMC12 after the AMC10 (or vice versa).  Originally the test was scheduled for the morning but we asked the host to change the time to late afternoon so that students could attend without missing class.  It was fun to bring together so many of the top students in Colorado for the AMC test.  Afterwards there were many animated discussions about the problems.

My only suggestion for next year is to allow students to specify the name of their actual school (rather than University of Colorado) so that they can be eligible for the various school awards.  We would grateful if this feature could be added.

[quote="coloradoarml"]My only suggestion for next year is to allow students to specify the name of their actual school (rather than University of Colorado) so that they can be eligible for the various school awards.  We would grateful if this feature could be added.[/quote]

I THINK that Bethel attempted to do this. At least, they know each student's regular school of attendance for contacting the student for results. I didn't actually see any of the official answer forms to know what school designation was put on those.
I just asked my son, one of the test-takers, and he said that the space on the answer form for the school address was filled in with Bethel's address, but I'm not sure that that prevents some other data entry for recognizing students for school awards. We were asked to provide CEEB numbers for each student, and I have memorized the AMC-assigned number for my testing group, so maybe that will be disaggregated in the AMC headquarters records.


In any triangle  ABC  prove  that
1/(l_a)^2 + 1/(l_b)^2 + 1/(l_c)^2 <= 1/(3r^2)
Crux  problem 2821  but  i  dont have  that  issue

I think that Nickolas is talking about this one: $\sum\frac{1}{l_{a}^{2}}\geq\frac{1}{3r^2}$

 Using the well-known inequality, $ l_a\leq\sqrt {p(p-a)} $ the inequality reduces to:

          $\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\geq\frac{p}{3r^2}$.
 We know that $(p-a)(p-b)+(p-b)(p-c)+(p-c)(p-a)=r(4R+r)$ and from here we deduce that $ \frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}=\frac{4R+r}{pr}$.Well, in this moment we only need to prove that $4R+r\geq p\sqrt{3} $, which is a well-known one. ;)

 I hope I haven't made any mistakes. ;)


Let\[y=x^{-\frac{1}{3}}\]

Now we have a quadratic. I'm not going to bother solving the quadratic. Find $y$. Cube it. Take the inverse. You're done.


AMC
USA(J)MO
USAMO
number theory proposed
number theory

I'm hoping I'm putting this in the right place, I couldn't find anywhere else it could possibly go though...

Anyway, the problem states:
Let n be a positive integer. Define a sequence by setting $ a_1 \equal{} n$ and, for each k > 1,
letting $ a_k$ be the unique integer in the range 0 ≤ a_k ≤ k − 1 for which $ a_1 \plus{} a_2 \plus{}$ · · · $ \plus{} a_k$
is divisible by k. For instance, when n = 9 the obtained sequence is 9, 1, 2, 0, 3, 3, 3, . . . .
Prove that for any n the sequence $ a_1, a_2, a_3, . . .$ eventually becomes constant.

I didn't actually take this exam, but suppose my solution proved the following:

Define the sequence $ b_k$ to be $ k^2 \minus{} a_k$. I show:
(1) $ b_k$ is strictly increasing
(2) $ a_k$ is constant when $ b_k \geq 0$
Note that this solution gives no indication as to what constant value is actually approached, whereas the solutions given on the AMC website do describe that constant. Assuming this was written clearly, etc., could this solution still be scored a 7, as the problem doesn't actually ask you to describe the constant?

Yes, you should not lose credit for not doing something the problem does not ask you to do.

As a note, this is actually problem 1, not problem 2.


How does one get a bar over two letters to indicate a line segment?  I want the effect that \widebar would have if it existed.

like this [tex]\overline{AB}[/tex]

Although I don't personally like it (I use just $AB$) because when you try to use $\overline{AB}$ in equations (like $\overline{AB}^2$) it doesn't look very good.

Thanks.  I use $\overline{AB}$ to refer to the line segment AB and $AB$ to refer to its length, so I won't run into problems with $\overline{AB}^2$.

[quote="hashproduct"](Something seems to be badly wrong with the [tex]\TeX[/tex] displayer.)[/quote]
That's because you should use [tex ] ... [/tex ] or dollar signs, but not both
[code][tex]\overline{AB}[/tex][/code]
gives [tex]\overline{AB}[/tex] and  [code]$\overline{AB}$[/code]
also gives $\overline{AB}$

So your post would read [quote="hashproduct"]I use [tex]\overline{AB}[/tex] to refer to the line segment AB and [tex]AB[/tex] to refer to its length, so I won't run into problems with [tex]\overline{AB}^2[/tex].[/quote]

[quote="albertfermat"]because when you try to use $\overline{AB}$ in equations (like $\overline{AB}^2$) it doesn't look very good.[/quote]
Does this look any better [tex]\left.\overline{AB}\right.^2[/tex] ? This uses invisible auto-sized brackets which have the effect of moving the 2 a little to the right. On the other hand, a small space would move the 2 down a little as in [tex]\overline{AB}\;^2[/tex]

[quote="hashproduct"]I use $\overline{AB}$ to refer to the line segment AB[/quote]

Isn't that generally denoted as [AB] ? Or are there different conventions again? :?

Maybe this looks better $\overline {AB^2}$, though I'm not sure.


modular arithmetic
number theory unsolved
number theory

Let $ p$ be a prime such that $ p \equiv 5 ( mod 12 )$. Prove that :
$ \prod_{i \equal{} 1}^{\frac {p \minus{} 3}{2}}(i^2\plus{}i\plus{}1) \equiv \minus{} 2 ( mod p )$
Sr :(

$ p\equal{}17 \implies 17 \nmid 7!\plus{}2$
in fact if

$ (\frac{p \minus{}3}{2})! \equiv \minus{}2  \pmod {p}$ then

$ (\frac{p \minus{}1}{2})! \equiv 1  \pmod {p}$

which is a contradiction with the following well-known congruency (it holds if $ p \equiv 1 \pmod {4}$)

$ ((\frac{p \minus{}1}{2})!)^2 \equiv \minus{}1  \pmod {p}$

[quote="conan6032"]Let $ p$ be a prime such that $ p \equiv 5 ( mod 12 )$. Prove that :
$ \prod_{i \equal{} 1}^{\frac {p \minus{} 3}{2}}(i^2 \plus{} i \plus{} 1) \equiv \minus{} 2 ( mod p )$
Sr :([/quote]
Nobody  :( ??? Can you help me  :wink:

at first note that if $ q | x^2\plus{}xy\plus{}y^2$ and $ q$ is a prime and also $ (x,y)\equal{}1;(xy,q)\equal{}1$ then $ q\equal{}3$ or $ q \equiv 1 \pmod {3}$
so there doesn't exist any $ i,j$ for which $ p | i^2\plus{}ij\plus{}j^2$ since $ p \equiv \minus{}1 \pmod {3}$
also if $ i\plus{}j \equiv \minus{}1 \pmod {p}$ then $ i^2\plus{}i\plus{}1 \equiv j^2\plus{}j\plus{}1 \pmod {p}$
which implies

$ \prod_{i\equal{}1}^{p\minus{}1} (i^2\plus{}i\plus{}1) \equiv \left(\prod_{i\equal{}1}^{\frac{p\minus{}3}{2}} (i^2\plus{}i\plus{}1) \right)^2 ((\frac{p\minus{}1}{2})^2\plus{}\frac{p\minus{}1}{2}\plus{}1)$

we multiply both sides of the congruency by $ \prod_{i\equal{}2}^{p\minus{}1} (i\minus{}1)$ or in other words $ (p\minus{}2)!$ which is congruent to $ 1$ modulo $ p$ :

$ 3 \times \prod_{i\equal{}2}^{p\minus{}1} \left((i\minus{}1)(i^2\plus{}i\plus{}1) \right)$

$ \equiv 3 \times \prod_{i\equal{}2}^{p\minus{}1} (i^3\minus{}1)$

$ \equiv \left(\prod_{i\equal{}1}^{\frac{p\minus{}3}{2}} (i^2\plus{}i\plus{}1) \right)^2 ((\frac{p\minus{}1}{2})^2\plus{}\frac{p\minus{}1}{2}\plus{}1) \times (p\minus{}2)!$

$ \equiv  \left(\prod_{i\equal{}1}^{\frac{p\minus{}3}{2}} (i^2\plus{}i\plus{}1) \right)^2 ((\frac{p\minus{}1}{2})^2\plus{}\frac{p\minus{}1}{2}\plus{}1) \pmod {p}$

but by what has been said in the first line we can see that for no $ 1 \le i,j \le p\minus{}1$ we can have $ i^3\minus{}1 \equiv j^3\minus{}1 \pmod {p}$

which means that $ \prod_{i\equal{}2}^{p\minus{}1} (i^3\minus{}1) \equiv (p\minus{}2)! \equiv 1 \pmod {p}$

so we get

$ 3 \equiv \left(\prod_{i\equal{}1}^{\frac{p\minus{}3}{2}} (i^2\plus{}i\plus{}1) \right)^2 ((\frac{p\minus{}1}{2})^2\plus{}\frac{p\minus{}1}{2}\plus{}1) \pmod {p}$

but we have $ (\frac{p\minus{}1}{2})^2\plus{}\frac{p\minus{}1}{2}\plus{}1\equal{} \frac{p^2\plus{}3}{4}$

by multiplying both sides of the congruency by $ 4$ we get

$ 4 \equiv \left(\prod_{i\equal{}1}^{\frac{p\minus{}3}{2}} (i^2\plus{}i\plus{}1) \right)^2 \pmod {p}$

so now we must show that the congruency:

$ 2 \equiv \prod_{i\equal{}1}^{\frac{p\minus{}3}{2}} (i^2\plus{}i\plus{}1) \pmod {p}$

isn't true.
too lazy to think about the rest....

[quote="shoki"]at first note that if $ q | x^2 \plus{} xy \plus{} y^2$ and $ q$ is a prime and also $ (x,y) \equal{} 1;(xy,q) \equal{} 1$ then $ q \equal{} 3$ or $ q \equiv 1 \pmod {3}$
so there doesn't exist any $ i,j$ for which $ p | i^2 \plus{} ij \plus{} j^2$ since $ p \equiv \minus{} 1 \pmod {3}$
also if $ i \plus{} j \equiv \minus{} 1 \pmod {p}$ then $ i^2 \plus{} i \plus{} 1 \equiv j^2 \plus{} j \plus{} 1 \pmod {p}$
which implies

$ \prod_{i \equal{} 1}^{p \minus{} 1} (i^2 \plus{} i \plus{} 1) \equiv \left(\prod_{i \equal{} 1}^{\frac {p \minus{} 3}{2}} (i^2 \plus{} i \plus{} 1) \right)^2 ((\frac {p \minus{} 1}{2})^2 \plus{} \frac {p \minus{} 1}{2} \plus{} 1)$

we multiply both sides of the congruency by $ \prod_{i \equal{} 2}^{p \minus{} 1} (i \minus{} 1)$ or in other words $ (p \minus{} 2)!$ which is congruent to $ 1$ modulo $ p$ :

$ 3 \times \prod_{i \equal{} 2}^{p \minus{} 1} \left((i \minus{} 1)(i^2 \plus{} i \plus{} 1) \right)$

$ \equiv 3 \times \prod_{i \equal{} 2}^{p \minus{} 1} (i^3 \minus{} 1)$

$ \equiv \left(\prod_{i \equal{} 1}^{\frac {p \minus{} 3}{2}} (i^2 \plus{} i \plus{} 1) \right)^2 ((\frac {p \minus{} 1}{2})^2 \plus{} \frac {p \minus{} 1}{2} \plus{} 1) \times (p \minus{} 2)!$

$ \equiv \left(\prod_{i \equal{} 1}^{\frac {p \minus{} 3}{2}} (i^2 \plus{} i \plus{} 1) \right)^2 ((\frac {p \minus{} 1}{2})^2 \plus{} \frac {p \minus{} 1}{2} \plus{} 1) \pmod {p}$

but by what has been said in the first line we can see that for no $ 1 \le i,j \le p \minus{} 1$ we can have $ i^3 \minus{} 1 \equiv j^3 \minus{} 1 \pmod {p}$

which means that $ \prod_{i \equal{} 2}^{p \minus{} 1} (i^3 \minus{} 1) \equiv (p \minus{} 2)! \equiv 1 \pmod {p}$

so we get

$ 3 \equiv \left(\prod_{i \equal{} 1}^{\frac {p \minus{} 3}{2}} (i^2 \plus{} i \plus{} 1) \right)^2 ((\frac {p \minus{} 1}{2})^2 \plus{} \frac {p \minus{} 1}{2} \plus{} 1) \pmod {p}$

but we have $ (\frac {p \minus{} 1}{2})^2 \plus{} \frac {p \minus{} 1}{2} \plus{} 1 \equal{} \frac {p^2 \plus{} 3}{4}$

by multiplying both sides of the congruency by $ 4$ we get

$ 4 \equiv \left(\prod_{i \equal{} 1}^{\frac {p \minus{} 3}{2}} (i^2 \plus{} i \plus{} 1) \right)^2 \pmod {p}$

so now we must show that the congruency:

$ 2 \equiv \prod_{i \equal{} 1}^{\frac {p \minus{} 3}{2}} (i^2 \plus{} i \plus{} 1) \pmod {p}$

isn't true.
too lazy to think about the rest....[/quote]
Nobody continue  :(

I avoided to read shoki's post, and finally I reached this nice proof (imho, I hope you like it too)  :D

[url=http://bboyjordan.wordpress.com/2009/10/31/every-number-m2m-nm-has-all-digits/]Problem 7- A strange divisibility[/url]


Prove that the system equation has infinitively many solutions$x^2+y^2+z^2=2t^2$and$x^4+y^4+z^4=2t^4$with$gcd(x,y,z,t)=1,x,y,z,t\in Z+$

This system equation is easy.First we have $(x^2+y^2+z^2)^2=2(x^4+y^4+z^4)\rightarrow (x+y-z)(x+z-y)(y+z-x)=0$if $z=x+y$,we have $x^2+xy+y^2=t^2$This equation is easy


function calculus derivative parameterization

compute the cubic function with rel max 2,3 and x intercept at 3,0

even a hint, anyone? I'm trying systems of equations right now and they aren't working.

[hide="HINT"]
Easily done with Calculus.
[/hide]

I know how to do it with calculus, but the directions on my paper are "Don't use a calculator or any calculus". It's for a calc 1 course, and I've already taken AP calc, so I know how to do it by taking derivatives, but I can't use calculus :(

[hide]Ok, Consider the equation $ g(x)=f(x)-3$. Since $ g(x)$ is $ 0$ and has a relative max at $ x=2$, it has a double root at $ x=2$.

Then, $ g(x) = k(x-c)(x-2)^{2}$

Also, $ g(3)=-3 \implies k(3-c)=-3 \implies k = \frac{3}{3-c}$

$ c$ can generally vary, but clearly cannot equal $ 3$ (division by $ 0$) or $ 2$ (triple root instead of double root, it is not a min).

Quick testing on the three intervals reveals that $ g(x)$ has a local minimum, not maximum, at $ x=2$ if $ 2 < c < 3$. However, it does in fact produce a relative maximum for the two extremal intervals.

Hence, $ \boxed{f(x) = \frac{3}{3-c}(x-c)(x-2)^{2}+3 \; \forall c \in \mathbb{R}^{\not \in (2,3)}}$

[i]Note that we are given three pieces of information: two points on $ f$ and one point on $ f'$. It takes four pieces of information to uniquely determine a cubic. As a result, there will be one parameter in the final answer.[/i]
[/hide]


induction
trigonometry
number theory proposed
number theory

Let $ A$ be infinite subset of the set of positive integers, let $ n > 1$ be an integer. Given that, for arbitrary prime number $ p$ that cannot divide $ n$, there exist infinite many elements that indivisable by $ p$ among the set $ A$. Prove that, for arbitrary integer $ m > 1, (m,n) \equal{} 1$, there exist finite distinct elements among the set $ A$, the sum of its elements $ S$ satisfying $ S\equiv 1(\mod m),$ and $ S\equiv 0(\mod n).$

We shall prove a slightly stronger result: we can find $ a_{i}$ so that $ \sum a_{i} \equiv nc (mod$  $ mn)$, for any $ c$ such that $ (c,m) \equal{} 1$. [This problem follows, by choosing $ c$ so that $ m | nc \minus{} 1$).

Let the prime factors of $ m$ be $ p_{1}, p_{2} ... p_{k}$.
Consider the sets $ S_{0}, S_{1} ... S_{mn \minus{} 1}$, that cover $ S$, where $ S_{i}$ is the set of all elements in $ S$ that are equivalant to $ i (mod S)$. Now, from these sets, we consider those sets $ S_{a_{1}}, S_{a_{2}} ... S_{a_{j}}$ which are infinite.
We know, that for any $ p | m$, that we can find infinitely many numbers in $ S$ that are not divisible by $ p$ (since if $ p|m$, then $ p$ cannot divide $ n$). Apply this to $ p \equal{} p_{l}$, we have: there are infinitely many numbers in $ S$ that are not divisible by $ p_{l}$. A number is not divisible by $ p_{l}$, iff it is part of some set set $ S_{m}$, where $ (m, p_{l}) \equal{} 1$. So at least one of the sets $ S_{m}$, where $ (m, p_{l}) \equal{} 1$ must be infinite (since their union is infinite - if they are all finite, their union can only be finite). This means for each prime $ p_{l}$ from the set $ {p_{1} ... p_{k} }$, there exists an $ a_{i}$ so that $ (a_{i}, p_{l}) \equal{} 1$.  Now consider $ b \equal{} (a_{1}, a_{2} ... a_{j}).$ If $ (b, p_{l}) > 1$ for some $ l$, then $ p_{l} | a_{i}$ for $ i \equal{} 1, 2 ... j$. However, this contradicts the fact that for each $ l$, there exists some $ i$ for which $ (a_{i}, p_{l}) \equal{} 1$. Thus $ b$ is relatively coprime with $ p_{1}, p_{2} ... p_{k}$, and so $ (b,m) \equal{} 1$.
Now we shall prove there exist integers $ d_{1}, d_{2} ... d_{j}$, so that $ a_{1}*d_{1} \plus{} a_{2}*d_{2} \plus{} ... \plus{} a_{j}*d_{j} \equiv nc ( mod$ $ mn)$. This will suffice, since then we can pick $ d_{i}$ distinct integers from each set $ S_{a_{i}}$, and then this set of $ \sum d_{i}$ integers will have the required property.
First it's pretty well known, all sufficiently large integers, that are multiples of $ b$, can be expressed as $ \sum a_{i}*d_{i}$ for some integers $ d_{i}$ - to prove this, it suffices to prove this for $ b\equal{}1$, and the conclusion follows since if $ x \equal{} \sum \frac{a_{i}}{b} d_{i}$, then $ xb \equal{} \sum a_{i}*d_{i}$. To prove for $ b\equal{}1$, let's use induction. The induction step: Let $ e \equal{} (a_{1}, a_{2} ... a_{n\minus{}1})$. By induction hypothesis, all sufficiently large multiples of $ e$ can be expressed as $ \sum_{1 \leq i \leq n\minus{}1} a_{i} * d_{i}$, so all numbers of form $ fe\plus{}g*a_{n}$, where $ f,g$ may be any integers, with $ f$ sufficiently large, can be expressed in the form $ \sum a_{i}*d_{i}$. But sine $ (e, a_{n}) \equal{} 1$, all sufficiently large numbers can be expressed as $ fe \plus{} g*a_{n}$, completing the induction. Now using this result, all sufficiently large numbers of the form $ bq$, with $ q$ integer can be expressed as $ \sum a_{i}*d_{i}$. Thus we want to find a solution to: $ bq \equiv nc (mod$ $ mn)$, for $ q$ a sufficiently large number. Since $ (b,m) \equal{} 1$, this equation is clearly solvable, so we are finished.


I know there is such a thing as the Toronto math circle, but thats about it. sorry. :(

ok, I'm not totally sure, but maybe you can e-mail Prof. Barbeau or Prof. Rice about it?

[quote="rem"]I am moving to Toronto in the end of September, and I would like to attend the Toronto Math Circles. Does anybody know when they start, which room they are held in, and at what time?
Thanks for your help.[/quote]

Send e-mail to Prof. Rice. I sent him mail and I just know
that it starts from Oct 2. and time is 5 to 7. He was not sure
about room number and so on. So send him email to make sure.

So r u in Toronto right now??
I hope to see you in Math circle on Oct 2.

And you know. You will be definitely in 2007 IMO Team LOL.
You are too smart.  :)

I am moving to Toronto September 30. :)
I will try really hard to make the team but I am not that smart. :lol:
btw I sent an E-mail to Prof Rice and Prof Berbeau several days ago but they never replied. pLease post here/pm me if you find out the info about room number.
I hope to see you on October 2 too. :)

YAY! Prof. Rice gave me the information about the Toronto Math Circles. :)  :omighty:
To those who want to know when and where they will be, here is the information:
[quote="Professor Larry Rice"]the Toronto Math circles will be at Macdonald CI in Scarborough from 5 to 7 pm on Mondays starting on Oct 2...Macdonald CI is in Scarborough on Pharmacy Avenue north of Sheppard ave south of Finch[/quote]

[quote="Kyle_k"][quote="chilly_snow"]That's my school...
I will be there too!
btw my name is Shirley.
See you on Monday!
rem, which school do you go to?[/quote]

Hello. My name is Kyle.
I will be in the Math Circles too.^^

Hm,, Actually I just have a question to ask you.
Since Prof. Larry Rice lost his email list, he asked me to
send email to ppl living in Metro Toronto.

However, everyone was asking me where's Macdonald CI.
So, is Macdonald CI = Sir John A. Macdonald CI??[/quote]

Yup!  :)

[quote="chilly_snow"]That's my school...
I will be there too!
btw my name is Shirley.
See you on Monday!
rem, which school do you go to?[/quote]
William Lyon Mackenzie CI

wow, that's a far move. why do that on ur senior year? if u don't mind me asking

[quote="rem"]i am in gr 11.[/quote]

yes... i apologize, too many frisbee to the head.  :D

Haha,  you are in Toronto now, Alex?
I may show up for math circle sometimes, once I get my G2.
You are coming to T of T too right?
[quote="rem"]
i am in gr 11.[/quote]
lol, yea, Alex is young and hopeful. You are making IMO for sure this year, Alex.
I feel old, really.

Nice to see you here, J-I-M-B-O, hopefully we can play frisbee again if I get invited back to Seminar this year!!

awesome thx alot  :)

hopefully i wont get lost in that big school lol

What if I'm in gr.9 would they be too hard? b/c i think that school level stuff is just way too boring :lol:

They have 2 sections -- one for seniors and one for juniors. The junior section should not be hard I think.

Is there math circles in Toronto tomorrow cuz its cancelled for us here in waterloo for this week..

is it just like professor rice talking about problems (like his euclid workshop thing he gives to all the schools, at my school he said the workshop was like the toronto math circle thing) and it's free right? lol

i was planning on going tomorrow (in toronto)
hopefully it's not cancelled

wow there was like so many asian ppl there

like only 2 non asian ppl i saw

LOL this "property" is common to every mathematics extracurricular group activity. :lol:

ya lol he's right..its like the same in Waterloo except sometimes half the ppl there are from WCI and ASIAN..lol

does anyone know if there is math circle at toronto(john a macdonald) next monday (easter monday)?

i didn't go last week so i don't know

no there is no meeting on April 9, and April 16 is the last class

omg , i'm so sad, i only joined for a small while
 :(
and it's over so soon

and this is my last high school year
i wish i had known about this earlier

they should let ppl in other schools know, nobody in my school knew about this math circle thing

Hmm, I might come on April 16. :)

Haven't been at Math Circles for a while.

[quote="SM4RT"]what about your math teachers? they should no..i gues..[/quote]

LOL

my math teachers

they're all talk, when it comes to defending and promoting the integrity of math it's everything to them

but when it actually comes to staying behind after school or actually come to math club to supervise, they are nowhere to be found!

our math teacher's pride themselves on being waterloo graduates and yet they know nothing about math circles or aops or the best resources for math

our math club is poorly organized (2 members? lol) and we have some relatively good math ppl at our school too, but they're all passionless and selfish ppl who rather do stuff on their own rather than help others or talk about it with great enthusiasm

oh and btw lena, are you that new girl from 2 weeks ago, from richmond hill?


function
geometric series
complex analysis

Find Laurent series for function $ f(z)\equal{}\frac z {(1\minus{}z)(z\minus{}i)^3}$ around $ z_0\equal{}1$, which convergences in $ z\equal{}\frac i 2$.

Thanks.

[hide="Hint"]Try writing $ f(z)$ in its partial fraction decompostion, but be careful to watch your radius of convergence.[/hide]

Thanks, [b]JRav[/b], but the decomposition is my problem. I am not sure that I am doing it wright. It is suppose to be done like this $ \frac {a\plus{}ib} {1\minus{}z} \plus{} \frac {c\plus{}id} {z\minus{}i} \plus{} \frac {e\plus{}if} {(z\minus{}i)^2} \plus{} \frac {g\plus{}ih} {(z\minus{}i)^3}$, wright? But it is horrible for a homework problem.

Of course, I could use fact that $ (\frac 1 {z\minus{}i})'\equal{}\minus{}\frac 1 {(z\minus{}i)^2}$, and so on... If the previous is correct, is there an easier way?

Thanks.

Often, it helps to try to get your function in terms of a geometric series but you have to be careful about how you structure it, to insure convergence in the right disk.

[quote="Солунац"]It is suppose to be done like this $ \frac {a \plus{} ib} {1 \minus{} z} \plus{} \frac {c \plus{} id} {z \minus{} i} \plus{} \frac {e \plus{} if} {(z \minus{} i)^2} \plus{} \frac {g \plus{} ih} {(z \minus{} i)^3}$, wright? But it is horrible for a homework problem.[/quote]  There's no reason at all to use 8 real variables in the numerators instead of 4 complex variables.  This requires a moderate amount of algebra, but nothing unusually intimidating.


geometry
rectangle
combinatorics proposed
combinatorics

The cells of a $11\times 11$ chess-board are colored in 3 colors. Prove that there exists on the board a $m\times n$ rectangle such that the four cells interior to the rectangle and containing the four vertices of the rectangle have the same color.

[i]Ioan Tomescu[/i]

A lemma: note that $(a+1)a+(b+1)b-(a+2)(a+1)-b(b-1)=2(b-a-1)$. If $b>a+1$, then $2(b-a-1)>0$ so $(a+1)a+(b+1)b>(a+2)(a+1)+b(b-1)$. This implies that, given a sum of the form
\[ \sum_{i=1}^n{a_i(a_i-1)} \]
with $a_1, a_2, ..., a_n$ variable nonnegative integers with fixed $a_1+a_2+...+a_n$, the given sum is minimized when there are no two $a_i$ that differ by $2$ or more.

Now, in the board, since there are $121$ squares, there must be $41$ of the same color. Let $a_i$ for each $i=1, 2, ..., 11$ be the number of squares of that color on the ith row. Then $a_1+...+a_n=41$, and the sum of the terms $\dfrac{a_i(a_i-1)}{2}$ is minimized when there are no two $a_i$ that differ by $2$ or more, i.e., when there are $8$ of them equal to $4$ and $3$ of them equal to $3$. In other words, the number of pairs of squares of that color in the same row is at least $\dfrac{8 \times 4 \times 3}{2}+\dfrac{3 \times 3 \times 2}{2}=48+9=57$. Since there are $\dfrac{11 \times 10}{2}=55$ pairs of columns, there must be two of those pairs of squares belonging to the same pair of columns, and these two pairs will form a rectangle.

The proof generalizes verbatim to a $(n^2 + n -1) \times (n^2 + n -1)$ table colored in $n > 1$ colors, and indeed it was used as an ulterior contest problem.


I noticed that whenever you refresh the page of your blog, or post an entry, the view count goes up.  Could you make it so that the count does not increase when you are just moving around your blog or refreshing the screen?  Thanks.

No, don't change it, that way your count gets higher with every view :D.

That's the view count. And it should get higher with any view. It's not inaccurate.


At a mall's food court, Crystal has $ \$7.50$ to buy a meal (one entree, one drink and one dessert). The table below lists Crystal's choices and their prices including sales tax. How many distinct possible meals can she afford to buy?

\[ \begin{tabular}{|c|c|c|} Entrees & Drinks & Desserts \\
\hline Pizza \$3.50&Lemonade \$1.50&Frozen Yogurt \$3.00\\
\hline Corn Dog \$2.50&Soda \$1.25& Cookies \$2.00\\
\hline Fish \& Chips \$3.50& & \\
\hline Fried Rice \$4.75& & \\
\hline \end{tabular} \]

[quote="GameBot"]At a mall's food court, Crystal has $ \$7.50$ to buy a meal (one entree, one drink and one dessert). The table below lists Crystal's choices and their prices including sales tax. How many distinct possible meals can she afford to buy?

\begin{tabular}{|c|c|c|} Entrees & Drinks & Desserts \\
\hline Pizza$ \$3.50$& Lemonade$ \$1.50$& Frozen Yogurt$ \$3.00$\\
\hline Corn Dog$ \$2.50$& Soda$ \$1.25$& Cookies$ \$2.00$\\
\hline Fish \& Chips$ \$3.50$& & \\
\hline Fried Rice$ \$4.75$& & \\
\hline \end{tabular}[/quote]

I see it now.

The combos are:

pizza-lemonade-cookies
pizza-soda-cookies
corn dog-lemonade-yogurt
corn dog-lemonade-cookies
corn dog-soda-yogurt
corn dog-soda-cookies
fish-lemonade-cookies
fish-soda-cookies

So the answer, as mewto said, is $ \boxed{8}$.


function
induction
TST
inequalities
additive function
algebra
functional equation

Given is a function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $|f(x+y)-f(x)-f(y)|\leq 1$.
Prove the existence of an additive function $g:\mathbb{R}\rightarrow \mathbb{R}$ (that is $g(x+y)=g(x)+g(y)$) such that $|f(x)-g(x)|\leq 1$ for any $x \in \mathbb{R}$

Thank you for clarifying some sources, but we are solving problems here... ;)

Myth, you should know that this kind of attitude does not impress me anymore for quite a long time.

And since just you solve problems here, then ignore the following solution:
   Consider a fixed $n$ and the sequence $ a_n= f(nx)$. Then we have $ s(a_{m+n}-a_m-a_n)\leq 1$ for all m,n. Here $s(x)$ is the absolute value of x. I showed some time ago (in an problem from Austrain Polih competition)  that for such a sequence there exists a number $t$ such that $ s(a_n-nt)\leq 1$ for all n (just use the fact that $ s(a_{mn}-ma_n)\leq m-1$ and $ s(a_{mn}-na_m)\leq n-1$ for all m,n -which are direct by induction- to find that $ s(ma_n-na_m)\leq m+n$; this shows that $ a_n/n$ is a Cauchy sequence and thus it converges to a number t; make m big in $ s(a_m/m-a_n/n)<1/m+1/n$ and you will find $ s(a_n-nt)\leq 1$). Thus, we can define $ g(x)$ as the limit of $ f(nx)/n$ and we aare done, since the aditivity is clear from $ s((f(n(x+y))-f(nx)-f(ny))/n)\leq 1/n$ and the fact that $ s(f(x)-g(x))\leq 1$ is also clear from $ s(f(nx)-nf(x))\leq n-1$ (which is true by induction).
   And, of course, obviously, I did not solve the problem.

I said "we are..." and I am not John Nash...

I know you said " we are", but how could I suggest the plural form in "just you"? Anyway, Myth, even if this does not make you feel good, you will have to support me on mathlinks. Too bad, I know.

I really don't understand this phrase  :?

$g_n(x)=\frac{f(2^nx)}{2^n}$

$|g_{n+1}(x)-g_n(x)|\leq 1/2^{n+1}$

$|g_{n+p}(x)-g_n(x)|\leq 1/2^n$

Cauchy criterion is verify $g_n$ converge to $g$

$|g_n(x+y)-g_n(x)-g_n(y)|\leq 1/2^n$ tend n to oo

$g(x+y)=g(x)+g(y)$

[quote="harazi"]I think we have discussed this one, but I am not sure. Anyway, what is clear is that it is far older than 2000 TST. I am sure we had this in one of our national olympiads in 11-th grade ( 1994 I think), but yet I think it is still far older than that.[/quote]

I don't know about when it appeared in competitions, but this lemma must go back to Poincare's construction of the rotation number of a map from circle $\to$ circle (circa 1900), and (for quadratic forms) also Tate's construction of the canonical height on elliptic curves (circa 1950-60's).   In connection with the latter it appears as an exercise in Serge Lang's large textbook on algebra.

Also, Artin (Ph.D. supervisor of Tate and Lang) used the same inequality for functions $\mathbb{N} \to \mathbb{Z}$ as a construction of the real numbers, in one of his textbooks.
(Considering two functions the same if their difference is bounded, the equivalence classes are those of $f(x) = [ax]$, with $a$ real.)

[quote="Moubinool"]Can you give precise references fleeting guest ?
thx[/quote]

I will look for the references.  As harazi mentioned, this genre of problem
is quite ancient.  Tate's idea to use it for constructing a canonical height was quite ingenious, though.

Moubinool,  the books I was seeking are not in the library, but I would suggest to search in the exercises of Serge Lang, Real Analysis, 2nd or (?) 3rd edition.

[quote="Omid Hatami"]The problem was sent months ago.[/quote]

I think he was asking about the history of this lemma and similar ones.
Silverman's book on elliptic curves should have a reference to Tate's construction of the height, but I don't have access to it right now.  The lemma for linear functions must be 100 years old, at least.

Moubinool,  I located the exercise in Serge Lang, [i]Algebra[/i], first edition (publ. 1965), end of the chapter "Structure of Bilinear Forms".
It is attributed to Tate in the generality of $f:E \to F$ a map of complete normed vector spaces with $f(x+y)-f(x)-f(y)$  bounded, and the quadratic generalization is given as an exercise.  I think that Tate had constructed the canonical height on elliptic curves well before 1965.

WLOG assume $f(0)=0$.
We have $|f(2x)-2f(x)|\leq 1$ for all $x$. Not hard to deduce that
$$\Big| \frac{f(2^mx)}{2^m}-\frac{f(2^nx)}{2^n}\Big| \leq \frac{1}{2^n}$$for all positive integers $m>n$ and real number $x$.
This implies that, for every real $x$, the sequence $\frac{f(2^nx)}{2^n}$ where $n\in \mathbb{Z}^+$ is a [url=https://en.wikipedia.org/wiki/Cauchy_sequence#In_real_numbers]Cauchy sequence[/url].
So we can define $g:\mathbb{R}\to \mathbb{R}$ by $g(x)=\lim_{n\to \infty}{\frac{f(2^nx)}{2^n}}$. I'll leave the verification process to the diligent readers.

This problem was given at the Bulgarian TST's for IMO 2020. There was additional assertion $g$ is unique, but it's easy. I [url=https://dgrozev.wordpress.com/2020/07/11/almost-additive-function-bulgarian-tst-2020/]commented it on my blog[/url]. The ideas used are different from those used in this thread.

Could you post the other problems from the Bulgarian TST 2020, if you have them?


[b] Let [/b] $ A,B \in M_{2} \mathbb{(R)}$ ,
$ x \in R , x \in [ - \pi \pi]$ .
[b] Prove That : [/b]
 $ (A + xB)^{2} = \frac {\pi^{2}}{3}.B^{2} + A^{2} + \sum_{k = 1}^{\infty} {[\frac {4.( - 1)^{k}.coskx}{k^{2}}.B^{2} - \frac {2.( - 1)^{k}.sinkx}{k}(AB + BA)]}$
[b]"..Do math to love math , love math must do math , do math ,love math , math is my life !!!"[/b]
[b]"..Math must have steel .. "[/b]

 :roll:  :roll:  :roll:

It's sufficient to calculate the Fourier sequences. The result is valid in any Banach algebra. That is of little interest.


I currently use the King James version, I love it for it's lyric power and grace.

KJV vs. NIV: Matthew:22:21 (there are many more places like this too)

KJV: Render therefore unto Caesar the things that are Ceasar's; and unto God the things that are God's.

NIV: Give to Caesar what is Caesar's, and to God what is God's

you should make an option called "None" because not all of us use Bibles

The KJV, because it:
1. has great vocabulary
2. has great math
3. is more verbose

[quote="solafidefarms"]...has great math...[/quote]

:?

None. All of them contradict each other and don't make much sense a lot of the time.

I can only place one vote?  :P
I usually use the NIV, simply because it's the one translation I always have around, but IIRC the NASB (for instance) is somewhat more accurate.  The KJV isn't bad, but one has to be careful reading it because many words don't necessarily mean the same things as they did 400 years ago.  (Plus, nice as it sounds, it can be difficult to get one's mind around the language sometimes.)

[quote="Buffalo"]what does "use bible" mean? read it?[/quote]
yes read it.

How could there be so many different versions of Bible(I guess all these are in English)? Is it because there are many branches like Catholicism, Christianism? Rumor says these branches dislike each other more than that to the other religions, like Islam, is it really a rumor?

[quote="Buffalo"]How could there be so many different versions of Bible(I guess all these are in English)? Is it because there are many branches like Catholicism, Christianism?[/quote] No -- the Roman Catholic and Eastern Orthodox Bibles differ from the protestant Bibles only in their inclusion of apochryphal books in the canon.  (e.g., Wisdom of Solomon, the books of Maccabees, etc.) This does not account for the many different "versions" of the Bible.  The reason that there are many versions is that all of them are translations, and are translated to convey the same passages with different primary goals.  Some translations try to be accurate word-for-word; some try to convey the most accurate possible meaning of the passage (neglecting whether specific words are used in each case); some try to be more lyrical; others more easily readable.  Of course, there are a number of translations which are rather poorly done (most paraphrases, for instance), but even without this fact there is still legitimate room for a number of translations.

[quote="Buffalo"]Rumor says these branches dislike each other more than that to the other religions, like Islam, is it really a rumor?[/quote]  Whatever it is, it isn't true.  Even the most disparate divisions of orthodox Christianity have much more in common than does Christianity with other religions.  Wackos like Jack Chick aside, the reason that many Christians so vocally criticize others is because they feel they are correcting their brethren -- not fighting an enemy.

[quote="Setofallsets"]...not fighting an enemy.[/quote]

 :? [i]I[/i] don't regard other religions as an enemy... To me, I just feel sorry for those who haven't been guided to the right path, those who have gone astray. The way you said that, it seems like Christians want to make war in order to convert by the masses. :( I hope that this is not the case. :noo:

What do you put if u dont use a bible, but something else? I dont see any options like: None, Koran (dif prints if there are), or Torah...

I'm not going to go into this in detail, having not read the Bible or any versions of it but from what I've heard - theres more than just translation differences. Even if there were just translating differences, - 1. why are some christians SO against people using other versions? and 2. The way you translate something defines to you what it means. So if you translate something to be more pleasing to read, then you are changing what it means. It changes the tone, it changes what it means to somebody.
Also, how can you include or not include something to say something else? Doesnt that say that you are changing a text? Why is changing a religious book to mean something totally different ok? Seems like in some cases the print is sugar-coating the story. I dont mean to be attacking any religious books, but..

Hmm, from my experience most Christians actually aren't too worried about what english translation of the (Christian) Bible others use, if you look at the example I gave, the message is the same, just worded in a slightly different style. None of the styles are of course completely accurate, but that's what you pay for not learning hebrew and greek  ;) .

I can read hebrew  :lol: What's the phrase?

[quote="chess64"] :? [i]I[/i] don't regard other religions as an enemy... To me, I just feel sorry for those who haven't been guided to the right path, those who have gone astray. The way you said that, it seems like Christians want to make war in order to convert by the masses. :( I hope that this is not the case. :noo:[/quote]
I suppose that my comment was worded rather poorly.  As a whole (I can't speak for everyone, so I won't say all), Christians do not consider the adherents of other religions to be enemies as such.  (In any case, we are instructed to "love your enemies", so it [i]ought[/i] to make little difference.)  However, I will say that orthodox Christians do consider, to varying degrees, opposing religions to be (in some sense) enemies.  To use an example; I don't consider Muslims to be enemies, but I do consider Islam, as a set of creeds, to be an enemy [i]insofar as it denies the truths put forward by Christianity[/i].
Only the fringe few in Christendom (as I suppose is the case with most other religions) consider it a good idea to try to impose the correct views by force, however.  I do not agree with these few, however, as I feel that an honest conversion is nigh impossible when Christianity is the persecutor.  We ought to have learned by now that enforcing Christianity by means of human power serves only to kill off the true faith.  We seek the eradication of false doctrine, but by conversion -- the work of the Holy Spirit -- not by force.
So why the contrast in my above post which you quote?  Because in the disagreements among Christians, though we desire to correct each other's errors, we recognize that it is not conversion from a false religion to the true one that is needed, but refinement of doctrine within the Church.

But this is getting a little off topic, and I don't want to derail the thread...

i use new international since its easy to read and understand.

I like NKJ because it is the easiest to understand and has flowery language.


Here's a cute one, even if rather easy :).

Given $R,r$, construct circles with radi $R,r$ s.t. the distance between their centers is equal to their common chord.

I will give only a sketch.

The problem is equivalent to that of the construction of triangles of which are given a side BC (proportional to the distance between the two centers) the altitude AH related to the given sides (proportional to the half length of the common chord) and the ratio od the other sides (ratio R/r).

The latter problem can be solved intersecting the parallel line to BC at distance AH with the locus of points P s.t.  PB/PC = R/r which is the apollonian circle of BC with the given ratio.

We obtain two (triangles) solution. To derive the solution of the original problem we need to scale down the figure  of the factor R/R' (or r/r') where R' (or r') is the length of side AB (or AC).

This is the solution I found at very first glance. I will try to find a different one.

Wow! That is exactly what I did. Thanks! :)


Let $ a,b,c,d\geq0$.Prove that:
$ 6(a^2\plus{}b^2\plus{}c^2\plus{}d^2)\plus{}(a\plus{}b\plus{}c\plus{}d)^2\geq 12(ab\plus{}bc\plus{}cd)$

Should RHS be $ 12(ab\plus{}bc\plus{}cd\plus{}da)$?

[quote="Brut3Forc3"]Should RHS be $ 12(ab \plus{} bc \plus{} cd \plus{} da)$?[/quote]
Try $ a\equal{}b\equal{}c\equal{}d\equal{}1.$ :wink:

[quote="Inequalities Master"]Let $ a,b,c,d\geq0$.Prove that:
$ 6(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \plus{} (a \plus{} b \plus{} c \plus{} d)^2\geq 12(ab \plus{} bc \plus{} cd)$[/quote]

If you bring everything over to the LHS, you have:

$ 6(a^2 \plus{} b^2 \plus{}c^2 \plus{} d^2) \plus{} (a \plus{} b \plus{} c \plus{} d)^2 \minus{} 12(ab \plus{} bc \plus{} cd)\geq 0$

Expanding and simplifying LHS, you have:
$ 7(a^2 \plus{} b^2 \plus{}c^2 \plus{} d^2) \minus{} 10ab \minus{} 10bc \minus{} 10cd \plus{} 2ac \plus{} 2ad \plus{} 2bd\geq 0$

This factorises into:
$ (2a \minus{} 2b \plus{} c)^2 \plus{} (b \minus{} 2c \plus{} 2d)^2 \plus{} (a \minus{} c \plus{} d)^2 \plus{} (b \minus{} c)^2 \plus{} 1/2(2a \minus{}b)^2 \plus{} 1/2(b \minus{} 2d)^2\geq 0$
(which must be true as it is a sum of squares)


Ten test papers are to be prepared for the National Olympiad. Each paper has 4 problems, and no two papers have more than 1 problem in common. At least how many problems are needed?

[hide="not sure"]
To minimize the number of unique problems, we want as many as possible to be shared. To do this, we have a problem be shared by only one pair.

The 1st test shares P1 with the 2nd test.
The 2nd test shares P2 with the 3rd test.
The 3rd test shares P3 with the 4th test.

continuing

The 9th test shares P9 with the 10th test. The 1st and 10th test then have two unique problems, followed by 20 more unique ones to fill the other half of each test.

Summing them all up $\Longrightarrow 31$
[/hide]

Let $n$ the minimal number of problems, and $P_1$, ... , $P_n$ the problems.
Let $A_1$, .. , $A_{10}$ the sets of problems in each test.
We have $|A_i \cap A_j| \leq 1$, so, each pair ($P_i$, $P_j$) can't occur more than one time in this sets.

now, we have ${n \choose 2} \geq \sum{|A_i| \choose 2} = 10 . {4 \choose 2} = 60$

so, $n \geq 12$.

see that the projective plane of order 3 satisfy the conditions (for n = 13):

http://home.wlu.edu/~mcraea/Finite_Geometry/NoneuclideanGeometry/Prob14ProjPlane/problem14.html

now, we have to prove that n = 12 can't occur.

From above, $n \geq 12$.  We show that $n = 12$ cannot work.  (Let #s be problem #s.)  Some problem, say 1, must occur at least 4 times. Then, the other three problems on each of these tests must be distinct. so we must have 12 distinct problems in addition to 1, showing $n \geq 13$ contradiction.

Here's an explicit construction for $n = 13$:

1,2,3,4
1,5,6,7
1,8,9,10
1,11,12,13
2,5,8,11
2,6,9,12
3,5,9,11
3,6,8,12
4,5,10,12
4,6,11,9

If I understand you correctly, you have proved the fact that one problem is there 4 times.
I want to give the trivial prove for this:
Imagine you have only 12 different problems and each problem is there max. 3 times. So you would have a number of problems less than 37 but you have 40 problems. So there are 13 different problems or one problem is there at least 4 times.

Naphthalin

Missed it completely, barely reduced the # at all  :oops: Thanks for the explanations :)

[quote="ThAzN1"]From above, $ n \geq 12$.  We show that $ n \equal{} 12$ cannot work.  (Let #s be problem #s.)  Some problem, say 1, must occur at least 4 times. Then, the other three problems on each of these tests must be distinct. so we must have 12 distinct problems in addition to 1, showing $ n \geq 13$ contradiction.

Here's an explicit construction for $ n \equal{} 13$:

1,2,3,4
1,5,6,7
1,8,9,10
1,11,12,13
2,5,8,11
2,6,9,12
3,5,9,11
3,6,8,12
4,5,10,12
4,6,11,9[/quote]

I think you did this wrong.
Note that you have two sets:
2,6,9,12
3,6,8,12
these two sets appear to have two common problems.


Let $A$ be a set of 20 integers chosen from the set $B = \{1,4,7,...,100\}$. Prove that there must be two distinct integers in $A$ with sum $104$.

We want to divide the set $B$ of $34$ elements into two distinct subset of $20$ and $14$ elements.
Now suppose to put numbers for which the sum is $104$ into different subset. There are $16$ couples.
This can be done for only $14$ elements since in the subset $A$ there will be at least $2$ distinct integers for which the sum is $104$.


Asymptote geometry conics ellipse hyperbola parameterization function

When making a picture in asymptote dealing with geometry, I have come across the problem of angles. Usually angles in a shape are denoted with a small arc. However, I find it very difficult to manipulate a bezier curve approximation to get the arc I want (proper position, inclination, etc...). Is there an easier way to do this than harsh trial and error?

Also, what is a good way of creating an ellipse or hyperbola in asymptote?

Why don't you guys read the manual :roll:?

There is an [b]arc[/b] command in asymptote with syntax [b]arc(A,r,a,b)[/b] that draws an arc centered at $ A$ of radius $ r$ from $ a$ to $ b$ degrees. That is helpful if you know your angles like in
[b]
size(100);
pair A=(0,0), B=(100,0), C=100*dir(65);
path P=arc(A,15,0,65);
fill(P--A--cycle,yellow);
draw(B--A--C,blue);
draw(P,red);
[/b]
which yields
[asy]size(100); pair A=(0,0), B=(100,0), C=100*dir(65); path P=arc(A,15,0,65); fill(P--A--cycle,yellow); draw(B--A--C,blue); draw(P,red);[/asy]
As to ellipses, they can be obtained by the [b]ellipse(A,a,b)[/b] command with $ A$ being the center and $ a$ and $ b$ being the semiaxes:
[b]
size(100);
draw(rotate(30)*ellipse((0,0),20,40),green);
[/b]
gives
[asy]size(100); draw(rotate(30)*ellipse((0,0),20,40),green);[/asy]
As far as I can tell, there is no standard command for hyperbolae in asymptote and you have to write a special routine for that. The same concerns the arcs that span the angle given by two lines whose directions are not known apriori. I can try to write these routines for you if you want but not right away :)

What's this manual you speak of? Everything I know about asymptote comes from AOPS Wiki. And I would greatly appreciate the hyperbola routine  :D

[quote="pkerichang"]Well, if you download asymptote for your home use, asy.pdf in right in your folder...[/quote]

Can't find it. I checked a folder called Asymptote in program files but all it had were "example" files, which isn't really a manual.

sorry, I think it's asymptote.pdf.  I'm 100% sure it's there, or else you can try downloading the newer versions, maybe (cuz I'm reading it right now...)

and yes it's right under the folder Asymptote in Program Files on mine too

One way to find it is to type [b]help;[/b] in the asymptote dialog window. If for some reason it is missing in your distribution, you can also find the manual on the web [url=http://asymptote.sourceforge.net/doc/]here[/url]. As to hyperbola, I think I'll have some free time today and tomorrow to play with programming, so how do you want it cooked? (I mean, what parameters do you want to use to describe your hyperbolae) :)

Meanwhile, here is the angle mark function. I hope the code is self-explanatory. All the guessing you'll have to do is to guess some angle before you want to start your arc and some angle after you want to finish it. Truncation to exact values is automatic. Works both with lines and with curves :)
[code]
path SA(pair A, path P, path Q, real r, real a, real b, bool f=false, pen p=lightyellow)
{
path T=arc(A,r,a,b);
path R=lastcut(firstcut(T,P).after,Q).before;
path S=buildcycle(P,T,Q);
if (f) fill(S,p);
return R;
}
[/code]
Example:
[b]
path P=(0,0)..(30,0)..(60,60), Q=(0,0)--(10,30);
draw(SA((0,0),Q,P,6,10,360,true),red);
draw(P^^Q,blue);
[/b]
yields
[asy]path SA(pair A, path P, path Q, real r, real a, real b, bool f=false, pen p=lightyellow) { path T=arc(A,r,a,b); path R=lastcut(firstcut(T,P).after,Q).before; path S=buildcycle(P,T,Q); if (f) fill(S,p); return R; } path P=(0,0)..(30,0)..(60,60), Q=(0,0)--(10,30); draw(SA((0,0),Q,P,10,10,360,true),red); draw(P^^Q,blue);[/asy]

[color=green]Unnecessary quote removed by moderator[/color]

Parameters?

both focii and the center point labelled, the transverse and conjugate axes drawn as well as both asymptotes. The asymptotes aren't that important, since you could just draw two straight lines as a separate command. You'll also likely need the vertices and two sets of points to actually be able to draw the hyperbola. I'm not too picky, don't worry, and I greatly appreciate the fact that you'd go out of your way to design this  :D

So you'll be happy with a function that takes in, say, two points A and B (foci), the difference $ d=|AX|-|BX|$ that is constant along the hyperbola and one extra parameter $ a$ that tells how large piece to draw and returns the path that is a part of the corresponding hyperbola? OK, I'll try to design something like that. The axes are easy to draw once you know the foci, so I'll just take care of the asymptotes :).

Pretty much. I thought the hyperbola constant was for any point $ p$ and focii $ c_{1}$ and $ c_{2}$, the length of the transverse axis is: $ |c_{1}p-c_{2}p|$ where cp is the distance from either focus to any point on the hyperbola.

In your form, that would be the equivalent of $ d = | | Ax|-|Bx | |$ Since if you have Ax < Bx, the value would be negative without the absolute, even though distance is always a positive value.

OK, try this
[code]
path[] hyperbola(pair A,pair B, real d, real a)
{
 real U=sqrt((A.x-B.x)^2+(A.y-B.y)^2), D=d/U, T=acos(D), f=180/3.1415926;
 guide H;
 for(real t=-T+0.03;t<T-0.03;t=t+0.01) H=H..A+(B-A)*dir(t*f)*(1-D^2)/2/(cos(t)-D);
 pair M=(A+B)/2;
 path C=circle(M,a*U);
 H=lastcut(firstcut(H,C).after,C).before;
 return H^^M--M+(B-A)*a*dir(-T*f)^^M--M+(B-A)*a*dir(T*f);
}
[/code]
So, the code
[b]
pair A=(0,0), B=(10,15);
draw(hyperbola(A,B,10,1.5),red);
draw(hyperbola(B,A,10,1.5)[0],green);
dot(A);dot(B);
[/b]
will draw
[asy]size(250); path[] hyperbola(pair A,pair B, real d, real a) { real U=sqrt((A.x-B.x)^2+(A.y-B.y)^2), D=d/U, T=acos(D), f=180/3.1415926; guide H; for(real t=-T+0.03;t<T-0.03;t=t+0.01) H=H..A+(B-A)*dir(t*f)*(1-D^2)/2/(cos(t)-D); pair M=(A+B)/2; path C=circle(M,a*U); H=lastcut(firstcut(H,C).after,C).before; return H^^M--M+(B-A)*a*dir(-T*f)^^M--M+(B-A)*a*dir(T*f); } pair A=(0,0), B=(10,15); draw(hyperbola(A,B,10,1.5),red); draw(hyperbola(B,A,10,1.5)[0],green); dot(A);dot(B);[/asy]
I made hyperbola return the path array the first (indexed by $ 0$) component of which is the hyperbola branch itself and two other components are the asymptotes so you can choose to draw it with or without each particular component. The code is "dirty" in the sense that if the parameters are out of range, it crashes the program but at this stage I'd like to see whether that's what you need or you prefer something else. I hope that the code is self-explanatory but, of course, you are welcome to ask questions if something is unclear :)

And, please, stop quoting the last post in full when answering it. It never makes sense :spam:

Well done! However, I included a pretty ugly picture attachment of what I'm looking to create. If you made a program that only works for horizontal hyperbolas, I could just rotate the image as needed.

Also, do you mind explaining a little how your code works? I generally don't like using programs I don't understand.

Excellent job  :o that's pretty much exactly what I wanted

Edit: Never mind this post, I was having an asymptote error where path E prevented my labels from working. The reason why is because the letters E,W,S,N are all used as compass directions, and as such those letters can't be used in path commands.

Actually they can but then you have to write [b]plain.E[/b] instead of [b]E[/b] in label commands.

Two quick questions:

1) Does the olympiad package work on this forum?

2) What's a good place to get asymptote packages? When I search, all I find is downloads for the asymptote program, which I obviously have.

1) Yes, and you even do not need to write the import olympiad; command on the site :)

2) What exactly are you looking for? All standard packages like graph, math, etc. are included in the standard distribution, so you must have them already. :?

a couple more queries

In the olympiad package, there is a command that can draw arcs. However, it always draws double arcs. How do I draw single arcs, triple arcs, etc... using this command?

Also, is there a command for creating overbraces or underbraces in asymptote?

I believe you are talking about [b]anglemark[/b] in your first question. I regret to say it but the code for this command in the olympiad package is flawed and results in a mistake when you have less than double arc. The fix is easy though: replace the code with

[code]
path anglemark (pair A, pair B, pair C, real t=8 ... real[] s)
{
 pair M,N,P[],Q[];
 path mark;
 real xs;
 int n;
 n=s.length;
 xs=currentpicture.xunitsize;
 M=scale(t/xs)*unit(A-B)+B;
 N=scale(t/xs)*unit(C-B)+B;
 for (int i=0; i<n; ++i)
 {
  P[i]=scale(s[i]/xs)*unit(A-B)+B;
  Q[i]=scale(s[i]/xs)*unit(C-B)+B;
 }
 mark=arc(B,M,N);
 for (int i=0; i<n; ++i)
 {
  if (i/2==floor(i/2))
  {
   mark=mark--reverse(arc(B,P[i],Q[i]));
  }
  else
  {
   mark=mark--arc(B,P[i],Q[i]);
  }
 }
 if (n/2==floor(n/2) && n!=0) mark=(mark--B--P[n-1]);
 else if (n!=0) mark=(mark--B--Q[n-1]);
 else mark=(mark--B--cycle);
 return mark;
}
[/code]
(I suggest that Valentin implement this fix to the Olympiad package on the site :))
Then you'll be able to write anglemark(A,B,C) or anglemark(A,B,C,r) for single arc (the first case corresponds to the default $ r=8$), anglemark(A,B,C,r1,r2) for double arc, anglemark(A,B,C,r1,r2,r3) for triple arc, and so on:

[asy]size(150); unitsize(1cm); path AM(pair A, pair B, pair C, real t=8 ... real[] s) { pair M,N,P[],Q[]; path mark; real xs; int n; n=s.length; xs=currentpicture.xunitsize; M=scale(t/xs)*unit(A-B)+B; N=scale(t/xs)*unit(C-B)+B; for (int i=0; i<n; ++i) { P[i]=scale(s[i]/xs)*unit(A-B)+B; Q[i]=scale(s[i]/xs)*unit(C-B)+B; } mark=arc(B,M,N); for (int i=0; i<n; ++i) { if (i/2==floor(i/2)) { mark=mark--reverse(arc(B,P[i],Q[i])); } else { mark=mark--arc(B,P[i],Q[i]); } } if (n/2==floor(n/2) && n!=0) mark=(mark--B--P[n-1]); else if (n!=0) mark=(mark--B--Q[n-1]); else mark=(mark--B--cycle); return mark; } pen p=green+linewidth(1.5); pair A=(0,0), B=(3,1), C=(1,3); draw(A--B--C--A,blue); draw(AM(B,A,C,9),p); draw(AM(C,B,A,9,11),p); draw(AM(A,C,B,9,11,13),p);[/asy]

There are two more things you should know about anglemark. First of all, you have to execute the [b]unitsize[/b] command somewhere in you code before you use [b]anglemark[/b] (otherwise you'll end up with "division by zero" error; that's not very clever if you ask me, but this is how the code is currently written :P). Secondly, the path returned by anglemark consists of the arcs and a few travels to the vertex (just look at the picture above to see what I mean). This is a matter of taste, of course, and it makes no difference if you draw everything in one color and line thickness. Still, if you draw in color, you may wish to draw the arcs only. Then you'll have to modify the code even further (in particular, you'll want it to return path[] instead of path, etc.) I leave the corresponding programming to you :)

I'm too lazy to search the manual for "braces" right now, so I'll assume that you tried to do it and found nothing. That means that we'll have to program the corresponding routine ourselves. Do you want me to do it for you again, or you want to try it by yourself this time? ;)

Thank you so much for the arc help. This is going to help me a great deal because I'm writing some tex documents about geometry theorems.

Anyway, here's the best I could do for the brace command. Unfortunately, I can't find a reasonable way to manipulate the length of the brace.

[asy]unitsize(50); pair A,B,C; A = origin; B = dir(30); C = (0.8660254038,0); draw(origin--dir(30)--(0.8660254038,0)--cycle); draw(incircle(A,B,C)); draw(circumcircle(A,B,C)); draw(rightanglemark(B,C,A,2)); draw(anglemark(A,B,C,3,4)); draw(anglemark(C,A,B,3,4)); label(rotate(30)*"$\overbrace { \phantom { 22222222} } $",(0,0),N);[/asy]

You are a genius! Is there anything you can't do? You should seriously make an asymptote package one day, the world would benefit greatly  :D

I'm trying to add your brace program into the olympiad.asy file in my asymptote directory but it doesn't seem to be working. I try to add it onto the end of the document like this:

//explanation of what brace does:
{

program is inserted here

      }

then whenever I have a file with import olympiad; , the file doesn't compile  :(. I'm not sure what I'm doing wrong because I fixed the anglemark problem easily by replacing the old code in the olympiad.asy file by the new one. However, using the same approach with the brace command, I run into problems. Any idea what's going wrong?

I just tried it on my computer and everything was fine :?. Looking at what you posted I wonder what the braces {} are doing there. The way to add the code to the package is

//explanation
code as it is (just copy and paste),

not

//explanation
{
code
}

[quote="Prime factorization"]weird... now it's working. I swear, I always get these weird one time problems with asymptote that drive me crazy. One time, I had to actually redo an entire picture because there was a bracket missing that I couldn't find.[/quote]
You can put the code in notepad; it matches left and right brackets.

[quote="Prime factorization"] One time, I had to actually redo an entire picture because there was a bracket missing that I couldn't find.[/quote]
Hmmm.... Asymptote's error messages about runtime errors may be perplexing sometimes but the syntax error messages are quite reasonable: at least, they give you the exact lines where the errors occur, so I cannot imagine the situation when one would have to retype the entire code because of a missing parenthesis :? Are you looking at the error reports at all when compiling the code?

GRRRRRR... this command is causing me problems.

draw((anglemark(dir(150),origin,dir(60),5)));

it's giving me the reflex angle for some reason  :mad:

If anglemark(A,B,C,...) gives you the "reflex angle", simply change it to anglemark(C,B,A,...) ;) :
[asy]unitsize(1cm); pair A=(0,0),B=(3,1),C=(1,3); draw(A--B--C,blue); draw(anglemark(A,B,C,8,10,12),red+linewidth(1)); draw(anglemark(C,B,A,14,16,18),green+linewidth(1) );[/asy]

But how can you tell without taking a 50-50 chance which angle you'll get? Cuz I get the reflex angle about 70% of the time and I can't seem to find a pattern for getting the intended result.

Also, I desperately need a minimal example of the pathticks command being used on a diagonal line. If I see that, I am sure I'll be able to understand how that command works because I followed the AOPS Wiki instruction and they aren't working for me.

I believe the arcs in anglemark(A,B,C) go from BA to BC counterclockwise.

Example for pathticks:
[asy]unitsize(1cm); path P=(0,0)--(20,20); draw(P); add(pathticks(P,10,0.6,30,70));[/asy]
The important thing to understand is that pathticks returns a picture, so you cannot use the [b]draw[/b] command on the result but have to use the [b]add[/b] command instead to see the image. Also, like with anglemark, you need to execute the [b]unitsize[/b] command somewhere before you can use pathticks.

Sorry to bother again, but it seems that whenever I use rightanglemark and anglemark in an asy picture on this forum, I get the "unparseable or potentially dangerous..." error message. How do I get anglemarks to work on the forum (such as in the following code)?

[code]
import olympiad;
unitsize(100);
pair A,B,C,D,F,G,H,O;
A = dir(0);
B = dir(30);
C = dir(60);
D = (1.3660254,0);
O = origin;
F = (0.1830127,-0.6830127);
G = (0.57735027,0);
draw(anglemark(D,O,C,18,20),red);
draw(anglemark(O,D,F,18,20),red);
draw(anglemark(O,B,F,18,20),red);
draw(anglemark(D,F,G,18,20),red);
draw(anglemark(O,C,B,12,14),green);
draw(anglemark(F,B,D,12,14),green);
draw(anglemark(F,G,D,8,10),magenta);
draw(anglemark(B,G,O,8,10),magenta);
draw(unitcircle); draw(O--C,blue); draw(O--D,blue); draw(O--F); draw(O--B,blue,MidArrow); draw(C--D,blue); draw(A--B); draw(F--D,MidArrow); draw(F--B);
label("\small{$O$}",O,W);
label("\small{$A$}",A,S);
label("\small{$B$}",B,NE);
label("\small{$C$}",C,NE);
label("\small{$D$}",D,E);
label("\small{$E$}",F,S);
label("\small{$F$}",(0.57,0),SW);
add(pathticks(F--G,1,0.5,1,5));
add(pathticks(G--D,1,0.4,1,5));
add(pathticks(G--B,2,0.5,1,5));
add(pathticks(O--G,2,0.5,1,5));
[/code]

Also, observe the picture below:

[asy]size(300); pair A,B,C,D,F,G,H,O; A = dir(0); B = dir(30); C = dir(60); D = (1.3660254,0); O = origin; F = (0.1830127,-0.6830127); draw(unitcircle); draw(O--C,blue); draw(O--D,blue); draw(O--F); draw(O--B,blue,MidArrow); draw(C--D,blue); draw(A--B); draw(F--D,MidArrow); draw(F--B); label("\small { $O$} ",O,W); label("\small { $A$} ",A,S); label("\small { $B$} ",B,NE); label("\small { $C$} ",C,NE); label("\small { $D$} ",D,E); label("\small { $E$} ",F,S);[/asy]

For line segments EB and OC, I want to mark both lines with a double mid-arrow to indicate parallelism. How would I do this? The asymptote guide doesn't mention anything about this  :( And finally, can I get a black mid-arrow to appear on a blue line?

The first question is easy: you do not need to write the [b]import olympiad;[/b] command in your post code. The olympiad package is added to your code automatically and [b]import[/b] is one of those words that the "safety parser" (or whatever it is called) reacts to by rejecting the code. I just removed this part from your quoted code and, as you see, it works now :).
[asy]unitsize(100); pair A,B,C,D,F,G,H,O; A = dir(0); B = dir(30); C = dir(60); D = (1.3660254,0); O = origin; F = (0.1830127,-0.6830127); G = (0.57735027,0); draw(anglemark(D,O,C,18,20),red); draw(anglemark(O,D,F,18,20),red); draw(anglemark(O,B,F,18,20),red); draw(anglemark(D,F,G,18,20),red); draw(anglemark(O,C,B,12,14),green); draw(anglemark(F,B,D,12,14),green); draw(anglemark(F,G,D,8,10),magenta); draw(anglemark(B,G,O,8,10),magenta); draw(unitcircle); draw(O--C,blue); draw(O--D,blue); draw(O--F); draw(O--B,blue,MidArrow); draw(C--D,blue); draw(A--B); draw(F--D,MidArrow); draw(F--B); label("\small{$O$}",O,W); label("\small{$A$}",A,S); label("\small{$B$}",B,NE); label("\small{$C$}",C,NE); label("\small{$D$}",D,E); label("\small{$E$}",F,S); label("\small{$F$}",(0.57,0),SW); add(pathticks(F--G,1,0.5,1,5)); add(pathticks(G--D,1,0.4,1,5)); add(pathticks(G--B,2,0.5,1,5)); add(pathticks(O--G,2,0.5,1,5));[/asy]
The easiest way to get double middle arrow on a straight line is to draw the path twice like here:
[asy]size(100); pair A=(0,0), B=(20,10); draw(A--0.1*A+0.9*B,MidArrow); draw(0.9*A+0.1*B--B,MidArrow);[/asy]
This is not the best picture but, I hope, the idea is clear. The same trick works if you want to get a black arrow on a blue line
[asy]size(100); pair A=(0,0), B=(20,10); draw(A--B,MidArrow); draw(A--B,blue);[/asy]
though you would, probably, prefer to have the arrow color to devour the line color and not vice versa.
Of course, if you are unhappy with these solutions, you can always write your own routines for everything (arrow is nothing but a triangle filled with color and you can easily draw a triangle of any size and color in any position you wish :)).

I'm now on vacation in Colorado, so my access to the web is quite limited :(. So, don't get surprised if I don't respond immediately :P.

For some reason, I am having the hardest time getting a "dotted arrow". It seems I can get one, or the other, but not both at the same time (I always get an error message). I've checked the manual, AOPS Wiki and I still can't figure it out. :mad:

Hope you have a great vacation  :D

Hmmm... Works just fine for me :?.
[asy]draw((0,0)--(20,10),dotted+linewidth(1),MidArrow);[/asy]

The following code won't work on the website.

[code]
pair A,B,C,D,F,G,O;
O = origin;
A = dir(80);
B = dir(230);
C = dir(-40);
D = 2.5dir(80);
F = 2.5dir(230);
G = 2.5dir(-40);
draw(anglemark(C,B,A,10,12),red);
draw(anglemark(A,C,B,10,12),green);
draw(anglemark(B,A,C,10,12),blue);
draw(anglemark(G,F,D,10,12),red);
draw(anglemark(D,G,F,10,12),blue);
draw(anglemark(F,D,G,10,12),green);
draw(circumcircle(A,B,C));
draw(circumcircle(D,F,G));
draw(dir(230)..dir(260)..dir(-40),blue+linewidth(1));
draw(2.5dir(230)..2.5dir(260)..2.5dir(-40),blue+linewidth(1));
draw(dir(-40)..dir(0)..dir(30)..dir(80),red+linewidth(1));
draw(2.5dir(-40)..2.5dir(0)..2.5dir(30)..2.5dir(80),red+linewidth(1));
draw(dir(80)..dir(120)..dir(150)..dir(180)..dir(200)..dir(230),green+linewidth(1));
draw(2.5dir(80)..2.5dir(120)..2.5dir(150)..2.5dir(180)..2.5dir(200)..2.5dir(230),green+linewidth(1));
draw(A--B--C--cycle);
draw(D--F--G--cycle);
dot((0,0));
draw(O--D);
draw(O--F);
draw(O--G);
add(pathticks(O--A,1,0.5,1,5));
add(pathticks(O--B,1,0.5,1,5));
add(pathticks(O--C,1,0.5,1,5));
add(pathticks(O--D,2,0.5,1,5));
add(pathticks(O--F,2,0.5,1,5));
add(pathticks(O--G,2,0.5,1,5));
[/code]

Is it too long? If so, how would I shorten it (scale an image maybe)? If not, what's the problem?

Execute unitsize() before using anglemark or pathtiks! It is a known "bug" and I intend to send Valentin the "fixed olympiad package" but I'd like to discuss some details with Maria Monks before making too drastic changes :lol:
[asy]unitsize(50); pair A,B,C,D,F,G,O; O = origin; A = dir(80); B = dir(230); C = dir(-40); D = 2.5dir(80); F = 2.5dir(230); G = 2.5dir(-40); draw(anglemark(C,B,A,10,12),red); draw(anglemark(A,C,B,10,12),green); draw(anglemark(B,A,C,10,12),blue); draw(anglemark(G,F,D,10,12),red); draw(anglemark(D,G,F,10,12),blue); draw(anglemark(F,D,G,10,12),green); draw(circumcircle(A,B,C)); draw(circumcircle(D,F,G)); draw(dir(230)..dir(260)..dir(-40),blue+linewidth(1)); draw(2.5dir(230)..2.5dir(260)..2.5dir(-40),blue+linewidth(1)); draw(dir(-40)..dir(0)..dir(30)..dir(80),red+linewidth(1)); draw(2.5dir(-40)..2.5dir(0)..2.5dir(30)..2.5dir(80),red+linewidth(1)); draw(dir(80)..dir(120)..dir(150)..dir(180)..dir(200)..dir(230),green+linewidth(1)); draw(2.5dir(80)..2.5dir(120)..2.5dir(150)..2.5dir(180)..2.5dir(200)..2.5dir(230),green+linewidth(1)); draw(A--B--C--cycle); draw(D--F--G--cycle); dot((0,0)); draw(O--D); draw(O--F); draw(O--G); add(pathticks(O--A,1,0.5,1,5)); add(pathticks(O--B,1,0.5,1,5)); add(pathticks(O--C,1,0.5,1,5)); add(pathticks(O--D,2,0.5,1,5)); add(pathticks(O--F,2,0.5,1,5)); add(pathticks(O--G,2,0.5,1,5));[/asy]


geometry
circumcircle
inequalities proposed
inequalities

Show that in any triangle $ABC$ of area 1 the following inequlaity holds: $MM_1 \cdot MM_2 \cdot MM_3 \leq \frac{2}{27 \cdot R}$, where $M$ is an arbitrary point inside the triangle, $R$ is the radius of the circumscribed circle of triangle $ABC$ and $M_1,M_2,M_3$ are the projections of $M$ on sides $BC,CA$ and $AB$ respectively.

Easy but really good one!! :)

See that $MM_1=\frac{2E_1}{a}$ where $E_1$ is the area of the triangle $MBC$

Similarly we obtain analogous relations for $MM_2$ and $MM_3$

So $MM_1 \cdot MM_2 \cdot MM_3=\frac{8E_1E_2E_3}{abc}$ and of course $E_1+E_2+E_3=E=1$

from AM-GM we have $\frac{8E_1E_2E_3}{abc}\leq\frac{8}{abc}(\frac{E_1+E_2+E_3}{3})^3=\frac{8E^3}{27abc}=\frac{8}{27\cdot4ER}=\frac{2}{27R}$

and the proof is complete!! :D  :D

yes, it is right ;).that's also my solution  :D


Our teacher talked about the fact that

1^2+ 2^2+ 3^2+ ...  + n^2    =  (n(n+1)(2n+1))/6

I just wanted a proof other than the one that appears on the side of this web page...(Its very visual)

Does any one know the proof or can find it?

Notice that $(n+1)^2-n^2=2n+1$. I know that has something to do with the proof.

[quote="JBL"]Close, RC.  One way to do it is this:  if you know an expression for the sum $1 + 2 + \ldots + n$ in terms of n, try writing $(n+1)^3-1$ as $((n+1)^3 - n^3) + (n^3 - (n-1)^3) + ((n-1)^3-(n-2)^3) + \ldots + (3^3 - 2^3) + (2^3 - 1^3)$.  Another is induction.  I believe this has been discussed many, many times on this forum before, so you might do a search for it if you want more information.[/quote]
Hi JBL, thx for pointing that out, I have given a proof not long time ago, below is the link to it:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=154420&highlight=#154420[/url]

In fact, my proof was just a slight modification of yours :)

is that the method of summation by differences? we are taught that at college (double maths only though)
you can use induction as well, or the fact that every square can be expressed as a sum of consecutive odd numbers.

Induction is a very powerful tool, but the only thing weak about it is that you must know what you are proving to use induction... you can't induct without knowing the answer.

However, sometimes you can use an induction-like method to help you find a formula/equation/whatever.


pigeonhole principle
Putnam
graph theory
algebra solved
algebra

Prove that within six people there are three people who each other or three who don't know each other !
There are different methods pigeonhole principle, Ramsey or graph theory in general.

this is equivalent to proving that there exists a monochromatic triangle in the graph of the 6 points .easy,use pigeonhole principle

This problem was given at a contest in Romania:

"Acad. Radu Miron", in Vaslui 8-10 november 2002

cheers!

Dear teacher orl,

Does this problem is related to problem 2 Krschak 1947
http://www.kalva.demon.co.uk/eotvos/kus1947.html ?

And Putnam 1953 A2
http://www.kalva.demon.co.uk/putnam/putn53.html

regards,

moubi

Dear Moubi and Arne,

indeed these problems are connected with these olympiad questions. The question I pose was kind of introducing problem and might be found in many sources, especially text books. Ramsey theory extends the pigeonhole principle. Ramsey theory will be covered as one of the topics in graph theory.
Actually it's an intriguing topic for any book teaching problem-solving, e.g. in the following German olympiad book one can find a nice introduction: Lsungsstrategien fr mathematische Aufgaben by Wolfgang Mayer. Nice English books are Pearls in graph theory by Nora Hartsfield and Modern Graph Theory by Bela Bollobas.
The problems actors, meeting and table tennis competition might be solved by applying the pigeonhole principle repeatedly. But sometimes Ramsey might be useful.

Represent the 6 people by 6 points in the space. The fact that two people know each other is denoted by a red edge. If they don't know each other we might colour the edge blue between the respective vertices. Choose one of the six vertices. Let's call it P. There are 5 edges which start at P. Applying the pigeonhole principle we know that at least three of them have the same colour. Assume these are three red egdes. These edges connect vertices P and A_1, A_2 and A_3 respectively. Is one of the triangle's edges A_1,A_2,A_3 red we have a red triangle. In the other case A_1,A_2,A_3 is a blue triangle. Hence in each case we obtain a red or blue triangle which represents a group of three people who know each other or don't.

There are nice references at the Wolfram Research web site:

The main site is:

http://mathworld.wolfram.com/topics/RamseyTheory.html

Related topics may be found at:

http://mathworld.wolfram.com/DirichletsBoxPrinciple.html
http://mathworld.wolfram.com/Erdos-SzekeresTheorem.html
http://mathworld.wolfram.com/DilworthsLemma.html
http://mathworld.wolfram.com/RamseyTheory.html
http://mathworld.wolfram.com/RamseyNumber.html
http://mathworld.wolfram.com/RamseysTheorem.html


How many triples $ (a,b,c)$ of even positive integers satisfy $ a^3 \plus{} b^2 \plus{} c \leq 50$?

Obviously, $ a\equal{}2$, so we must now have $ b^2\plus{}c\le42$. Now, if $ b\equal{}6$, then $ c\le6$ with $ 3$ solutions, if $ b\equal{}4$, $ c\le26$ with $ 13$ solutions, and if $ b\equal{}2$, $ c\le38$ with $ 19$ solutions, for a total of $ 3\plus{}13\plus{}19\equal{}\boxed{35}$ solutions.


inequalities
geometry
inradius
geometry proposed

Prove for any triangle $\triangle ABC$  with inradius $r$ and semiperimeter $p,$ the following inequality:
\[1225 \left ( \frac{p}{r} \right)^2+729 \left ( \frac{r}{p} \right)^2 \ge 33102\]

Ok, here is my solution:

From $2p^2 \ge 27Rr$ and Weitzenbock $p^2+4Rr+r^2 \ge 4\sqrt{3}p^2,$ we have

$\frac{27}{4}p^2+27Rr+\frac{27}{4}r^2+2p^2 \ge 27\sqrt{3}rp+27Rr$

$\Longrightarrow \ 35  \cdot \frac{p}{r} +27 \cdot \frac{r}{p}  \ge 108 \sqrt{3}$

$\Longrightarrow \ 1225 \left ( \frac{p}{r} \right)^2+729 \left ( \frac{r}{p} \right)^2 \ge 33102.$


Suppose that $ f(x)$ is a function with $ f(78)\equal{}22$ and $ f'(78)\equal{}2$. Estimate $ f(81)$.

First, how is this physics?
Second, $ f(x)\approx f(a)\plus{}f'(a)(x\minus{}a)$ so $ f(81)\approx f(78)\plus{}f'(78)(81\minus{}78)\equal{}22\plus{}6\equal{}28$.

Haha.. looks like I accidentally posted these to the wrong forum. I could how sworn I selected new thread from the Calculus forum. Can a moderator can move them?


calculus derivative vector trigonometry rotation

This is presumably AP-level, since the concepts I use here are calculus-based, despite the fact that I live under a different system, where one goes through two years of college [i]before[/i] university, after high school.

I know that induced EMF for a fixed nmber of loops (i.e. when dN/dt = 0) is equal to the derivative of magnetic flux = d(phi)/dt = S cos (theta) dB/dt + B cos (theta) dS/dt - BA sin (theta) d(theta)/dt

where (phi) = magnetic flux = BA cos (theta)
t = time
B = magnetic field (in teslas)
S = surface (in square meters)
theta = the angle between the surface vector and the magnetic field
N = the number of loops

Would a dN/dt (a variation in the number of loops) generate an EMF by itself or a shift in the magnetic flux (in other terms, a d(phi)/dt) is necessary to induce an EMF?

I'm not sure what you mean.  By what mechanism does $ N$ change?  Is the coil becoming wound tighter by a motor, for example (which would also change $ S$)?  Is wire being somehow created?  (I think a physically meaningful answer depends on a physically meaningful mechanism for $ N$ to change.)

[quote="CEM_College_Champ"]
I live under a different system, where one goes through two years of college [i]before[/i] university, after high school.
[/quote]

Same system here!

[quote="CEM_College_Champ"]
Would a dN/dt (a variation in the number of loops) generate an EMF by itself or a shift in the magnetic flux (in other terms, a d(phi)/dt) is necessary to induce an EMF?

[/quote]

As far i have understood your question, the variation in the number of turns, $ \frac{dN}{dt}$ will affect the flux and consequently the EMF.

Since, $ \phi \equal{} NBA\cos \theta$

Hence,

$ e \equal{} \minus{}\frac{d\phi}{dt}
\\
\\
\equal{}\minus{}\frac{d}{dt} NBA\cos \theta
\\
\\
\equal{}\minus{}\left(BA\cos \theta \frac{dN}{dt} \plus{} NB\cos \theta \frac{dA}{dt} \plus{} NA\cos \theta \frac{dB}{dt} \minus{} NBA\omega \sin \theta \right)$

[quote="CEM_College_Champ"]
The coil would simply still be wound, though its radius hasn't changed one bit.

[/quote]

Then the term $ \frac {dA}{dt} \equal{} 0$, and since the coil does not rotate, $ \frac{d\theta}{dt} \equal{} \omega \equal{} 0$


Given a set of lattice points, we can perform one of the following operations (note that we still keep the original point in each case):
1. $(x,y) \rightarrow (x+1,y+1)$ (note that we still keep $x,y$
2. If x and y are both even, $(x,y) \rightarrow (x/2,y/2)$
3. $(x,y), (y,z) \rightarrow (x,z)$
If we start with $7,29$, can we get to $3,1999$?

[hide="Big Hint"]Invariants[/hide]
Masoud Zargar

[hide=" :maybe: Maybe..."]
Applying rule 1: Difference of terms remains the same.
Applying rule 2: Difference of terms is halfed.
Applying rule 3: Difference of terms can be added.

The original difference of the terms is 22.
This can only be halfed once, forming 11.

Since all differences that can be created by rules 1 and 2 must be a multiple of 11, adding these differences with rule 3 results in a difference that is a multiple of 11.

1999-3 = 1996, which is not a multiple of 11, so the point cannot be formed.
[/hide]


function
geometry
3D geometry
sphere
search
real analysis
real analysis unsolved

Be $A\subset R^{n}$ a set with positiv measure.

Exists there a function $f: R^{n}\to R^{n}$ in that way that $f(A)=S_{n}$ ??? ($S_{n}$ is the unit sphere)

Do we have some restrictions on the function? Otherwise both $A$ and $S_{n}$ have cardinality $c$ (I think).

There're no restrictions. The existence in most general case must be proved. Though of course can be search for continous functions, but this restriction for problem is not necesarry.


geometry
trigonometry
trapezoid
AMC
AIME
angle bisector

In the diagram below, angle $ABC$ is a right angle.  Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$.  Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10.$  Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG.$

[asy]
size(250);
pair A=(0,12), E=(0,8), B=origin, C=(24*sqrt(2),0), D=(6*sqrt(2),0), F=A+10*dir(A--C), G=intersectionpoint(E--F, A--D);
draw(A--B--C--A--D^^E--F);
pair point=G+1*dir(250);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));
label("$G$", G, dir(point--G));
markscalefactor=0.1;
draw(rightanglemark(A,B,C));
label("10", A--F, dir(90)*dir(A--F));
label("27", F--C, dir(90)*dir(F--C));
label("3", (0,10), W);
label("9", (0,4), W);[/asy]

[quote="4everwise"]In the diagram below, angle $ABC$ is a right angle.  Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$.  Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10.$  Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG.$

[img]6604[/img][/quote]

[hide]$\frac{3}{10}=\frac{EG}{\sqrt{91}-EG}$, $EG=\frac{3\sqrt{91}}{13}$.

$\frac{12}{37}=\frac{BD}{35-BD}$, $BD=\frac{60}{7}$.

$[DCFG]=35\cdot 6-12\cdot \frac{30}{7}-\frac{3\sqrt{91}}{2}+\frac{9\sqrt{91}}{26}$

The closest integer to $[DCFG]$ is something around $\boxed{148}$  :| [/hide]

$[ABC]=x=210$, then $[AEF]=\frac{5}{74}x$, and $[AGF]=\frac{10}{13}[AEF]=\frac{25}{481}x$, $[ADC]=\frac{37}{49}x$, so we have $ECFG=210(\frac{37}{49}-\frac{25}{481})=\frac{497160}{3367}\approx 147.65...$ WTF stupid computation...this takes a few seconds with a calculator, but on the aime....omg  :mad:

[quote="ch1n353ch3s54a1l"]$\frac{3}{10}=\frac{EG}{\sqrt{91}-EG}$, $EG=\frac{3\sqrt{91}}{13}$. [/quote]
That part is incorrect. $\angle AEF \ne \pi/2$ so the $\sqrt{91}$ isn't correct. I don't know how you got 148 as your answer. I made  the same mistake and the answer came out to be 149...

Blargh.

[hide]$BD=\frac{60}{7}$ and $DC=\frac{185}{7}$ by the Angle Bisector Theorem and $BC=35$. $[AEF]=\frac{525}{37}$ because $\sin A=\frac{35}{37}$. Since $[AEG]$ and $[AGF]$ have the same height, $\frac{[AEG]}{[AGF]}=\frac{EG}{GF}=\frac{3}{10}$ and the combined area of the two triangles is $\frac{525}{37}$ so then you can solve: $[AEG]=\frac{825}{74}$ so the area of the quadrilateral is $\frac{12\cdot 35}{2}-\frac{\frac{60}{7}\cdot 12}{2}-\frac{525}{37}+\frac{825}{74}\approx 156$ which apparently is wrong but I don't know why and I'm not going to go through that again.[/hide]

so...is my post right...

p.s. i love seeing my old posts turn up months later

[quote="davidyko"]I'm reviving this...
Well, the answer is $148$, and I got this right assuming $\angle AEF$ was right (no pun intended), but how would you get the answer otherwise?
Altheman, pardon my stupidity, but how did you get $[AEF]=\frac{5}{74}x$?[/quote]

$\frac{[AEF]}{[ABC]}=\frac{AE*AF}{AB*AC}=\frac{3*10}{12*37}=\frac{5}{74}$

I would assume it's alright to revive as this is a contest problem thread.
[hide="Cheating the System"]
It is easy to see that $BC = 35$ by Pythagoras Theory. Then, by Angle Bisector Theory and a bunch of computation, we get that $BD = \frac{60}{7}$ and $CD = \frac{185}{7}$.

Now here comes the part where we have to gamble a bit.
Let's assume $\triangle AEF \sim \triangle ABC$, even though it is clearly not. This will result in a minute change in the answer, hopefully less than 1.

Then, this means $GFCD$ is a trapezoid with height 9 and bases $\frac{185}{28}$(similarity) and $\frac{185}{7}$.
We have the area of $GFCD$ to be $\frac{1}{2} \cdot \frac{185}{28} \cdot \frac{185}{7} = 148.66$.
Rounding this down (since the area of the actual $GFCD$ seems to be a bit less than the $GFCD$ in our situation), we get the answer to be $\boxed{148}$.
[/hide]

Sorry for revive, but:
Why is $\dfrac{[AFG]}{[AFE]} = \dfrac{10}{13}$. It doesn't seem to follow from $\frac{1}{2}ab\sin C$. Can someone please explain?

By the Angle Bisector Theorem on triangle $\triangle AFE,$ we get $\dfrac{EG}{GF} = \dfrac{3}{10},$ so $\dfrac{EF}{GF} = \dfrac{EG + GF}{GF} = \dfrac{EG}{GF} + 1 = \dfrac{13}{10}.$ (Can you take it from there?)

^ Use multiplication and long division for a couple digits

[quote="forthegreatergood"]Does anyone have a solution to this which does *not* involve a calculator?[/quote] The issue is that the exact answer is \[ \frac{497160}{3367} = 147 + \frac{2211}{3367}. \] So it seems brutally difficult to try and avoid computation without doing something really ad-hoc. Quite nasty for an AIME problem...

[quote=Math1331Math][quote=leequack]/bump

I'm trying to solve this problem with mass points, but it doesn't seem like it is working.
I gave C a mass of 10, A a mass of 27, which gives B a mass of 9. This gives D a mass of 19, E a mass of 36, and F a mass of 37. We see the the mass of D is unbalanced, therefore meaning I did something wrong.

I cannot find what I did incorrectly, could someone help me?[/quote]

$EF$ isn't a cevian, in other words it doesn't have a vertex as one of the endpoints, so mass points doesn't work[/quote]

Ohhhhh that makes more sense. Thanks!

From the Angle Bisector Theorem, we can obtain that $BD=\frac{60}{7}, CD=\frac{185}{7}$. We also have $\frac{[AFE]}{[ABC]}=\frac{[AFE]}{210}=\frac{AE \cdot AF}{AB \cdot AC}=\frac{5}{74}$, so $[AFE]=\frac{525}{37}$. Next, we have $\frac{EG}{EF}=\frac{3}{13}$ and $\frac{GF}{EF}=\frac{10}{13}$ from Angle Bisector, so $[AGF]=\frac{525}{37} \cdot \frac{10}{13}=\frac{5250}{481}$. We want $[ABC]-[ABD]-[AGF]=210-\frac{360}{7}-\frac{5250}{481}$, which after computation rounds to $\boxed{148}$.

[quote]mass points doesn't work[/quote]Barycentrics do.

By the Angle Bisector Theorem, $BD=\frac{60}{7}$ and $CD=\frac{185}{7}$. Thus, $[ABC]=\frac{(3+9)\left(\frac{60}{7}+\frac{185}{7}\right)}2 =210$. Now we can barybash.

Set $\triangle ABC$ as the reference triangle and let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=\left(0:185:60\right)=\left(0,\frac{185}{245},\frac{60}{245}\right)$ and $F=(27:0:10)=\left(\frac{27}{37},0,\frac{10}{37}\right)$ and $E=(9:3:0)=\left(\frac34,\frac14,0\right)$.Since $G\in \overline{AD}$, we can parameterize $G$ by $G=\left(t:185:60\right)$. But since $E,G,F$ are collinear, we should have $$\begin{vmatrix}9&3&0\\ t&185&60\\ 27&0&10\end{vmatrix}=0 \Rightarrow t=717.$$ So now we have $G=\left(\frac{717}{962},\frac{185}{962},\frac{60}{962}\right)$. Now, by the area formula, $$[DCFG]=[CFG]+[CGD]=[ABC]\left(\frac{[CFG]}{[ABC]}+\frac{[CGD]}{[ABC]}\right)=210\left(\begin{vmatrix}0&0&1\\ 27/37&0&10/37\\ 717/962&185/962&60/962 \end{vmatrix}+\begin{vmatrix}0&0&1\\ 717/962&185/962&60/962\\ 0&185/245&60/245 \end{vmatrix}\right)=\frac{497160}{3367}$$ which is closest to $148$.


geometry
parallelogram
geometry unsolved

let E,F is the midpoints of AB and AD in parallelogram ABCD .CE and BF meet at K and M is on EC so that BM and KD are parallelo .Prove that area(KDF)=area(KBMD)

please post your solutions at http://www.mathlinks.ro/Forum/viewtopic.php?t=78435 .

  dg


inequalities
calculus
calculus computations

Can the initial condition of a linear ODE be an inequality? Like $f(x\geq 0)\geq 0$? I we say: $f(x)$ defined for $x\geq 0$, can we the reckon $f(x)\geq 0$ as the initial condition of the equation?
Boy, this is a silly question...

It's more like a boundary condition if you think of $x$ as a time variable. Anyway, what's in a name? you can use this inequality to rule out some (if not most) of solutions, just as you would with an initial condition.

Here's the dilemma: In a thread on he Unsolved and proposed problems board, Neumann was shown how to reduce $f^{2}(x)+Ch(x)=g^{2}(x)$ to $f(2\frac{df}{dt}-C)=2g\frac{dg}{dt}$, or $f(\frac{df}{dt}-\alpha)=g\frac{dg}{dt}, \alpha=\frac{1}{2}C$ Thus, we've eliminated dependancy on $h(x)$, but now we need an initial condition, which represents a new unknown. So, we've still got three unknowns. My thought was that wen we defined $f(x)$ and $g(x)$ for $x\geq 0$, the initial condition was already known. (The whole point of the problem is to reduce the equation to a form where only $f(x)$ and $g(x)$ are unknowns.)
Help..?


modular arithmetic
algebra unsolved
algebra

Show that the equation $ x^x \equal{}y^3 \plus{}z^3$ has infinitely many natural solutions.

___________________________
Azerbaijan Land of the Fire   :ninja:  :roll:  :rotfl:


trigonometry
function
search
superior algebra
superior algebra unsolved

(a) The subject line says it all: find the degree of the algebraic number $\cos(\pi/n)$. :)

But (b): let ${\wp : = z^{-2}+\sum_{m,n \neq (0,0)}\Big( \frac{1}{(z+m+in)^{2}}-\frac{1}{(m+in)^{2}}}\Big)$ be the Weierstrass function for the Gaussian lattice $\mathbb{Z}+i \mathbb{Z}$. For every $n$, prove that $\wp \Big( \frac{1+i}{2n}\Big)$ is an algebraic number, and find its degree.

Dear Vess-ok, you will say I have something with you, probably-there exist some nice forums called superiour algebra, advanced topics, etc, in which such problems would be much more appropriate. We discussed lots of times this aspect: if one says olympiad section, then olympiad section shall be.

OK, maybe you're right. I hesitated where to put it... If you [or any other moderator] think it is more appropriate for "advanced section" (part (b) probably is :)), then please move it there. (I can't move it myself, I'm not a moderator...) At any rate, part (a) could be "olympiad level" :) I just remembered a conversation with ZetaX in this thread... http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2078713379&t=119143

[quote="bilarev"]For the first one it's $\frac{\varphi(2n)}{2}$ where $\varphi(k)$ is Euler's totien function. See also the problem http://www.mathlinks.ro/Forum/viewtopic.php?t=149153[/quote]
Why $\frac{\phi(2n)}{2}$?
Obviosly for $\sum_{k=-n}^{n-1}cos\frac{k\pi }{n}=0$,  because $1+cos \frac{-n\pi}{n}=0$ we get
$\sum_{k=1}^{n-1}cos\frac{k\pi }{n}=0$. But for $k>\frac{n}{2}$ we have $cos\frac{k\pi }{n}=-cos\frac{\pi (n-k)}{n}$ and $cos\frac{\pi}{2}=0$. It give for $x=cos\frac{\pi}{n}$ equation $P(x)=0$ with integer coefficients and degree=$[\frac{n-1}{2}]$.
In other hand for $\theta =exp(\frac{\pi i}{n})$ we have equation $Q(\theta )=$ with degree $\phi(2n)$. From these equation we get equation for x with degree $\frac{\phi(2n)}{2}$ ($\phi(2n)$ is even for n>1).
It is easy to show, that $\frac{\phi(2n)}{2}\le [\frac{n-1}{2}]$ for n>1.

[quote="Rust"]for $\theta =exp(\frac{\pi i}{n})$ we have equation $Q(\theta )=$ with degree $\phi(2n)$. From these equation we get equation for x with degree $\frac{\phi(2n)}{2}$.[/quote]

It still remains to explain the irreducibility of the equation you have found! :)

Let $\alpha = \cos(\pi/n) = \zeta+\zeta^{-1}$, where $\zeta=e^{\pi i/n}$ is a primitive $2n$-th root of unity.  Since $\zeta$ satisfies $\zeta^{2}-\alpha \zeta+1=0$, we have $[\mathbb{Q}(\zeta): \mathbb{Q}(\alpha)]\leq 2$.  Since $\alpha$ is real and $\zeta$ isn't, we have $[\mathbb{Q}(\zeta): \mathbb{Q}(\alpha)]=2$.  It's a standard result that $\zeta$ has degree $\varphi(2n)$ over $\mathbb{Q}$, so $\alpha$ must have degree $\varphi(2n)/2$.