AMC
AIME
probability
geometry
USA(J)MO
USAMO
arithmetic sequence

For those of you that get in the double digit aime score, how do you manage you time. How long do you spend on one problem? When is it time to skip a problem? How long do you check your work? How many times do you check your work? etc.

[quote="Calculus"]For those of you that get in the double digit aime score, how do you manage you time. How long do you spend on one problem?[/quote]
Under 10 minutes for problems 1-10. Usually the first 10 are done before the 1 hour mark.
[quote] When is it time to skip a problem?[/quote]
When you can't solve it in 10 minutes (but always go back to it later if you have time).
[quote]How long do you check your work?[/quote]
I don't... but it takes about 3 extra minutes per problem to make sure your calculations are accurate.
[quote] How many times do you check your work? etc.[/quote]
0 :) It's unlikely that you'll run into time trouble on the AIME. You shouldn't worry about it too much.

Assuming you don't try to brute force problems that aren't meant to be brute-forced, you shouldn't run out of time, you just won't know how to do some of the problems.

lol, I'm sure that'll make the guy feel better about his time management, probability.  :P

Honestly tho, all the first 10 AIME problems are quick 5 minute problems as long as you don't head into a dead end. Some of them are really really quick problems even if you brute force it... an example would be one that asked for the coefficient of $x^2$ in $(1-x)(1+2x)(1-3x)\ldots(1-15x)$. Multiplying all this out takes 2 minutes and you can get the answer really quickly.

The next 5 are basically just being tricky with your skills...that's what really takes up your time.

Be sure to check your answers, though.  What kills you is when you know exactly how to do a problem but have a computational error.  Since time isn't usually a problem, you can afford to check you answers.  Like the above people have said, most of the 1-10 are fairly fast and easy solves if you know enough, so it won't cost much more time, and could potentially save you a whole point.

[quote="Elemennop"]an example would be one that asked for the coefficient of $x^2$ in $(1-x)(1+2x)(1-3x)\ldots(1-15x)$. Multiplying all this out takes 2 minutes and you can get the answer really quickly.[/quote]

That one was pretty easy, since all you need to do is keep the first three terms (i.e., don't go past $x^2$), since they'll just keep increasing and won't affect the $x^2$ value anymore.

10 minutes for problem 10? OMG PROBABILITY

10 minutes for problem 1!!! OMG.

And while we're on the subject of brute forcing, I just tried brute forcing number 15 on 2000 today. Try writing out all the cards and crossing them out one by one (there were 2000). Yeah, I was desparate to get the last one.

My goal is to spend 1 minute on problem one, 2 on problem 2, 3 on problem 3, etc... and 15 on problem 15.  Then I could get a perfect score in lets see.... 15*16/2= 120 minutes= 2 hours.

Haha, 10 minutes. Pretty INTENSE.

But that's a pretty good rule of thumb I guess, 10 minutes for first 10 is definitely doable. Though most people might need a bit more on at least one or two of 7-10. I don't really work very fast, but I make it a goal to finish 8, and at the very least 7 in the first hour. Then I can relax a bit more on the 2nd hour, doing 4 or 5 most of the time. By the third hour, it's basically stuff I either won't figure out or am to impatient to figure out, but I usually get 1 more on the third hour and then check over my first 7 or 8. So, it's basically cut to a 2 hour exam for the most part. With this, I generally solve 12-14 problems, and don't make more than 1 mistake.

I actually have timed myself and ended up having finished #9, with calculations all checked and all in about 30 minutes on an AIME...
The first 5 are very straightforward, should take about a minute or two each.
The next 5 are a bit trickier...maybe 5-10 minutes each?  10 minutes?  Depends.
That's if you're going very quickly.  But let's say you get stuck in 6-10, and end up taking about 1.5 hours there.  That leaves you with maybe 75 minutes for the end, just to be safe.  Okay, not much time.  Just go slowly and try to do them as accurately as possible.  Even if you can do a lot, it isn't going to count for anything if you make careless mistakes.

What I would suggest is stop after the first 10 or 12 or so to go back and check.  I would suggest alloting a good 20-30 minutes to check.  I actually did an AIME at one point, and going back and checking i found like 4 careless mistakes...even afterwards, there was one more careless mistake i missed.

I'm not exactly a double digit AIME scorer but I'd like to think that for most people, it takes about a solid hour and a half to do the first nine questions or so.  Maybe a little less.  This is the point that separates alot of test takers.  I would personally check all of my calculations for a good 20-30 minutes so that I could tackle the last 4-5 I'm going to attempt with a fairly easy mind.

Hopefully, I'll alot my time like this:
9 out of first 10: 1.5 hours
Check work at this point: .5 hours
Tackle more difficult problems (10-14): 1 hour

That should hopefully take me far on the AIME...

Yeah I've found I miss obvious things on problems when I get nervous and blank out.  Thus, it would probably be better to secure earlier problems before moving on to later ones.

The thing about me is that if I can't find a solution to a problem right away, and skip it; by the time I'm working on other problems, I keep thinking about that problem, lowering my efficiency. And by the time I come back to the problem I skipped, I get so intimidated by it, I have no hope of solving it. This always occurs on easy problems for me. Usually its just a tiny thing I miss

Like these three problems (both less than #10 on the 2003 AIME, which I got an 8 on). I couldn't solve it right away, even though its really easy. By the time I got back to it, I had no hope of solving it.
Here they are (try them for extra practice!):
-Find the sum of the areas of all the triangles formed by choosing vertices on a unit cube. (kept missing half of a particular set of triangles)
-Find the volume of the set of points that are inside or within one unit of a rectangular 3 x 4 x 5 box. (messed up on the edges)
- 0 < a < b < c < d are integers such that a, b, c is an arithmetic progression, b, c, d is a geometric progression, and d - a = 30.
 Find a + b + c + d.  (Kept trying to find an algebraic approach)

I [i]know [/i]how to solve these problems; its just details tripping me up.

Whoa, sorry...... digression from the subject.

Anyways, this is how I manage my time:
-Randomly solve any problems I can during the first 170 minutes, check right after solving a problem.
-Spend last 10 minutes checking.

I would finally just suggest be very careful checking, going through every step carefully.  Also, if you can, try doing a given problem two different ways (when you're checking it).  If you get the same answer both ways, then you probably haven't missed anything.  It would be difficult to mess the same thing up twice in two different methods.

[quote]
Honestly tho, all the first 10 AIME problems are quick 5 minute problems as long as you don't head into a dead end.
[/quote]

The first 10 AIME problems!?

x_x

Strangely, that makes me feel a little better. They're not supposed to be impossible and time consuming then...I guess that makes me feel better. But I thought the national average for the AIME was like a 4?

I think the average is even lower... remember to consider all of the people that qualify for the AIME but really don't care how well they do.  (for example most of the people who qualify from my school only took the AMC for extra credit)

I think a fair number of people here are underestimating the difficulty of problems 6-10. First of all, these are definitely NOT all easy problems. Unless you are working on an 80's AIME (or you are a past USAMO Winner or something), these problems CAN take anywhere in the range of 10-20 minutes to figure out how to solve it the right way or to brute force it. Sure there will be some number 10's that you can figure out in 2 or 3 minutes, but chances are that you won't be able to solve all of them that quickly. Time management is still key though; I suggest 1.5-2 hours for the first 10 problems (relatively confident) and then see what you can do with the last 5.

To answer the original questions: I outlined time per question above - 10 minutes max each for 1-5 (30 minutes total), 20 minutes max for 6-10 (1-1.5 hours total), see what you can get in 11-15. It's time to skip a problem when you get frustrated by it, or you reach one of the max time limits. Check your work often, right after you do a problem is good. Otherwise, before you move on to 11-15. I try to check my work at least once, hopefully twice. I'm very good at reading problems incorrectly or writing the wrong number down; that's my reason for caution.

Of course we must keep in mind that these are timeframes for hopeful double-digit AIME scorers, a small minority of the population of AIME test-takers.

Looking it up on the AMC site, the average last year was 2.75 on the AIME I, and 2.43 on the AIME II.

In 2004: 2.25 on I, 1.96 on II
In 2003: 3.06 on I, 3.05 on II
In 2002: 2.06
In 2001: 2.65
In 1999: 2.52 (I had a 14 that year)

You can expect the average to be between about 2 and 3, and I don't think anyone can predict more precisely than that. It also depends on how many students qualify; that 1999 average is inflated by the unusually hard AHSME that year, with less than 6000 qualifiers.

The key to time management: If you get stuck on a problem, even for a minute or two, move on to something else. Make sure you look at all of the problems; they're not strictly ordered by difficulty. You should have time to come back to all of the problems you skipped the first time through. If you still can't solve them, it's not the end of the world; a 5 or 6 is a fine score.

My AIME scores, in sequence: 6,7,14,13.
What happened between 10th and 11th grade? MOP. I was dazed and overwhelmed for a month, but I learned an enormous amount. It took a long time to digest that, but the results showed on the AIME. Unfortunately, my USAMO scores stayed constant- HM three years in a row.

[quote="jmerry"] Unfortunately, my USAMO scores stayed constant- HM three years in a row.[/quote]

I'm not sure that the word 'unfortunately' can legally go in that sentence :P

Eh hmm you should get to #9 in the first hour. Get to #12 in the second hour. Last hour is your call, I should really check over my work, but I'm too tempted to try the last two problems I never do check. Actually time is really not that much of an issue as long as you skip and come back later, the most difficult part of the AIME is careless mistakes, I make 1 careless mistake in every 4 problems :( :(.

[quote="1234567890"][quote="K81o7"][quote="beta"] ...the most difficult part of the AIME is careless mistakes, I make 1 careless mistake in every 4 problems :( :(.[/quote]

Heh, I took a practice AIME...
I made careless mistakes on numbers 2, 5 and 7, and got everything else right...[/quote]

That sucks.... What year? I find the 90's have tricky middle questions.[/quote]

96 i believe.

[quote="yif man12"][quote="K81o7"][quote="beta"] ...the most difficult part of the AIME is careless mistakes, I make 1 careless mistake in every 4 problems :( :(.[/quote]

Heh, I took a practice AIME...
I made careless mistakes on numbers 2, 5 and 7, and got everything else right...[/quote]

That's only 1 in 5. Beta still wins.[/quote]

...Stating the obvious doesn't really help...
The point was supposed to be that I messed up on easy problems... (or maybe just early problems)

i'm probably the least qualified to give advice on this issue, but

i would say the best strategy is to do as many problems as possible in the first 2 hrs(solve problems), and spend the last hr to check the work of the problems. if applicable. :)

My strategy (not that I'm any good - I've never gotten double digits on a real AIME) is to spend the first 2 hours working on 1-10, and 11-15 if there is time or you get desperate.  Then in the last hour, check work and look through 11-15 for any problems you think are approachable.

[quote="yif man12"][quote="K81o7"][quote="beta"] ...the most difficult part of the AIME is careless mistakes, I make 1 careless mistake in every 4 problems :( :(.[/quote]

Heh, I took a practice AIME...
I made careless mistakes on numbers 2, 5 and 7, and got everything else right...[/quote]

That's only 1 in 5. Beta still wins.[/quote]

By winning you mean losing?

[quote="yif man12"]You call it losing, I'll call it winning.[/quote]

On some AIMEs I've made about 6/11 careless mistakes even...not that that's a good thing...
But that was like 2 years ago, when I wasn't totally sure about what I was doing with the problems anyways.


Hi all,

I'm having a bit of trouble with the Pigeonhole principle. Can you guys help me prove these three problems?

1.Prove that the last two digits of the Fibonacci sequence are periodic.

2.Prove that among nine lattice points in three dimensional space, you can always chose two so that the midpoint of the corresponding line segment is also a lattice point.

3.Prove that among 18 people, there are always 4 who no each other or three that don't know each other.

Thanks

Hint for number 1:

[hide]If it is not periodic, then there are infinitely many possibilities for the last two digits of any two consecutive Fibonacci Numbers. Are there infinitely many possibilities for the last two digits of any given Fibonacci Number?[/hide]

Solution to number 1:

[hide]Consider two arbitrary consecutive Fibonacci numbers $ F_N$ and $ F_{N\plus{}1}$. Since there are only 100 possible sets of two last digits, there are 10000 possibilities for the last two digits of any given consecutive Fibonacci numbers $ F_N$ and $ F_{N\plus{}1}$. Consider the first 10001 pairs of Fibonacci numbers. By the pigeonhole principle, there must be a repeated set. Given the recurrence that yields the Fibonacci numbers, the last two digits of the Fibonacci numbers must therefore be periodic.[/hide]

Solution To Number 2:
[hide]If the midpoint of $ (a,b,c)$ and $ (x,y,z)$ is a lattice point, then $ a$ and $ x$ are either both even are both odd, and same with $ b$, $ y$; and $ c$, $ z$. Since, if you consider modulo $ 2$, there are only $ 2^3\equal{}8$ ways, one way must repeat. Hence two the midpoint of two lattice points among nine lattice points, is a lattice point.[/hide]

I think more info is required for number 3...

[hide="Solution to 3"]We assume that for any three people there must be a pair who know each other (1)
and
for any four people there must be a pair who don't know each other (2)

By (2) there must be a pair among the 18 people who don't know each other. Choose arbitrarily two such people.

Call those two teachers.
Call everyone else students.

For any student, consider him, and the two teachers. By (1), the student must know one of the teachers at least.

There are total 16 students and each of them knows at least one teacher.

By Pigeon hole principle there must be one teacher who knows at least 8 students. call this teacher [a].

By (2) there must be a pair among those 8 students who don't know each other. Choose such a pair arbitrarily.

Call them captains.
Call others other-students.

Like before, among the 6 other-students each must know at least one captain by (1).

So there is one captain who knows 3 other-students by (1) and PHP. Call this captain [b].

Among these three other-students two must know each other by (1). call them [c] and [d].

Now [a],[b],[c] and [d] knows each other. Contradicting (2)

Hence. at least one of (1) and (2) must be false. We are done.[/hide]

That 18 in the question can be replaced by 9, which is the least (best) constant.
But that would make the problem much more difficult.
Check out [url=http://mathworld.wolfram.com/RamseyNumber.html]Ramsay Numbers[/url] for more information.


A subset $ C$ of $ \mathbb{R}^d$ is called convex if for every $ x$ and $ y$ in $ C$, and every real $ t$ with $ 0\leq t\leq 1$, we have $ tx\plus{}(1\minus{}t)y\in C$. Show that the intersection of an arbitrary collection of convex sets is convex.
Also, show that a convex subset of $ \mathbb{R}^d$ is connected.

[hide="first part"]$ \{ C_\alpha \}_{\alpha \in \Delta}$ be an arbitrary collection of convex sets

assume $ X\equal{}\cap_{\alpha \in \Delta} C_\alpha$ is nonempty

$ x,y \in X \to x,y \in C_\alpha \forall \alpha$

$ C_\alpha$ convex $ \to r_t \equal{} tx\plus{}(1\minus{}t)y \in C_\alpha \forall \alpha$

$ \to X$ convex[/hide]

[hide="pathwise connected implies connected"]

so quite obviously the easiest path to connect any two points is the straight one between them, which we have

[/hide]


Let $ABC$ be a triangle. Let $A'$ and $A_{1}$ be respectively the foot of $A$ altitude and the midpoint of $BC$. Construct the semi-circle $(A_{1})$ with diameter $BC$ so that it lies on the side containing $A$. Construct the circle through $A'$ and $A_{1}$ that touches the circle $(A_{1})$ at $M$. Let $MU$ be the bisector of $\measuredangle{BMC}$, here $U$ lies on $BC$.

Define $V$ and $W$ analogously. Show that the orthocenter of $UVW$ lies on the Euler line of $ABC$.

I only get that $M$ is the intersection of circle $A_{1}$ and the altitude of $BC$. Then after constructing the angle bisectors, the whole figure become very complex, and I fail to see what to do next... ...

That 's what this problem is about. Have fun!!!   :lol:

Noone has any idea??  :)

Another way to state this problem is that the orthocenter of the cevian triangle of the square root of the orthocenter $H$ of $ABC$ lies on the Euler line of $ABC$.

The solution to a generalization by Paul Yiu is at http://forumgeom.fau.edu/FG2006volume6/FG200631index.html

Maybe use complex numbers to find a solution ? :maybe:


algebra
polynomial
inequalities
superior algebra
superior algebra unsolved

[color=darkblue]Prove that the set of algebraic numbers in $ \mathbb{Q}$ is countable?
the set of bijection $ f: \mathbb{N}\to\mathbb{N}$ is coutable?Or uncountable?[/color]
thanks in advance!

Are you sure that the first question is really what you want? Let $ A$ be the set of algebraic integers in $ \mathbb{Q}$.

$ A\subset \mathbb{Q}$, so $ |A|\leq |\mathbb{Q}|$ and the cardinality of the rationals is countable.

Presumably, the question is supposed to be "Prove that the set of algebraic numbers in $ \mathbb{C}$ is countable".

[hide="Hint"]
This is true since every polynomial with rational coefficients has finitely many roots, and there are countably many such polynomials.
[/hide]

The set of bijections $ f : \mathbb{N} \to \mathbb{N}$ is uncountable.
[hide="Hint"]Try to define, for each subset $ A$ of the natural numbers, a bijection $ f_A$ that depends on $ A$ in such a way that different sets yield different bijections. What can you conclude regarding the cardinality of the set of bijections?
[/hide]

Look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=214083]here[/url] for the second problem.

It is uncountable end equals continuum.

It can be easily seen from Riemann theorem about conditionally convergent series. Every real number can be obtained by permutation of such series.

Another easy proof: If $ \sigma: \mathbb{N}\to\mathbb{N}$ is permutation. Let $ F(\sigma)(i): \equal{}0$ when $ \sigma(i)\geq i$ and $ F(\sigma(i)): \equal{}1$ otherwise. F is suriection onto set of all 0,1 sequences.
(Similar way can be proved that set of all permutations without fixed points is uncountable).

P.S. Opposite inequality is easy since cardinal number of set of all mappings $ \mathbb{N}\to\mathbb{N}$ is continuum.


calculus
integration
function
real analysis
real analysis unsolved

Let $ f: [0,\infty)\rightarrow (0,\infty)$ be an increasing differentiable function with continuous derivative.

Prove that:

$ \int_{0}^{\infty}\frac{dx}{f(x)+f'(x)}<\infty$ if and only if $ \int_{0}^{\infty}\frac{dx}{f(x)}<\infty$.

we only need to prove that $ \int_{0}^{\infty}\frac{dx}{f(x)+f'(x)}-\int_{0}^{\infty}\frac{dx}{f(x)}<\infty$

that is

$ \int_{0}^{\infty}\frac{f'(x)dx}{f(x)(f(x)+f'(x))}<\infty$

we notice that f is increasing, $ f'>0$, so we have

$ \int_{0}^{\infty}\frac{f'(x)dx}{f(x)(f(x)+f'(x))}\leq \int_{0}^{\infty}\frac{f'(x)dx}{f(x)^{2}}\leq \int_{a}^{\infty}\frac{dx}{x^{2}}<\infty$


using the numbers 1, 3, 4, and 6 once with the four math symbols (* / + -) make 24.

Yes, but I guess we are allowed to use it twice. The problem said "using the numbers 1, 3, 4 and 6 once", but it didn't say anything about how many times you can use +, - etc.

I don't think that is possible. (Not completely sure though)

the directions in my book don't say whether or not to use each symbol once.

Hmm... how could it be possible? You have only four numbers, but four symbols also, and the symbols have to go between the numbers. Unless the $-$ is used for a negative number, there's no way you can use all four.

What about (14-6) * 3 = 24? Aren't parentheses allowed?

    3
1  * 4 * 6 = 24    {That is 1^3 = 1*1*1 = 1}
But, this also uses two '*'.


Ring Theory
superior algebra
superior algebra unsolved

I'm trying to prove Z[i] is UFD and this is where I am:
I'm trying to show right now that every nonzero, non-unit in Z[i] can be factored as unit times the product of irreducibles.  help appreciated : )


search
combinatorics unsolved
combinatorics

let p be a prime, and let $ a_1,a_2,a_3...a_{p \plus{} 1}$ be distinct positive integers.Prove that there are indices i,j such that
$ \frac {max(a_i,a_j)}{gcd(a_i,a_j)} \ge\ p \plus{} 1$

You can see the solution here:
http://www.mathlinks.ro/viewtopic.php?search_id=610519861&t=287610


Let $ a,b,c$ be the sidelenghts of a triangle with $ a\plus{}b\plus{}c\equal{}1.$ Prove that
$ \sqrt{2a\plus{}b^3}\plus{}\sqrt{2b\plus{}c^3}\plus{}\sqrt{2c\plus{}a^3} \ge \sqrt{2}\plus{}1.$
:)

[quote="can_hang2007"]Let $ a,b,c$ be the sidelenghts of a triangle with $ a \plus{} b \plus{} c \equal{} 1.$ Prove that
$ \sqrt {2a \plus{} b^3} \plus{} \sqrt {2b \plus{} c^3} \plus{} \sqrt {2c \plus{} a^3} \ge \sqrt {2} \plus{} 1.$
:)[/quote]
the inequality than stronger with the same condition:
$ \sqrt {2a \plus{} b^3} \plus{} \sqrt {2b \plus{} c^3} \plus{} \sqrt {2c \plus{} a^3} \ge\ \sqrt {\frac {19}{3}}$

Are you sure? Try $ a \equal{} b \equal{} \frac {1}{2},c \equal{} 0.$ ;)

Some other inequalities of this type is:

1) (Phan Thanh Nam) If $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$, then
$ \sqrt [3]{a \minus{} b \plus{} c^3} \plus{} \sqrt [3]{b \minus{} c \plus{} a^3} \plus{} \sqrt [3]{c \minus{} a \plus{} b^3} \le 1.$

2) (my inspiration from Phan Thanh Nam's one) If $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$, then
$ \sqrt {a \plus{} \frac {(a \minus{} b)^2}{2}} \plus{} \sqrt {b \plus{} \frac {(b \minus{} c)^2}{2}} \plus{} \sqrt {c \plus{} \frac {(c \minus{} a)^2}{2}} \ge \sqrt {3}.$
:)

[quote="can_hang2007"]1) (Phan Thanh Nam) If $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$, then
$ \sqrt [3]{a \minus{} b \plus{} c^3} \plus{} \sqrt [3]{b \minus{} c \plus{} a^3} \plus{} \sqrt [3]{c \minus{} a \plus{} b^3} \le 1.$
[/quote]
Hint:
Assume that $ c\equal{}\min\{a,b,c\}$, then we can prove the chain inequality
$ \sqrt [3]{a \minus{} b \plus{} c^3} \plus{} \sqrt [3]{b \minus{} c \plus{} a^3} \plus{} \sqrt [3]{c \minus{} a \plus{} b^3} \le \sqrt[3]{a\minus{}b\plus{}c^3}\plus{}\sqrt[3]{b\minus{}a\plus{}c^3}\plus{}\sqrt[3]{a^3\plus{}b^3\minus{}c^3} \le c\plus{}c\plus{}\sqrt[3]{a^3\plus{}b^3\minus{}c^3} \le 1.$
:)

[quote="can_hang2007"][quote="can_hang2007"]1) (Phan Thanh Nam) If $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$, then
$ \sqrt [3]{a \minus{} b \plus{} c^3} \plus{} \sqrt [3]{b \minus{} c \plus{} a^3} \plus{} \sqrt [3]{c \minus{} a \plus{} b^3} \le 1.$
[/quote]
Hint:
Assume that $ c \equal{} \min\{a,b,c\}$, then we can prove the chain inequality
$ \sqrt [3]{a \minus{} b \plus{} c^3} \plus{} \sqrt [3]{b \minus{} c \plus{} a^3} \plus{} \sqrt [3]{c \minus{} a \plus{} b^3} \le \sqrt [3]{a \minus{} b \plus{} c^3} \plus{} \sqrt [3]{b \minus{} a \plus{} c^3} \plus{} \sqrt [3]{a^3 \plus{} b^3 \minus{} c^3} \le c \plus{} c \plus{} \sqrt [3]{a^3 \plus{} b^3 \minus{} c^3} \le 1.$
:)[/quote]
Note that to do this, we need to have $ A\plus{}B \ge 0, B\plus{}C \ge 0,C\plus{}A \ge 0$ where $ A\equal{}a\minus{}b\plus{}c^3,B\equal{}b\minus{}c\plus{}a^3,C\equal{}c\minus{}a\plus{}b^3$ ;) :)


function
integration
real analysis
real analysis solved

Find all continuous functions f:R->[1,00) for which there exists a in R and k in N s.t. f(x)*f(2x)*..*f(nx)<=a*n k for all real x and for all n in N*.

[i]Romanian NMO/final round/Alba Iulia 1999/problem 1/12'th grade/proposed by Radu Gologan[/i]

We note a>0 since f(x) \geq 1

f(x)=1 identically constant 1 is a solution.
If f(x) = C > 1 constant  is solution then C^n/n^k \leq a for any n \in N
wich is absurd since lim_{n->+oo} C^n/n^k=+oo
The constant solution are f(x)=1

It is not finish, I just looked constant solution

Since f(x) \geq 1 >0, product it not easy, sum is better here.

 \sum _{k= 1 to n }lnf(kx) \leq lna + k.lnn for any x real, any n
take x= 1/n we see sum Riemann appears
 1/n.\sum _{k= 1 to n }ln(f(k/n))  \leq (lna + k.lnn)/n tend n->+oo
 \int _[0,1]ln(f(x))dx  \leq 0
f(x) \geq 1=>ln(f(x)) \geq 0=> \int ln(f(x)) dx  \geq 0

So \int ln(f(x)) dx = 0,
x-> ln(f(x)) is positive continuous on the compact [0;1]

We have ln(f(x))=0=> f(x)=1

I am going to write only the most important things:

You apply ln to  f(x)*f(2x)*..*f(nx)<=a*n^k  and rewrite the problem like so:
sum(i=1,n) ln(f(ix))) <=ln(a)+kln(a) (a>0)
for a number b real<>0 let x=b/n. so the condition becomes:
sum(i=1,n) b/n ln(f(i*b/n))<=(blna+kblna)*1/n
the sum in the left is a Riemann sum for f on [0,b]. Apply lim to it and get that  \int (0,b) ln(f(x)) dx=0 and because b was chosen arbitraly we get that ln(f(x))=0 for all x>=0 and because f is continuos we get that f(x)=1 for all x>=0
In the same manner you get that for any b>0 \int (0,b) ln(f(-x)) dx=0 so
\int (0,-b) ln(f(x)) dx=0 and get f(x) for all x>=0
cheers! :D :D

I didn' t realise that I only prove that lnf(x)=1 only on [0,1]
thx Lagrangia

I did exactly the same thing. What you did for [0,1] works perfectly for [0,x], so the initial solution was fine if you ask me :D

I wonder if this can be successfully approached without integrals.


calculus
integration
inequalities
function
algebra
polynomial
complex analysis

I would like to see a solution to $ \int_{0}^{\infty}\frac{dx}{1\plus{}x^{n}}$ through residue theory and contour integration.

Based on the below picture

1. By inequality the integral on $ A_{R}$ is $ 0$ when $ R\rightarrow\infty$
2.  all of poles of $ \frac {1}{1 + z^{n}}$ are simple poles ( they are the roots of the equation $ 1 + z^{n}$ so choose $ R$ is big enough the the only simple pole which is in interior of the picture is $ e^{\frac {i\pi}{n}}$ so the residue is $ \frac { - 1}{n}e^{\frac {i\pi}{n}}$
3. Parametrizing $ L'_{R}$ by $ te^{\frac {2i\pi}{n}}$ where $ 0\leq t\leq R$ then we have $ \int_{L'_{R}}\frac {1}{1 + z^{n}} = - e^{\frac {2i\pi}{n}}\int_{L_{R}}\frac {1}{1 + z^{n}}$

the we have $ 2i\pi\frac { - 1}{n}e^{\frac {i\pi}{n}} = \int_{L'_{R}} + \int_{L_{R}} = (1 -  e^{\frac {2i\pi}{n}})\int_{L_{R}}\frac {1}{1 + z^{n}}$

from the above find the integral, then let $ R\rightarrow\infty$ and use the definition of $ sinx =\frac{e^{ix} - e^{ - ix}}{2i}$ we have $ \int_{0}^{\infty}\frac {1}{1 + z^{n}} = \frac {\pi/n}{sin(\pi/n)}$

Can you carry out that residue calculation?  When I attempted to do it, I got something really ugly.

How can a residue of a simple rational function give something ugly¿  :wink:

Use the following fact

let $ F \equal{} \frac {P_{n}(z)}{Q_{m}(z)}$, where $ P_{n}, Q_{m}$ are polynomials , the residue of $ F$ at simple pole $ z_{0}$ is $ \frac {P_{n}(z_{0})}{Q'_{m}(z_{0})}$

Okay, I forgot about that formula.  I was doing a much uglier calculation using the coefficients of the Laurent series expansion.


geometry
geometric transformation
geometry unsolved

It seem be the problem of Jack Garfulkel propose in Crux  :D , but Jack's problem need to prove $AM,BN,CP$ are concurrent when $MNP$ max
maybe it is still open problem! Did you get a solution ?

This problem was given in of  the romanian team selection test  between 90-94.So it must have a solution.I looked for one but no success.So if anyone has an idea, try to solve this problem.

Come on!No geometer can solve this problem??

[quote="mateivld"]We are given a triangle ABC and M,N,P points on sides BC,CA,AB such that MNP is an equilateral triangle.Determine the maximum area of all such triangles  MNP. :lol:[/quote]

You can get a result very close to this in

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107596[/url]

It's on spanish, the thing is that the the angle in the spiral similarity should be the greatest you have to do some calculations


Find $n$ and $m$ integers with $n>m$ such that:
$a^{m}+x^{m}|a^{n}+x^{n}$ for all $x$ and $a$ positive integer.

Do you mean $a^{n}+x^{n}$ in the second one?

If not, the condition is equivalent to

$a^{m}+x^{m}| (a^{n}-a^{m})$

And there only exists trivial solutions since we can choose $x$ such that this is impossible.

If so, we obviously have the solutions where $m | n$ and $\frac{n}{m}\equiv 1 \bmod 2$.


Hello!
I have seen this subject for the 10th grade but no teacher was able to tell me what is about, maybe u can.
Bye!
tnx!
Silvi

recurrences of the form xn+1 = ( axn + b ) / (cxn + d ).

Usually one studies the convergence of the sequences defined in this matter, using the properties of the function f(x) = (ax+b) / (cx+d).

Can you tell me more about this kind of function?
Can you tell me a type of problem wich contains this matter?
Tnx!

anyway where did you come up with this kind of type of problems? if you are thinking about meeting the curriculum for the 10th grade NMO then don't worry about this, :)

try working usual recurrence sequence, with 2nd degree terms.

Thanks!
I was curious because no one was able to tell me about this and I wanted to know a little about it, just in case.
For the second degree recurencies I solved pb from the 11th grade manual by Ganga.
This is all for now but in a few days i'll be back with more bugging questions.
Bye!


induction
linear algebra
matrix
linear algebra unsolved

Prove that polynom $a_1x^{p_1}+...+a_kx^{p_k}$ is not diviseble by $(x-b)^k$ for any real b
here all the  a_i are real

No, it's not. Did I post the problem in the properate forum topic at all?

[quote="Orif"]Rustam bu misolni qattan topding.Qiyinku(a_i lar butun emasmi)[/quote]
I find that problem from the book Fadeev,Sominskiy"Zbornik zadach po visshey algebre" there was a hint:
use induction but i cant. please help me

Oh, I find the solution yesterday using ther hint.
For n=1  the problem is clear.
Assume that it works for all the k-polinoms.
Lets be given the polinom $a_1x^{p_1}+...+a_kx^{p_k}+a_{k+1}x^{p_{k+1}} =: f(x)$ the it has the same non zero roots as the polinom $b_1+...+b_kx^{q_k}+b_{k+1}x^{q_{k+1}}: =d(x)$ where d(x) is just the polinom divided to the smallest power of x  polinom f(x). Assume to the cotrary that f(x) and obviously the d(x) is divisible by $(x-\alpha)^{k+1}$ then the d'(x) is diviseble by $(x-\alpha)^{k}$ but the d'(x) is the polinom i form $a_1x^{p_1}+...+a_kx^{p_k}$ and is not diviseble to $(x-\alpha)^{k}$ for any $\alpha$. Contradiction . The proof is finished.

So easy but get me a lot of time . Sorry that i've posted it to the Matrix Algebra(because of I foud it in the book metioned above). :?


an electrical device uses 84 kilowatts per hour. The device is turned off every 30 seconds of being on, for 10 seconds. How many kilowatts will it use in 3 hours and 38 minutes? round any decimals in the problem to the nearest hundredth.

EDIT: after solving i realized above problem is too easy.. changing it so that it uses 98.6 kilowatts per hour. Turned off every 26 seconds for 15 seconds. still easy, i know..

[hide]I think its 1.94 kilowatts. Just convert 3h38m to minutes (218) then multiply by .75, because it's only on 3/4 of the time. Then just divide by 84.[/hide]

[i]please use hide[/i]


How many  :ninja: numbers are there if a  :ninja: number cannot have more than 10 digits, and can have 0's in the leftmost digits?

For example: 0004 is a  :ninja: number, while also being different than just 004, 04 or 4.

If this isn't middle school level, I'm sorry  :o


$ x\equal{}\underbrace{111...11}_{n\ digits}$ is prime. Prove that $ n$ is prime.

[quote="pelao_malo"]$ x \equal{} \underbrace{111...11}_{n\ digits}$ is prime. Prove that $ n$ is prime.[/quote]

If $ n\equal{}ab$, $ \underbrace{111...11}_{n\ digits}$ is divisible by $ \underbrace{111...11}_{a\ digits}$ or $ \underbrace{111...11}_{b\ digits}$

[quote="pelao_malo"]$ x = \underbrace{111...11}_{n\ digits}$ is prime. Prove that $ n$ is prime.[/quote]
You can write $ 11...1$ as $ (10^n - 1)/9$ now $ n=ab$ where $ ab$ is a nontrivial factorization then write ${ (10^{ab}-1})$ as $ (10^a)^b - 1$ and use the factoring identity for $ x^k - y^k$.


Solve the $ x^{2}\minus{}3y^{2}\equal{}1$ diophantine equation.

Pell's Equations are not HSB material, and there is no extremely basic way to solve them because their solution set is not easily described.  Consult

http://en.wikipedia.org/wiki/Pell's_equation
http://mathworld.wolfram.com/PellEquation.html

[url]http://en.wikipedia.org/wiki/Pell%27s_equation[/url]
@t0rajir0u:
Be careful about those apostrophes, they'll screw up your link.

The original poster forgot to specify [b]integer[/b] solutions.

They said Diophantine Equations though  :wink: .


Λοιπον επειδη βλεπω οτι τωρα τελευταια ανεβηκαν πολυ οι μετοχες του Fermat
στο μαθηματικο χρηματιστηριο.....Θελω να βαλω σαν ασκηση μια αποδειξη που ειχα βρει για μια ειδικη περιπτωση (παρολα αυτα ειναι αρκετα γενικη....)
Και λεει:
Νδο η εξισωση
$x^{n}+y^{n}=z^{n}$ δεν εχει λυση στους θετικους ακεραιους $(n\geq 3)$ με
$0\leq x\leq n$ και $0\leq y\leq n$

Την αποδειξη την ειχα βρει στην Α λυκειου,οποτε αναζητηστε μια ευκολη λυση+οχι με θεωρια ελλειπτικων καμπυλων και δεν συμαζευετε..... :P

 :D  :D  :D

Edit:H συνθηκη $n\geq 3$ ειναι περιττη! Δεν χρειαζετε...(Μου φαινεται οτι ολο και γενικευεται η αποδειξη... :P )

μηπως το θεωρημα Zsigmondy's ?

Φίλε μου Ράλλη όλο κι το παμε εδώ κι το πάμε εκεί μη΄πως θες να βρούμε άλλη απόδειξη του ΤΘΦ ???? :rotfl:  :rotfl:  :D  :rotfl:  ΓΙατί αν είναι έτσι πρεπι να μοιράστούμε 1000000 Δολλάρια τότε κι εγώ συμφωνώ τελος πάντων θα κάνουμε ό,τι μπορούμε
Χαιρετώ

[quote="ambros"] να μοιράστούμε 1000000 Δολλάρια [/quote]
ισχύει ακόμα?? δεν τα πήρε όλα ο Wiles??

ρε παιδια υπαρχει και η γενικευση του τελευταιου θ. του Φερμα. δινει 100.000 δολλαρια  για μια αποδειξη ή αντιπαραδειγμα :D

Το αξιοπεριεργο ειναι οτι ο κυριος Wiles ασχοληθηκε με το θεωρημα αυτο απο ηλικια περιπου 10-12 ετων!Εντυπωσιακη η επιμονη του.Εγω θα του εδινα αλλο ενα μεταλλιο για την επιμονη του! :)

τι γενίκευση εννοείς Κώστα????

[quote="RAKAR"]
Νδο η εξισωση
$x^{n}+y^{n}=z^{n}$ δεν εχει λυση στους θετικους ακεραιους $(n\geq 3)$ με
$0\leq x\leq n$ και $0\leq y\leq n$

[/quote]

Φυσικά πρέπει να πάρουμε $x,y > 0$ γιατί αν $x=0$ η εξίσωση έχει άπειρες λύσεις για κάθε $n$  .

Επίσης για $x=y$ η εξίσωση έχει λύσεις μόνο αν $n=1$ .

Έστω λοιπόν $n \geq x>y>0$ .

Τότε $z \geq x+1 \Rightarrow z^{n}\geq (x+1)^{n}= x^{n}+n x^{n-1}+...+1 > x^{n}+nx^{n-1}> x^{n}+y^{n}$

[quote="silouan"]Φίλε Ράλλη φαινεται ότι σου αρέσει πολύ αυτή η ασκηση....Την είχες ξαναπροτείνει εδώ....... :P  και την είχα αποδείξει
Για του λόγου του αλειθές http://www.mathlinks.ro/Forum/viewtopic.php?t=86337[/quote]

 :maybe:  :ninja:

[quote="Ramanujan"]

 :maybe:  :ninja:[/quote]

Τι εννοείς ??

[quote="silouan"]

Τι εννοείς ??[/quote]

Απλά κατάλαβα γιατί κάτι μου θύμιζε... :D


logarithms
real analysis
real analysis unsolved

[quote="Frozen"]Solve the following equation in real numbers :

$1^x + 4 ^x + 8^x + 9^x=2^x + 3^x + 7 ^x + 10^x$[/quote]
Very cute problem. :)

There are three real solution that are readily checked, namely $x = 0, \, x = 1, \, x = 2$. Equivalently, one can look at the equation
\[ 1 + t^2 + t^3 + t^\alpha - t - t^\beta - t^\gamma - t^\delta = 0 \]
with $t = 2^x, \, \alpha = \frac{\ln 9}{\ln 2} = 2\beta, \, \beta = \frac{\ln 3}{\ln 2}, \, \gamma = \frac{\ln 7}{\ln 2}, \, \delta = \frac{\ln 10}{\ln 2}$. We are now interested in positive solutions and want to prove that $t = 1, \, t = 2, \, t = 4$ are the only solutions.

Let $f(t) =$ left hand side of this equation, and set $g(t) = t^{1-\beta}f(t)$. The strategy is to show that $g'''(t) < 0$ everywhere. This implies that $g''$ can change its sign at most once, $g'$ can have no more than two zeroes, and $g$ cannot have more than three zeroes.

A calculation shows that
\[ g'''(t) = a_1t^{-\beta-2} + a_2t^{-\beta} + a_3t^{1-\beta} + a_4t^{\beta-2} - a_5t^{-\beta-1} - a_6t^{\gamma-\beta-2} - a_7t^{\delta -\beta-2} \]
with suitable coefficients $a_1, \dots, a_7$ that satisfy
\[ a_1 < 0, \, a_2 < 0, \, a_3 > 0,\, a_4 = -a_1> 0, \, a_5 > 0, \, a_6 > 0,\, a_7 > 0 \, . \]
For $t \le 1$, it turns out that $a_1t^{-\beta-2} + a_4t^{\beta-2} \le 0$ and $a_2t^{-\beta} + a_3t^{1-\beta} - a_5t^{-\beta-1} - a_6t^{\gamma-\beta-2} - a_7t^{\delta -\beta-2} \le 0$, and for $t > 1$, a similar combination of terms also shows that $g'''(t) < 0$. Thus $g$ has at most three zeroes, and we've found them all.

[b][color=blue]Related problem:[/color][/b] Let $F(x) = \sum_{i=1}^N a_i b_i^x$, where $a_i \in \mathbb{R}$ and $b_i > 0$. Show that the equation $F(x) = 0$ has at most $N-1$ real solutions. ;)

another proof goes here:

$x \in A=\{0,1,2\}$ verify the equation. We prove that these are all the solutions.
Suppose $x \notin A$ is a solution.
Denote $f(u) = (u + 1)^x - u^x, u \geq 1$, the equations can be rewritten: $f(3) - f(1) = f(9) - f(7)$,
using Lagrange, $f'(c_1) = f'(c_2), c_1 \in (1,3), c_2 \in (7,9)$ and from here $f''(c_3) = 0, c_3 \in (c_1, c_2), c_3 \in (1,9)$
But, $f''(u) = x(x-1)[(u+1)^{x-2} - u^{x-2}] \neq 0, \forall x \notin A, \forall u \geq 1$. Contradiction :).


inequalities
geometry
circumcircle
trigonometry
inequalities proposed

Let $ a, b, c$ be the sidelengths of a triangle, $ R$ its circumradius and $ S$ its area. If $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 4$ holds, prove that $ 6R^{2}\plus{}S^{2}\ge 3$.

This is the same as $ K^{2}\minus{}3K\plus{}\frac{3}{8}*(abc)^{2}$.

Homogenize and do $ 256*f(a,b,c)\minus{}256*f\left(a,\sqrt{\frac{b^{2}\plus{}c^{2}}{2}},\sqrt{\frac{b^{2}\plus{}c^{2}}{2}}\right)$

to get $ (b^{2}\minus{}c^{2})^{2}*(4b^{4}\plus{}4b^{2}c^{2}\plus{}4b^{4}\minus{}a^{2}\minus{}4a^{2}b^{2}\minus{}4a^{2}c^{2})$.

If $ a\le b\le c$, then we have that line is nonnegative.

Then it suffices to prove that f$ \left(a,\sqrt{\frac{4\minus{}a^{2}}{2}},\sqrt{\frac{4\minus{}a^{2}}{2}}\right)$ where $ \sqrt{\frac{4\minus{}a^{2}}{2}}>a$

but this is equivalent to: $ \frac{a^{4}*(3a^{2}\minus{}4)^{2}}{256}$, so the inequality is true.

An instance of equality is when $ a\equal{}b\equal{}c$.

it can be shown that the inequality is equivalent to

$ 2\sin^2A\sin^2B\sin^2C\geq 3\cos A\cos B\cos C(\sin^2A \plus{} \sin^2B \plus{} \sin^2C)$

it follows that it's enough to prove it for an acute angle triangle.

using that $ \sin^2A \plus{} \sin^2B \plus{} \sin^2C \equal{} 2 \plus{} 2\cos A\cos B\cos C$ this inequality is equivalent to

$ \sum \cos^2B\cos^2C\geq 4\cos^2A\cos^2B\cos^2C \plus{} \cos A\cos B\cos C$.

using the well-known inequality $ \sum \cos^2B\cos^2C\geq \frac {3}{2}\cos A\cos B\cos C$ it reduces to show that

$ \frac {1}{2}\cos A\cos B\cos C\geq 4\cos^2A\cos^2B\cos^2C$

which follows from $ \frac {1}{8}\geq \cos A\cos B\cos C$


quadratics
number theory unsolved
number theory

Let $ p\equiv\minus{}1,11(mod 60)$ be a prime.Prove that there are no positive integers $ x,y$ satisfying $ y^{\frac{p\minus{}1}{2}}\equiv{x^2\plus{}4}(mod p!)$

Just observe that $ y^{\frac{p\minus{}1}{2}} \equiv \minus{}1,0,\minus{}1 \bmod{p}$

Then you have to prove that $ \minus{}3$, $ \minus{}4$, and $ \minus{}5$ are not quadratic resiudes $ mod p$  (It is useful to use quadratic reciprocity law, and tha fact that
$ p\equiv 3 \bmod{4}$
$ p\equiv 2 \bmod{3}$
$ p\equiv \plus{}/\minus{}1 \bmod{5}$)


[quote="Pain rinnegan"]Prove that for all reals a,b,c we have $ a^4 \plus{} b^4 \plus{} c^4 \plus{} a^3b \plus{} b^3c \plus{} c^3a\geq 2ab^3 \plus{} 2bc^3 \plus{} 2ca^3$[/quote]
it is one old problem of Vasile Cirtoaje :
 $ ineq\Leftrightarrow \sum (a^{2} \minus{} ab \plus{} ac \minus{} c^{2})^{2}\geq 0$

I have a solution for $ a,b,c>0$. In this case, the inequality is equivalent with:

$ [a^2(a\minus{}c)^2\plus{}b^2(a\minus{}b)^2\plus{}c^2(b\minus{}c)^2]\plus{}$

$ \plus{}[ab(a\minus{}b)(a\minus{}c)\plus{}ca(c\minus{}b)(c\minus{}a)\plus{}bc(b\minus{}a)(b\minus{}c)] \ge 0$

where the second inequality can be deduced using Schur


quadratics
modular arithmetic
number theory solved
number theory

Prove that : there are unbounded prime numbers in the form : $ 4k+1 $

Just take the divisors of $(n!)^2+1$. They have to bve primes of the form $4k+1$ ($-1$ is not a quadratic residue $\pmod p$ when $p$ is a prime of the form $4k+3$), and it's easy to see that there are infinitely many.


Let $ p > 2007$ is a prime number and $ n$ is an integer exceeding $ 2006p$, Prove that
$ C^{2006p}_{n}\minus{}C^{2006p}_{k}$ is a multiple of p, where $ k \equal{} [\frac{n}{p}]$ is the largest integer which is not larger than $ \frac{n}{p}$

This is double-posted and likely a contest problem; moderator, please lock or delete.


MATHCOUNTS trigonometry algebra polynomial AMC AIME

Yeah...read the title and description

give school, grade, and placings...

it was today

Hyde Middle:

6th-2nd
7th-2nd
8th-2nd

and one individual prize (not mine)

 :lol:

:winner_second: individual

The funniest thing is, 2 people didn't show up for our team, leavinng us with 2 sixth graders and one 8th (me).  Because of me, our team was counted as an 8th grade team.  We still came in third :rotfl: , and would have been higher if I hadn't made my team put down a stupid estimation  (40 tesselations)

[quote="star99"]yeah...the top 3 eigth teams had all the same scores, the estimation made the difference

i dont like the estimation.  nothing to do with math.  they could put a hard problem or something for tiebreaking[/quote]
Yeah, I don't know why they came up with that   silly   rule...

Who got the individual award from Hyde? congrats to archimedes again! Was this MathCounts style?

its not like math counts

individual: 20 probs, no calc, 45 min, and there were some tie breaker questions

team (5-7 people, but some had less because of no-shows):  12 probs, in a 2-3-3-4 (do one sheet, turn it in, get another sheet), 1 hour (tho the high school students who were handing out sheets wasted 10 minutes of peoples time), with calc, no tie breaker! (stupid estimation. cant they come up with anything better?)

and the estimation wasnt explained well.  the guy outside said one thing, and the organizer lady said another thing

argh

[quote="star99"]the high school students who were handing out sheets wasted 10 minutes of peoples time)
argh[/quote]

 :rotfl:   That reminds me of something that happened to the team sitting next to us:
The proctor walked by them about .002 seconds after they finished their problem sheet.  They were standing up and screaming but the proctor didn't hear and he kept walking...  It was about 5 min. before he finally came around again.  I feel sorry for that team.

I really hate the team format.  I made my 6th graders rush on the 2nd sheet because I thought that the problems would obviously get progressively harder and more time-consuming.  But no, the last two sheets were much easier and we had 20 minutes left at the end to check our last page, and we couldn't go back to the second page which we rushed and screwed up on.  :mad:

I got a perfect on the team score, and I made some silly mistakes on the individual that could've gotten me to third.

Next year...

[quote="archimedes1"]

I really hate the team format.  I made my 6th graders rush on the 2nd sheet because I thought that the problems would obviously get progressively harder and more time-consuming.  But no, the last two sheets were much easier and we had 20 minutes left at the end to check our last page, and we couldn't go back to the second page which we rushed and screwed up on.  :mad:[/quote]

I understand.  This was the first math competition for our sixth graders.  We (8th graders) constantly reminded them to keep track of time.  they had 25 min to do the last sheet

[quote="happyme"]

I got a perfect on the team score,

[/quote]

miller kid...good job

[quote="archimedes1"]NO, I'm not done with ranting about Harker Invitational.

The test is like,  "Hey there!  Did you take Trigonometry?  Good for you!  You get a free extra point for solving this Trig textbook problem!
Have you already learned polynomial division?  My, my!  You must be REALLY smart to be so far ahead in math!  Why don't you take another extra point for this trivial problem?  Not many kids are smart enough to be taking Alg.2/Trig in middle school!  You deserve those 2 points more than the other kids!"

I also don't like the fact that they ever so quietly banned all State qualifiers...[/quote]

I completely agree with you, except for the banning states qualifiers...

the contest used to be on the same day as states, so that this would be a contest for the people who didnt make it there.  now they do it before states, so that people who are going to states can use it for experience.

tho i have a slightly biased side because i didnt make it to states, but i tried to come to as fair of a conclusion as possible

Actually, they did ban State Qualifiers.

Actually State people could participate, but they had a big red NE and a cross on their I.D sticker.  "noneligible."  So they couldn't win any prizes, but could use the contest for "practice."

sorry, that is what i meant, i'm just kind of jusitifying them allowing state qualifiers

Aw man I wanna see that test. Whatever, mathcounts doesn't even test your abilities that well. I mean c'mon the first 20 questions on the sprint are basically like computation. The last ten are usually good though. And then comes the target. The only good target round questions are 5, 6, 7, and 8. The first 4 are just bogus (like 2007 chapter). Oh well, time for AIME, SMT, and AMCs prep.

chapter target this year was unbelievably easy for the first four, and the last four were harder.

@sonny: how did you do at states?

[quote="star99"]@sonny: how did you do at states?[/quote]
Don't ask

[quote="star99"]well, then, how did our chapter do at states?[/quote]

To sum it up the third and fourth place in the whole state were from Santa Clara. However none of the top 4 from our chapter made it to nats. This wasn't a great year for anyone. I focused too much on AIME and did really bad on Mathcounts. But in turn the AMC people decided to write an AIME that focuses a lot on casework and all (my worst subject) so even though I took lots of efforts I failed there as well. Oh well next year.......maybe....................next year.......maybe


A group of tourists arrived in a famous landmark. Three tourists are then chosen to be volunteers in a certain activity. If at least one boy and one girl are chosen as volunteers, there are exactly 49 ways to pick a group of 3 volunteers. Compute the number of tourists.

[hide="Solution"]We want a number of tourists $ n$ so that $ \binom{n}{3} \minus{} \binom{b}{3} \minus{} \binom{g}{3} \equal{} 49$ where $ b$ is the number of boys and $ g$ is the number of girls.

Obviously $ \binom{n}{3} > 49$ so the first we check is $ \binom{8}{3} \equal{} 56$. No solutions are found so we try 9. We find that 7 boys and 2 girls works. So $ 9$ is our answer.[/hide]


Fie $a_{1},a_{2},a_{3},a_{4}$ numere intregi nenule. Sa se construiasca o matrice patratica de ordinul 4 care sa aiba pe prima linie elementele $a_{1},a_{2},a_{3},a_{4}$ si al carei determinant sa fie $c.m.m.d.c.(a_{1},a_{2},a_{3},a_{4})$. Generalizare.

(Multumesc perfect ratio!... Am fost neatent)

Enunţul problemei n-are sens, dar cred că ai vrut să propui ceva de [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=determinant&t=90402]genul ăsta[/url].

Nu am stiut ca problema a mai fost postata...

[quote="Daniel Stanica"]Fie $a_{1},a_{2},a_{3},a_{4}$ numere intregi nenule. Sa se construiasca o matrice patratica de ordinul 4 care sa aiba pe prima linie elementele $a_{1},a_{2},a_{3},a_{4}$ si al carei determinant sa fie $c.m.m.d.c.(a_{1},a_{2},a_{3},a_{4})$. Generalizare.

(Multumesc [b]perfect ratio[/b]!... Am fost neatent)[/quote]

Cred ca e perfect_radio!

vai de mine, cum sa pocesti numele lui perfect radio ?  :|  :huh:  :roll:


I didn't really understand how the poster got the answer:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=767730#767730

Could someone explain?

What don't you understand??? Picking special cases and evaluating saves time...

In a slightly different form, you can think of it as this.... assume that a is either positive or negative and go from there.

First assume that a is positive. If so than log(b) is also positive. and let log(b)=c  so you have 10^c=b. Now lets look at the range of c. Technically it can go from just barely greater than zero to infinity. Considering these two cases as a range we obtain b>1 now assume it's n. Then you obtain at the min left side of equation(assuming a is 1) B^2=log(b) or 10^(b^2)=b. Possible? I don't think so. Now try a is negative. Follow the same argument and you will find the answer.


How many different  4 digit numbers be formed with digits 0-7 if none can be repeated, how many are even, how many are less than 10000 and how many are less than 5000?

[quote="Clifford"]How many different  4 digit numbers be formed with digits 0-7 if none can be repeated,[/quote]
$7*7*6*5=49*30=1470$ seven choices for thousands digit, each yielding seven choices for hundreds digit, each yielding six choices for tens digit, each yielding five choices for ones digit
[quote="Clifford"]how many are even,[/quote]
$3*6*6*5+1*7*6*5=(18+7)30=750$ three choices for ones digit that give six choices for thousands digit (2,4,6), and one choice (0) that gives seven choices for thousands digit
[quote="Clifford"]how many are less than 10000[/quote]
All of them, i.e., 1470
[quote="Clifford"]and how many are less than 5000?[/quote]
$4*7*6*5=940$ four choices for thousands digit, etc.

I appologize if my explanations are slightly incoherent.  I was more focused on typing them down than on making sentences.

This was a simple combinatorics exercise.  I answered the last two parts of the question on the assumption that they were referencing the first part rather than the second.

Just a quick note, there should be four choices for the ones digit for even numbers, since any number ending in 0, 2, 4 or 6 is even.

Also 4 * 7 * 6 * 5 = 840 rather than 940, but that was probably just a typo.

There are.  I split it into two subsets, because if zero is used, it leaves a different number of choices for the thousands digit than if 2, 4, or 6 is used. (The thousands digit cannot be zero.)


Prove that for all $a,b,c$ sides of a triangle, the following inequality holds:

\[1 \leq \frac{a^{2}}{b^{2}+bc+c^{2}}+\frac{b^{2}}{a^{2}+ac+c^{2}}+\frac{c^{2}}{b^{2}+ba+a^{2}}< 2\]
This is also the best possible bound.

It should be an easy exercise, (LHS is trivial) but I don't seem to succeed in proving the upper bound.
See http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1439136895&t=48960 for the trivial lower bound of this expression.

I have a very ugly solution...
The idea is as follow :
WLOG suppose $a\leq b \leq c$ so that, $b=a+x$ and $c=a+y$ for some $x, y \geq 0$. After expanding and using the inequalities $a+x \geq y$ and $a+y \geq x$ we can see that the inequality holds and equality case is $a=0$... Ok I did i using mathematica, it is not very interesting... but it is a solution


i came across this interesting qn in pugh's real analysis, i have no clue abt it

[b]An ant walks on the floors,ceiling and walls of a cubical room. What metric is natural for the ant's view of its world? what metric would a spider consider natural? If the ant wants to get from a point P to another pt Q, how could it determine the  shortest path?[/b]

i think the author should clarify what exactly he means by "natural". I'd appreciate if someone could give some reasonable intrepretation of that

A metric can be considered natural if the distance between $A$ and $B$ is proportional to the minimal time it takes the insect to get from $A$ to $B$. However, with this interpretation there is no natural metric for a spider. :huh:


algebra
floor function
algebraic identities
number theory
IMO Shortlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

First of all, notice that one of the numbers is $0$, then we have a solution, no matter what the other number is, so let's assume that they're both non-$0$.

Another thing worth noticing is that the condition is equivalent to $a\{nb\}=b\{na\},\ \forall n\in\mathbb N^*$. Clearly, from this condition we get $a\in\mathbb Q\iff b\in\mathbb Q$.

Let's consider the case when both our numbers are irrational. We have $\frac ba=\frac{\{nb\}}{\{na\}}$. $\{na\}$ is dense in $(0,1)$, so if we choose a sequence $n_k$ s.t. $\{n_ka\}\to 1$, and then we get $\{n_kb\}\to 1$, so $\frac{\{n_kb\}}{\{n_ka\}}\to 1$, meaning that $\frac ba=1\iff a=b$, and we can see that, conversely, all pairs $(a,b)=(a,a)$ satisfy the equation.

Assume now one of the numbers is an integer. We quickly find the other one to be an integer too, so this is another type of solution: $(a,b)\in\mathbb Z^2$.

Finally, the last case: the numbers are rational but not integral. Let $a=\frac{a'}q,\ b=\frac {b'}{q'}$, where $q,q'\in\mathbb N^*,a',b'\in\mathbb Z$, and $(a',q)=(a',q')=1$. By making $n=q$, and then $n=q'$, we find $q=q'$. We can use the relation $\frac ba=\frac{\{nb\}}{\{na\}}$. We can find $n$ s.t. $\{nb\}=\frac{q-1}q$, so $\frac ba\ge 1$. At the same time, we can show in exactly the same way that $\frac ab\ge 1$, so $a=b$.

Combining everything, the solutions are either of the form $(a,a)$ with $a$ being any real number, or $(a,b),\ a,b\in\mathbb Z$, or $(0,b),(a,0)$.

[color=blue]Lemma[/color].$[nx]=[x+\frac{1}{n}]+[x+\frac{2}{n}]+\dots+[x+\frac{n-1}{n}]$

if $a\not =b$from this lemma we have $a[2b]=b[2a]\rightarrow a[b]+a[b+\frac 12]=b[a]+b[a+\frac 12]\rightarrow a[b+\frac 12]=b[a+\frac 12]$from this we can reach to$\{a\},\{b\}<\frac 12$
now try for $n=4,8,\dots,2^k$ we will recieve to $\{a\},\{b\}<\frac{1}{2^k} \ for\ every\ k\in\ \mathbb{N}$

then the solutions are:
$a=b\in \mathbb{R}$
$a,b\in \mathbb {Z}$

Can anyone tell me if this solution is correct?

Note that $a\lfloor bn \rfloor=b\lfloor an \rfloor \Longrightarrow a(bn - \{bn\})=b(an - \{an\}) \Longrightarrow a\{bn\}=b\{an\}$ $(*)$

This means $a,b \in \mathbb Q$ or $a,b \notin \mathbb Q$. Moreover, if one of them is an integer, then also the other one is. Clearly, equality holds in this case.

So we may assume neither $a$ nor $b$ is an integer.

It follows that we may also assume $a=(X,a_1a_2\cdots)_2$ and $b=(Y,b_1b_2\cdots)_2$ where $a_i, b_i \in \{0,1\}$ but not all $0$. If $a, b\in \mathbb Q$, sequences $a_i$ and $b_i$ are finite.

In the equality $(*)$, take $n=2^k \Longrightarrow a\cdot(0,b_kb_{k+1}\cdots)_2=b\cdot(0,a_ka_{k+1}\cdots)_2$. Multiply both sides by $2 \Longrightarrow a(b_k,b_{k+1}\cdots)_2=b(a_k,a_{k+1}\cdots)_2$

Take $n=2^{k+1} \Longrightarrow a\cdot (0,b_{k+1}b_{k+2}\cdots)_2=b\cdot (0,a_{k+1}a_{k+2}\cdots)_2$

Subtracting these two equalities, we get $a\cdot (b_k)_2 = b\cdot (a_k)_2 \Longrightarrow a\cdot b_k = b\cdot a_k$

Since there is an $a_m\neq 0$, taking $k=m$ gives $a=b$ if $b_m=1$ and $a=0$ if $b_m=0$.

Doing the same for $b_n$, we obtain all the solutions are $a=b, a=0, b=0$ or $a, b \in \mathbb Z$.

Storage (essentially Kronecker/Eray's proof)
[quote=1998 N2]Determine all pairs $(a,b)$ of real numbers such that $a \lfloor bn \rfloor =b \lfloor an \rfloor $ for all positive integers $n$. (Note that $\lfloor x\rfloor $ denotes the greatest integer less than or equal to $x$.)[/quote]

Firstly, if one of them is an integer, so is the other, and all such pairs work. If one of them is zero, that's also clear. If the two are equal, then that works.
Plugging $n=q$ will give ${an}=0$, forcing $a$ to be rational. So either both $a, b$ are rational or neither of them is.
If both are rational, then their denominators are equal (plug $n$ to be the smaller one). Say $a=p/q$ and $b=r/q$ with $q>0$ and $p>r$. Plug $n$ such that $rn \equiv -1 \pmod{q}$ to get a contradiction. Thus, if both non-integer rationals, then both are equal.

Otherwise, either one can finish by Kronecker, or do the following:

Let $a=u+.u_1u_2\dots$ and $b=v+.v_1v_2\dots$ in binary, with $u, v$ integers, and $u_i, v_i \in \{0, 1\}$ for all $i$ with not all of them as 0 (or 1). Plugging $n=2^k$ we get that $\frac{.u_k\dots}{.v_k\dots}=\frac{.u_{k+1}\dots}{.v_{k+1}\dots}=\frac{a}{b}$ hence $bu_k=av_k$ for all $k \ge 1$ and $bu=av$. Pick $k$ such that $u_k=1$ hence $b=av_k$ so either $b=0$ or $a=b$. The result follows.

[b]Remark.[/b] It is somewhat annoying that the above argument does not work so nicely with other bases (I spent 15mins doodling base 10 :()

Clearly, $a=b$ is a valid solution. If $a$ were an integer, then $a \lfloor b \rfloor =b \lfloor a \rfloor $ implies $b$ is an integer. Note that $a,b$ both integers is also a valid solution. Also, if $a=0$ or $b=0$ the equation holds. Now, suppose distinct $a,b \neq 0$ are non-integer. Then, \[\frac{b}{a}=\frac{\lfloor b \rfloor}{\lfloor a \rfloor}=\frac{\{b\}}{\{a\}}\] where $\{x\}=x-\lfloor x \rfloor$ is the fractional part of $x.$ Clearly, if $\frac{\lfloor b \rfloor}{\lfloor a \rfloor}=1$ then this reduces to $a=b$ so WLOG, assume $\lfloor a\rfloor >\lfloor b\rfloor.$ Now, we have \[\frac{b}{a}=\frac{\{bn\}}{\{an\}}=\frac{\{b\}}{\{a\}}\] for sufficiently large $n$ this pattern no longer holds, so the only answers are when two are equal out of $(a,b,0)$ or $a,b$ integers.

Suppose that one of $a,b$ is equal to $0$. Clearly this works, so this is a solution.

Now, suppose that $a,b\neq 0$. Then we have $a\{bn\}=b\{an\}$. If $a=b$ we are done, which gives another solution.

Next, note that if $a$ is an integer, then $b$ must also be an integer. This works, and gives a third solution set.

Finally, I claim that no other solutions exist. Note that $a,b$ must have the same sign, so WLOG let $0<a<b$. Then if $a$ is irrational then the RHS can become arbitrarily close to $b$, a contradiction. Thus $a$ is rational, so the RHS is periodic. Then $b$ is also rational. Since both the LHS and RHS have a common period, we can set $a=\frac{x}{d}$ and $b=\frac{y}{d}$. But now the maximum values of both sides are the same, so that $\frac{x}{d}\cdot \frac{d-1}{d}=\frac{y}{d}\cdot \frac{d-1}{d}\implies x=y$, contradiction. We are done. $\blacksquare$

ISL Marabot solve

Clearly, $(a,b) = (0,x)$ or $(x,0)$ works where $x\in \mathbb R$. Also, $a,b \in \mathbb Z$ works.

Now, let's assume that $a,b\notin \mathbb Z$. And we will prove that $a=b$.
Let, $a_i, b_i$ be the $i$th digits of $a,b$ when written in base-$2$. And here, we are considering the representation where there is no infinite tail of $1$s. [If there is $\dots 0111\dots$ then we can replace that with $\dots1000\dots$]
Here, $a \lfloor bn \rfloor =b \lfloor an \rfloor  \implies \dfrac a b = \dfrac{\lfloor an \rfloor}{\lfloor bn \rfloor}$.
Now, $b\notin \mathbb Z$ so, we have $k$ such that $b_k =1$. But, $$\dfrac a b = \dfrac{\lfloor 2^{k-1}a \rfloor}{\lfloor 2^{k-1}b \rfloor} = \dfrac{\lfloor 2^ka \rfloor}{\lfloor 2^kb \rfloor} = \dfrac{2\lfloor 2^{k-1}a \rfloor+a_k}{2\lfloor 2^{k-1}b \rfloor+b_k}$$
$$\implies \dfrac a b = \dfrac{a_k}{b_k}$$
Therefore, $a=b$ as, $a \neq 0$ and we are done.
So, the solution are, $a,b \in \mathbb Z$ or $(a,b) = (x,0)$ or $(0,x)$ or $(x,x)$ where $x \in \mathbb R. \ \ \ \ \blacksquare$

The condition can be rewritten as $$a\{bn\}=b\{an\}.$$

Note that $a$ and $b$ are either both rational or both irrational. \\

Case 1: $a$ and $b$ are both irrational. If $a=b$ then obviously it works. Otherwise WLOG $a>b.$ Then, pick $n$ such that $\{bn\}$ is as close to 1 as needed (possible due to irrationality). Then, $a\{bn\}>b\{an\},$ contradiction.\\

Case 2: $a$ and $b$ are both rational. Then, let $a=x/y$ and $b=z/w$ such that $\gcd(x,y)=\gcd(z,w)=1.$ We then have $$\frac{x}{y}\{\frac{z}{w}n\}=\frac{z}{w}\{\frac{x}{y}n\}.$$ Letting $n=y$, we have that $$\frac{x}{y}\{\frac{z}{w}y\}=0,$$ so either $x=0$ (or just $a=0$) or $w|y.$ Similarly, plugging in $n=w$ gives either $b=0$ or $y|w$. Therefore, unless either of $a$ or $b$ is 0 (giving solutions $(0,b),(a,0)$), $y=w.$\\

If $y=w$, then we have $$\frac{x}{y}\{\frac{z}{y}n\}=\frac{z}{y}\{\frac{x}{y}n\}.$$ Multiplying by $y^2$ gives $$x(zn\pmod y)=z(xn\pmod y).$$ Note that $z$ and $x$ are both relatively prime to $y$ by definition, so if we sum over $0\leq n\leq y-1$, each residue occurs exactly once, so we have $$x(\frac{y(y-1)}{2})=z(\frac{y(y-1)}{2}),$$ which means that either $y=1$ (which means $a$ and $b$ are both integers) or $x=z$ (which means $a=b$ since $y=w$). Therefore, we have either $(a,b)=(a,0),(0,b)$, or $a,b$ are both integers, or $a=b$.\\

Putting this all together, the solution set is:\\

(1) when $a=b$,\\

(2) when $a$ and $b$ are both integers,\\

(3) when at least one of $a$ and $b$ is 0.

The answer is $(a,a)$, $(a,0)$, $(0,a)$, and $(p,q)$, where $a \in \mathbb R$ and $p,q \in \mathbb Z$. Assume that $a \neq b$ and $ab \neq 0$ and $a$ and $b$ aren't both integers. Rewrite the condition as $a \{ bn \} = b \{ an \}$. Taking $n=1$ implies $\text{sgn} a = \text{sgn} b$. Assume $|a|>|b|$. Next note that taking $n=1$ gives $\frac{a}{b} = \frac{\lfloor a \rfloor}{\lfloor b \rfloor}$ is rational.  If $a$ and $b$ are irrational, note that $\{ bn \}$ is dense in $[0,1]$ by the irrationality of $b$. Choose $n$ so that $\{ bn \} \in  \left( \frac{b}{a},1 \right)$. Then $\{an\} = \frac{a\{bn\}}{b} > 1$ and we're done with this case. If $a$ and $b$ are both rational, note that $an \in \mathbb Z \equiv \{an\} = 0 \equiv \{ bn \} = 0 \equiv b\in \mathbb Z$. Thus $a=\frac{x}{z}$ and $b=\frac{y}{z}$ where $|x|>|y|$ and $\gcd(x,z)=\gcd(y,z)=1$. By assumption, $z>1$. Now take $n$ so that $xn \equiv 1 \pmod z$ which exists by the $\gcd$ condition. Then $\{ bn \} = \frac{b \{ an \} } {a} = \frac{b}{az} < \frac{1}{z}$. Thus $\{ bn \} = 0$ and $z \mid n$, an obvious contradiction. This concludes the proof.

bruh this could actually be a real analysis problem

The answer is all pairs of integers, $a=b$, $a=0$, and $b=0$. These obviously work.

Let $p=\lfloor a\rfloor$, $q=\lfloor b\rfloor$, $a=p+r$, and $b=q+s$. Note that if $a$ or $b$ is a nonzero integer, obviously the other is an integer as well, so assume $r,s\ne 0$.

[b]Claim.[/b] $r=s$.
[b]Proof.[/b] Suppose not. Note that
\[(p+r)\lfloor qn+sn\rfloor = (q+s)\lfloor pn+rn\rfloor\rightarrow (rq-sp)n=(q+s)\lfloor rn\rfloor-(p+r)\lfloor sn\rfloor.\]
Let the LHS be $x_n$ and the RHS be $y_n$. $x_n$ is clearly an arithmetic sequence. But consider some $k$, which gives $y_{k+1}-y_k=rq-sp$. This common difference can't be the common difference every time though, since there obviously exists $n_1$ and $n_2$ such that
\[\lfloor r(n_2+1)\rfloor-\lfloor rn_2\rfloor=\lfloor r(n_1+1)\rfloor-\lfloor rn_1\rfloor\]
but
\[\lfloor s(n_2+1)\rfloor-\lfloor sn_2\rfloor\ne \lfloor s(n_1+1)\rfloor-\lfloor sn_1\rfloor\]
since all four of these (LHS or RHS) are binary sequences, and them always changing at the same time would imply that $\lfloor rn\rfloor=\lfloor sn\rfloor$ for all positive integers $n$, so $|r-s|<\varepsilon$ for all $\varepsilon>0$, so $r=s$.

Now $n=1$ gives $(p+r)q=(q+r)p$, so $r=0$ or $p=q$ as desired. $\blacksquare$

Note that trivially all integers work, and so does when $a = b$. Furthermore, when $a = 0$ or $b = 0$, then it also works. So assume $a \neq b$ and $a, b \neq 0$. I will show all other cases fail.

First, if $a$ is irrational and $b$ is not, obviously this fails.

Second, if $a$ and $b$ are irrational and $a/b$ is irrational, this also fails. But if $a/b$ is rational, then $a/b = p/q$ for relatively prime integers $p$ and $q$ and we must have
\[ \frac{p}{q} = \frac{\lfloor prn \rfloor}{\lfloor qrn \rfloor} \]
for some irrational $r$. But then we can choose $n$ such that $p \nmid \lfloor prn \rfloor$, which is bad.

Thus, we just need to resolve the case when both are rationals $p/q$ and $r/s$ where $(p, q) = 1$ and $(r, s) = 1$. Note we can assume that $(q, s) = 1$ because if $\gcd(q, s) \nmid n$ then the floor function makes those cases irrelevant. This means that we only need to look at $n \pmod {qs}$ because after that it'll just repeat. Thus, we must have
\[ \frac pq \lfloor \frac rs n \rfloor = \frac rs \lfloor \frac pq n \rfloor \implies ps \lfloor \frac rs n \rfloor = rq \lfloor \frac pq n \rfloor \implies ps\{\frac{rn}{s}\} = rq\{\frac{pn}{q}\} \]
But from the assumption that $(q, p) = 1$ and $(q, s) = 1$ then $q \mid \{ \frac{rn}{s}\}$. This means that $\frac{rn}{s}$ must always be an integer, so $s \mid r$, contradiction. $\blacksquare$

Nice Problem!
We claim the answer is $(a,b)=(a,0);(0,a);(a,a),(a,b) \quad a,b \in \mathbb{Z}$
Note that the given condition is equivalent to $\dfrac{\{an\}}{\{bn\}}=\frac{a}{b}$
Assume $a,b \neq 0$ also if either one is integer then the other is forced to be integer which works. If $a$ and $b$ are rational, let $a = \frac{c}{d}$, $b = \frac{e}{f}$, then $d=f$ by putting $n = d$, then choose $\{cn\} = \frac{d-1}{d}$ to get $a=b$
If $a,b$ are irrational then, choosing ${an} \rightarrow 1$ makes RHS arbitarily close to $b$, contradiction!.


Let $ a,b \in \mathbb{R}$ so that $ a \cdot b >0.$ Prove that:

\[{ \boxed{a^2+b^2=10ab \Leftrightarrow \left| \frac{a-b}{a+b} \right|=\frac{\sqrt{6}}{3}}}\]

$ (a\minus{}b)^2 \equal{} a^2\plus{}b^2\minus{}2ab\equal{}10ab\minus{}2ab\equal{}8ab$
$ (a\plus{}b)^2 \equal{} a^2\plus{}b^2\plus{}2ab\equal{}10ab\plus{}2ab\equal{}12ab$
So, $ \left|\frac{a\minus{}b}{a\plus{}b}\right| \equal{} \left|\frac{\sqrt{8ab}}{\sqrt{12ab}}\right| \equal{} \frac{\sqrt{2}}{\sqrt{3}} \equal{} \frac{\sqrt{6}}{3}$


[url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2006]Hopefully, the AMC10 he's talking about[/url]

[hide="Click to reveal hidden text"]Expand to get $7^{14}x^{14}=14^7x^7$

Divide by $x^7$ to get $7^{14}x^7=14^7$

Change $14^7$ into $7^72^7$ and divide both sides by $7^7$:

$7^7x^7=2^7$ Change it into the form $(7x)^7=2^7$

Take the 7th root of both of them to get $7x=2$

Solve to get $\boxed {x=\frac{2} {7}}$[/hide]

[hide="Or even faster"]
Take the seventh root immediately, to obtain

$(7x)^2=14x$
$49x^2=14x$
$49x=14$
$x={14\over 49}={2\over 7}$[/hide]

[hide="Or possibly even faster"]Plug in the answer choices into $x$ until you find the right one.  :lol: [/hide]

YES! I got that one!  :P but I just used algebra


Let ABCD be a convex quadrilateral for which the circle
with diameter AB is tangent to the line CD. Show that the circle with
diameter CD is tangent to the line AB if and only if the lines BC and
AD are parallel.

I don't think that the "...if and only if ..." is correct. A counter example is the case $ CD//AB$ but $ CD\not\equal{}AB$.


trigonometry
inequalities
geometry
inradius
circumcircle
symmetry
inequalities unsolved

hello, do you mean
$\frac{3-(cosA+cosB+cosC)}{(3+cosA+cosB+cosC)^2}\ge 2(sinA+sinB+sinC)^2$
Sonnhard.

[quote][color=darkred]Prove that : $\boxed{\ \frac{\left(3+\cos A+\cos B+\cos C\right)^2}{3-\left(\cos A+\cos B+\cos C\right)}\ \ge\ 2\cdot\left(\sin A+\sin B+\sin C\right)^2\ }$ .[/color][/quote]

[color=darkblue][u][b]Proof.[/b][/u] Using the well-known identities in a triangle : $\left\|\ \begin{array}{ccc}
{\color[rgb]{0,0,0.5}\blacktriangleright} & \cos A+\cos B+\cos C=1+\frac rR \\ \\ {\color[rgb]{0,0,0.5}\blacktriangleright} & \sin A+\sin B+\sin C=\frac sR\ \end{array}\right\|$ the inequality becomes :

$\frac{\left(4+\frac rR\right)^2}{2-\frac rR}\ \ge\ 2\cdot\frac {s^2}{R^2}\ \iff\ \left(\frac {4R+r}R\right)^2\cdot\frac R{2R-r}\ \ge\ \frac {2s^2}{R^2}\ \iff\color{white}{..}$ $\boxed{\ s^2\ \le\ \frac {R(4R+r)^2}{2(2R-r)}\ }$ .

The last inequality can be obtained from the remarkable distance $\boxed{\ H\Gamma^2=4R^2\left[1-\frac{2s^2(2R-r)}{R(4R+r)^2}\right]\ }$ , where

$H$ is the [u][b]orthocenter[/b][/u] and $\Gamma$ is [u][b]Gergonne[/b]'s point[/u] of the given triangle .
[/color]

[quote="Sergic Primazon"][quote="kid1412"]ABC be triangles.Prove that:
[tex] \frac\ {(3-(cosA+cosB+cosC))}{(3+cosA+cosB+cosC)^2}\ge 2(sinA+sinB+sinC)^2[/tex].[/quote]

I think you mean : $\frac{(3+cosA+cosB+cosC)^2}{3-(cosA+cosB+cosC)} \ge 2(sinA+sinB+sinC)^2$[/quote]
Here is another proof: (It is obviously true that this proof is valid for the [b]Mateescu Constantin's[/b] inequality.)

We will  use the identities which Mateescu Constantin mentioned in the previous post.

In addition to these identities, we know that

$\frac{abc}{4R}=ur=\sqrt{u(u-a)(u-b)(u-c)}$  where $a,b,c$ are side-lengths of the triangle $ABC \; , \; u=\frac{a+b+c}{2} \; , \; r$ and $R$ are inradius and circumradius of the triangle $ABC$ respectively.

We know also that $a,b,c$ are side-lengths of a triangle if and only if there exist positive real numbers $x,y,z$ such that

$a=y+z \; , \; b=z+x \; , \; c=x+y$

By using all these identities, our inequality is equivalent to (this part is nothing but the substitution, expansion and some nice arrengements)

$x^3(xy+xz-yz)(y-z)^2+y^3(xy+yz-zx)(z-x)^2+z^3(zx+yz-xy)(x-y)^2 \geq 0$

Now by symmetry, we assume that $x \geq y \geq z$

$i) \: zx+yz-xy \geq 0$

In this case $xy+xz-yz \geq xy+yz-zx \geq zx+yz-xy \geq 0$ and we are done.

$ii) \: zx+yz-xy < 0$

In this case $xy+yz-zx \geq 0$ and it is sufficient to show the following inequality.

$x^3(xy+xz-yz)(y-z)^2 \geq z^3(xy-zx-yz)(x-y)^2$

Actually the following one is also sufficient.

$x^3(xz-yz)(y-z)^2 \geq z^3(xy-zx)(x-y)^2$

The last one is equivalent to

$f(z)=(x-y)z^2+x^2z-x^2y \leq 0$

$f''(z)=2(x-y) \geq 0$ and therefore $f(z)$ gets its maximum value at the endpoints, namely

$f(z) \leq \max \{f(0),f(\frac{xy}{x+y})\}$ since $0 < z < \frac{xy}{x+y}$

$f(0)=-x^2y<0$

$f(\frac{xy}{x+y})\}<0 \: \Longleftrightarrow$

$(x-y)\frac{x^2y^2}{(x+y)^2}+\frac{x^3y}{x+y}-x^2y<0 \: \Longleftrightarrow$

$(x-y)x^2y^2+x^3y(x+y)<x^2y(x+y)^2 \: \Longleftrightarrow$

$x-y<x+y$ which is obviously true.


inequalities
linear algebra
matrix
number theory unsolved
number theory

Let a,b,c are three integer numbers different 0 satisfy:$ \left\{\begin{matrix}
\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\in Z & \\
 \frac{a}{c}\plus{}\frac{b}{a}\plus{}\frac{c}{b}\in Z
\end{matrix}\right.$.
Prove that: $ \frac{3a^4}{b^2}\plus{}\frac{2b^4}{c^2}\plus{}\frac{c^4}{a^2}\minus{}4\left | a \right |\minus{}3\left | b \right |\minus{}2\left | c \right |\geq 0$

I don't know if it will help or not, but there is a result that says, if $ a,b,c$ are integers and $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$ is an integer, then $ abc$ is a perfect cube.

[quote="duythuc_lqd"]Let a,b,c are three integer numbers different 0 satisfy:$ \left\{\begin{matrix} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\in Z & \\
\frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{b}\in Z \end{matrix}\right.$.
Prove that: $ \frac {3a^4}{b^2} \plus{} \frac {2b^4}{c^2} \plus{} \frac {c^4}{a^2} \minus{} 4\left | a \right | \minus{} 3\left | b \right | \minus{} 2\left | c \right |\geq 0$[/quote]

That's obviously wrong : just choose as counter-example $ a\equal{}b\equal{}c\equal{}1$

[quote="duythuc_lqd"]Let a,b,c are three integer numbers different 0 satisfy:$ \left\{\begin{matrix} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\in Z & \\
\frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{b}\in Z \end{matrix}\right.$.
Prove that: $ \frac {3a^4}{b^2} \plus{} \frac {2b^4}{c^2} \plus{} \frac {c^4}{a^2} \minus{} 4\left | a \right | \minus{} 3\left | b \right | \minus{} 2\left | c \right |\geq 0$[/quote]
I remember that I solved a similar one before. We can prove that $ |a|\equal{}|b|\equal{}|c|$ and this problem seems quite tricky.


I don't know how to do this so any help will be appreciated.

  Can someone orthogonally diagonalize the matrix $ A\equal{}\begin{bmatrix}3&\minus{}2&4\\\minus{}2&6&2\\4&2&3\end{bmatrix}$  giving an orthogonal matrix $ P$ and a diagonal matrix $ D$.

We know that the eigenvalues are $ 7, \minus{} 2$. Thank you.

let $ \lambda_1, \lambda_2, \lambda_3$ be the three eigenvalues of $ A$, and let $ \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ be corresponding eigenvectors which are orthogonal and have norm 1. (this is always possible when $ A$ is symmetric.) then
\[ A \equal{} \begin{pmatrix}\mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3\end{pmatrix}\begin{pmatrix}\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3\end{pmatrix}\begin{pmatrix}\mathbf{e}_1^{\prime} \\
\mathbf{e}_2^{\prime} \\
\mathbf{e}_3^{\prime}\end{pmatrix}\]
gives the desired diagonalization. (in the first matrix, the eigenvectors are the columns; in the last, they're rows.) this works for any real symmetric matrix and is called the spectral decomposition.

You (new_member) forgot to mention that 7 is an eigenvalue with multiplicity 2. For the eigenspace corresponding to $ \lambda \equal{} 7$ we get $ N(A\minus{}7I) \equal{} \langle(1,\minus{}2,0), (0,2,1) \rangle$. Orthogonalize using Gram-Schmidt and then normalize, to get $ N(A\minus{}7I) \equal{} \langle(1/\sqrt{5},\minus{}2/\sqrt{5},0), (4/(3 \sqrt{5}),2/(3 \sqrt{5}),\sqrt{5}/3) \rangle$. For the eigenspace corresponding to $ \lambda \equal{} \minus{}2$ we get $ N(A\plus{}2I) \equal{} \langle (2,1,\minus{}2) \rangle$, which normalizing gives $ N(A\plus{}2I) \equal{} \langle (2/3,1/3,\minus{}2/3) \rangle$. Your matrices $ P$ and $ D$ are therefore

$ P \equal{} \begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{4}{3 \sqrt{5}} & \frac23 \\ \minus{} \frac{2}{\sqrt{5}} & \frac{2}{3 \sqrt{5}} & \frac13 \\ 0 & \frac{\sqrt{5}}{3} & \minus{} \frac23 \end{bmatrix}$,

$ D \equal{} \begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & \minus{}2 \end{bmatrix}$,

provided my calculations are correct.


[hide]So basically you have to look at it from a corner...

But since you can't count each unit-cube-thing more than once, you can't do $3*(11^2)$

so to find the number of cubes seen, do...$(11*11)+(11*10)+(10*10)$ which equals...$331$ (D)[/hide]

[hide]I am not sure of this but let me try my solution.
You can view the maximum three sides of the cube.

The surface are of the cube is (11^2)(3 sides) = 363.  But you have the three (11) corners.

There is 1 cube (the centermost with three faces shown) that we counted three times. So we should [b]subtract 2[/b] from our total count.

There are 30 cubes (with two faces shown) weve counted twice so we subtract 30.

So, 363-2-30 = 331?[/hide]

[quote="4everwise"]An $11x11x11$ wooden cube is formed by gluing together $11^3$ unit cubes. What is the greatest number of unit cubes that can be seen from a single point?

$\text{(A)}\ 328 \qquad \text{(B)}\ 329 \qquad \text{(C)}\ 330 \qquad \text{(D)}\ 331 \qquad \text{(E)}\ 332$[/quote]
[hide]
This reminds me of a recent AIME problem (last 2-4 years).
From one point, you can see 3 faces (at max).
From one point, you can see one face completely. This contributes $11^2$ cubes.
The next face has $11^2$ cubes, but you have already counted one row of cubes so that has to be subtracted out to get $11^2-11$. The last face has 2 rows counted, but 1 cube is counted twice so we have $11^2-11-11+1$
Summing, we get: 331, or [b]D[/b].
[/hide]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=73670]Click to reveal hidden content[/url]

[quote="IntrepidMath"][quote="4everwise"]An $ 11x11x11$ wooden cube is formed by gluing together $ 11^3$ unit cubes. What is the greatest number of unit cubes that can be seen from a single point?

$ \text{(A)}\ 328 \qquad \text{(B)}\ 329 \qquad \text{(C)}\ 330 \qquad \text{(D)}\ 331 \qquad \text{(E)}\ 332$[/quote]
[hide]
This reminds me of a recent AIME problem (last 2-4 years).
From one point, you can see 3 faces (at max).
From one point, you can see one face completely. This contributes $ 11^2$ cubes.
The next face has $ 11^2$ cubes, but you have already counted one row of cubes so that has to be subtracted out to get $ 11^2 \minus{} 11$. The last face has 2 rows counted, but 1 cube is counted twice so we have $ 11^2 \minus{} 11 \minus{} 11 \plus{} 1$
Summing, we get: 331, or [b]D[/b].
[/hide][/quote]
Which AIME problem are you talking about?

[quote="TZF"]I once asked my math team to [i]prove[/i] that the most you can see is 3 faces worth. I got some really interesting results, with some creative ideas, but it was generally unsuccessful. If anyone here has the time to prove something really trivial, I'd be very interested in seeing what they come up with. Sometimes things that appear obvious in mathematics can be tricky to prove![/quote]

It's impossible to see 2 parallel sides. Suppose we can see 4 cube's sides, but there exist a pair of sides which are parallel - contradiction. It wasn't so hard for me  :)

That's not a proof. Why can't you see two parallel sides? They do not entirely block eachother unless you are looking directly at one face (which is anyway suboptimal). It's not like you have to see the whole side, you only need to be able to see part of at least one more cube.

Technically, if the cube is small enough, you can see 2 parallel sides because you see the left face with your left eye and the right face with your right eye.
Specific example where this works:
unit=1/11 of an inch
look directly at an edge, having it paralell to the ground. You will see 4 faces.

There was an AIME problem that said that there was a rectangular prism, and there are something like $ 231$ cubes that cannot be seen that make up the prism.  What's the largest possible volume of this prism.

Or at least something of that sort.

[hide="An attempt"]Well, extend a face of the cube $ F$ into a plane $ A$. This plane divides three dimensional space into the side with the rest of the cube (the "cube side") and the side without the cube (the "empty side"). Represent your eyes as a point $ P$ outside of the cube such that a line may be drawn from $ P$ to $ F$ without intersecting any other faces of the cube. If $ P$ is on the empty side of $ A$, a line may always be drawn from $ P$ to any point on $ A$ without intersecting any part of the cube (since no part of the cube is in the empty side, there is nothing to intersect...).

If $ P$ is on the cube side of $ A$, we have defined it to be outside of the cube. This situation is equivalent to putting a cube on the floor and attempting to view the bottom face. Obviously, it's impossible, so by definition $ P$ cannot be on the cube side of $ A$.

By the same logic, in order to view the opposite face of the cube, you must be on the empty side of it. However, since the empty sides of the two faces do not overlap, there is no point from which you can view opposite sides of a cube.[/hide]


Let $ ABC$ is a triangle, let $ M$ is a point lie on side $ AB$ and let $ N$ is a point lie on side $ AC$ so that $ BM \equal{} CN$. Let $ D,E$ are midpoints of $ BN$ and $ CM$ respectively. Let $ DE$ meet $ AB, AC$ at $ H, K$ respectively. Prove that $ AHK$ is an isosceles triangle.

[quote="livetolove212"]Denote I, J the midpoints of MN, BC. We know that IJ// A-bisector of triangle ABC. So AHK is isosceles iff $ IJ\perp DE$. (*)
But JDIE is lozenge therefore (*) is true. We are done.[/quote]
What is lozenge?


Please mods try to close this topic. This is a recent contest in the Math Magazine in Vietnam. Also I hope  Arqady and April can hide your solutions for a while. Thank you.

[quote="friendlist"]Let $a;b;c$ be non-negative numbers satisfying $a+b+c=3$.Prove that:
$\frac{a^{2}}{a+b^{2}}+\frac{b^{2}}{b+c^{2}}+\frac{c^{2}}{c+a^{2}}\geq \frac{3}{2}$[/quote]

Let me do!!

Use AM_GM :

$\sum{\frac{a^{2}}{a+b^{2}}}=\sum{a}-\sum{\frac{ab^{2}}{a+b^{2}}}\geq\sum{a}-\sum{\frac{ab^{2}}{2b\sqrt{a}}}=\sum{a}-\sum{\frac{b\sqrt{a}}{2}}$

And we have:

$\sum{\frac{b\sqrt{a}}{2}}=\sum{\frac{\sqrt{b.ab}}{2}\leq\sum{\frac{b+ab}{4}}=\frac{a+b+c}{4}+\frac{ab+bc+ca}{4}\leq\frac{3}{4}+\frac{3}{4}=\frac{3}{2}}$

Cause :

$ab+bc+ca\leq\frac{(a+b+c)^{2}}{3}=3$

So we've done

ML is missing a "n00b" smilie...

  darij

[quote="April"]Hmm, I was deleted my solution, but someone think that my post is a spam post.  :)[/quote] :rotfl:
who is (refering to someone)


calculus
limit
integration
Putnam
real analysis
real analysis unsolved

Since there are marathons in most of the other sections of the site, I thought it would be fun to start a calculus marathon.

The rules are simple, if you solve a problem, post a new one, and the marathon continues.

[size=167]Problem 1[/size]

Evaluate:  $\lim_{k\to\infty} e^{-k}\int_0^k\int_0^k \frac{e^x-e^y}{x-y} \ dxdy$

Source: 1981 Putnam A3

If that's the level of question you want to ask, then let's move it here.


[quote="moldovan"]If $ a,b,c > 0$, $ abc \equal{} 1$, and $ k \in \mathbb{N}$ then:

$ (3)$ $ a^{3k} \plus{} b^{3k} \plus{} c^{3k} \ge 3$[/quote]
By Cauchy's ineq
 $ a^{3k} \plus{} 3k \minus{} 1 \geq 3k\sqrt [3k]{a^{3k}} \equal{} 3ka$
Silimiar, we have:$ b^{3k} \plus{} 3k \minus{} 1 \geq 3kb$ and $ c^{3k} \plus{} 3k \minus{} 1 \geq 3kc$
Hence, $ LHS \geq3k(a \plus{} b \plus{} c) \minus{} 9k \plus{} 3 \geq 9k \minus{} 9k \plus{} 3 \equal{} 3$

 By ineq Chebychev's, we have
 $ 3(a^{3k \plus{} 2} \plus{} b^{3k \plus{} 2} \plus{} c^{3k \plus{} 2} \ge (a^2 \plus{} b^2 \plus{} c^2)(a^{3k} \plus{} b^{3k} \plus{} c^{3k}) \ge 3(a^2\plus{}b^2\plus{}c^2)$

The ineq is proved silimiar


algebra
polynomial
function
calculus
derivative
search
real analysis

So I was reading a paper, and I came across the phrase "Taylor expansion for multivariate polynomials". However, I can't seem to find reference that lists the formula for the Taylor expansion. Anyone here know about one?

It's pretty annoying but managable for the first few orders.  It's contained in my lecture notes, attached below.

One thing I do in a Fourier Analysis class I teach is introduce multi-index notation.

Assume that we are talking about functions on (an open subset of) $ \mathbb{R}^n$, and that $ n$ is understood and fixed.

An $ n$-tuple of integers is a multi-index. Write $ \alpha \equal{} (\alpha_1,\alpha_2,\dots,\alpha_n).$

If $ x\equal{}(x_1,x_2,\dots,x_n)\in\mathbb{R}^n,$ define $ x^{\alpha}\equal{}x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}.$

That monomial expression would make sense even if some of the $ \alpha_i$ were negative. For derivatives, we usually don't want that.

Write $ \alpha\ge 0$ as a shorthand for $ \alpha\ge (0,0,\cdots,0)$ to mean that $ \alpha_i\ge 0$ for $ 1\le i\le n.$ More generally, write $ \alpha\ge\beta$ iff $ \alpha_i\ge\beta_i$ for $ 1\le i\le n.$ (That's a partial order, not a total order.)

The degree of a multi-index is $ |\alpha|\equal{}\alpha_1\plus{}\alpha_2\plus{}\cdots\alpha_n.$

For $ \alpha\ge 0,$ define the partial differential operator $ D^{\alpha}$ by
\[ D^{\alpha}f\equal{}\frac{\partial^{|\alpha|}f}{dx_1^{\alpha_1} dx_2^{\alpha_2}\cdots dx_n^{\alpha_n}}.\]
For $ \alpha\ge0,$ define $ \alpha!\equal{}\alpha_1!\alpha_2!\cdots\alpha_n!.$ Note that $ \frac{|\alpha|!}{\alpha!}$ is then a multinomial coefficient.

With all that notation set up, we now write a $ k$th degree Taylor polynomial of a sufficiently smooth function as
\[ f(x)\equal{}\sum_{\alpha\ge0,\,|\alpha|\le k}\,\frac1{\alpha!}\,D^{\alpha}f(a) (x\minus{}a)^{\alpha}\plus{}o(\|x\minus{}a\|^k).\]
=========

This is exactly what JRav wrote in (1) and (2) in his handout posted above, although you may have to do a little work to convince yourself of that. In particular, he has the coefficient $ \frac{1}{|\alpha|!}$ while my notation has the coefficient $ \frac{1}{\alpha!},$ which is not the same thing. But in JRav's notation, he has included, for instance, both $ \frac{\partial^2f}{\partial x_1\partial x_2}$ and $ \frac{\partial^2}{\partial x_2\partial x_1}$ as separate terms, while I have combined them into the same term. The count of how many terms I have so combined is precisely the multinomial coefficient $ \frac{|\alpha|!}{\alpha!},$ which is precisely what we need to make the two forms the same.

This is all just a different notation; the notion of Taylor series remains the same.

I would think that is you do a search on the term "Hessian" on this site, you'll turn up some topics dealing with classifying critical points for smooth multivariable functions.

OK, so the above equations are valid for a multivariate function $ f$ that maps points from $ \mathbb{R}^3\rightarrow\mathbb{R}$, what if we had to use the Taylor Expansion for a function $ f$ that maps points from $ \mathbb{P}^3\rightarrow\mathbb{P}$?


search
number theory
inequalities unsolved
inequalities

Problem 6:
If $ x, y, z$ are positive real numbers, prove that
$ (x\plus{}y\plus{}z)^{2}(yz\plus{}zx\plus{}xy)^{2} \leq 3(y^{2}\plus{}yz\plus{}z^{2})(z^{2}\plus{}zx\plus{}x^{2})(x^{2}\plus{}xy\plus{}y^{2})$.

[quote="makar"]Problem 6:
If $ x, y, z$ are positive real numbers, prove that
$ (x \plus{} y \plus{} z)^{2}(yz \plus{} zx \plus{} xy)^{2} \leq 3(y^{2} \plus{} yz \plus{} z^{2})(z^{2} \plus{} zx \plus{} x^{2})(x^{2} \plus{} xy \plus{} y^{2})$.[/quote]
$ 3(y^{2} \plus{} yz \plus{} z^{2})(z^{2} \plus{} zx \plus{} x^{2})(x^{2} \plus{} xy \plus{} y^{2}) \geq \frac {3^4}{4^3}(x \plus{} y)^2(y \plus{} z)^2(z \plus{} x)^2 \geq (x \plus{} y \plus{} z)^{2}(yz \plus{} zx \plus{} xy)^{2}$

@mods : move the topic from the number theory unsolved problems forum.


AoPSers,

To round out our offerings of CDs with all of the AMC contests, we are in the process of compiling the questions and solutions for all of the AIMEs and USAMOs.  We have discovered a few years' USAMOs for which we do not have full solutions.  We would like your help in writing those solutions.  We are posting the questions on a [url=http://www.unl.edu/amc/a-activities/a7-problems/USAMOchallenge.shtml]special page[/url] on our website.

Solutions should be submitted as follows:
1) Discussion of solutions should occur on AoPS to develop full and complete solutions.  This is to reduce the number of submissions to our office and ideally get the best solution possible.  There may be multiple approaches to solving the problem, in which case we would like to include those as well.
2) Once consensus on a solution has been gained, designate one individual who worked on the problem to type up the solution (preferably in LaTeX).
3) At the end include a sentence stating all who worked on the solution.
4) Submit the solution to amcinfo@maa.org with the subject line: [Year] USAMO Problem #n Solution.  If the email does not have that subject, it will likely be lost to spam filters.

We will include the credit line in the solutions on the CD which will be produced as soon as all solutions are submitted, so your work will be recognized.

Thanks and happy solving.

Elizabeth Claassen
AMC Senior Accounting Clerk

Great!  I'm going to post some of the problems now.

[quote="nsato"]I have to ask the obvious question: Why not draw on the solutions in [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=3010]USA Mathematical Olympiads 1972-1986[/url]?[/quote]

Because we were misinformed about the status of such a book.  Thanks for pointing it out, since we did miss it.   :blush:

Sorry to revive, but when is this CD coming out?

Again, apologies for reviving an old thread, but it is the appropriate place to answer:

The CD with all AIMEs (1983-2008) and all USAMOs (1972-2008) is now available.
(Interestingly enough, we had to do quite a bit to track down and install all the USAMOs and solutions. that's why it took longer than expected.)

Same ordering location as for all AMC materials.

Steve Dunbar
MAA Director, American Mathematics Competitions


inequalities
vector
calculus
integration
limit
inequalities proposed

Isn't this the second Carlson inequality? (Arthur Engel, [i]Problem Solving Strategies[/i], New York 1998, page 176.)

  Darij

I do not know whether it is the first or the second inequality, I just know that it is absolutely remarkable.

eh - I have absolutely no idea what to do with this $\pi$ ? I'll be greatful for any hint ...  :maybe:  It must have some obvious meaning which I cannot see now, haven't it  ?

I bet it comes from $\frac{\pi^2}6=\frac 1{1^2}+\frac 1{2^2}+\frac 1{3^2}+\ldots$ :).

I can't agree with you, grobber! Give us some another reasons!
For example, $\pi$ comes to
\[\pi\sum_{1\leq i\leq n}a_i^2\geq\sum_{1\leq i,j\leq n}\frac{a_ia_j}{i+j}\]
from some non-evidence reasons.

It was just an innocent guess (even if I used the words "I bet" :)). I guess it's pointless to argue abut this as long as we don't have a solution..

well I think grobber's guess could be true. Using his assumption I've ended with:

$6(\sum \frac{1}{i^2})(\sum i{x_i}^2)^2 \geq (\sum x_i)^4 $

which looks much better

@Megus!
You used Cauchy-Bunyakowski for non-collinear vectors, so you lost precision in inequality, moreover your new inequality is false. Just take $x_i=1/i$.

[quote="harazi"]Looks much better, but it is nothing in fact! Well, it seems that this problem is doomed to long discussion and no solution...[/quote]

sorry harazi - I'll try to find solution and not just speak unnecessary  :maybe:

Myth: yeah, I forgot about those "reals" in problem  :blush:

For darij: yes it is...but its proof is hard!

In fact, there is a 3 lines solution using Cauchy, but that solution is so damn hard to find!

Somehow  Pi/2>Sum t/(t^2+k^2) for every t.

I dont know if this can be done by integration or how? Harazi, please post your cauchy Solution!

[quote="pascual2004"]Somehow  Pi/2>Sum t/(t^2+k^2) for every t.

I dont know if this can be done by integration or how?[/quote]

\[
\lim_{t \to \infty} \sum_{k=1}^{t} \frac{t}{t^2+k^2}
= \lim_{t \to \infty} \frac{1}{t} \sum_{k=1}^{t} \frac{1}{\left(\frac{k}{t}\right)^2+1}
= \int_{0}^{1} \frac{dx}{x^2 + 1}
= \arctan x |_{0}^{1}
=\frac{\pi}{2}
\]

harazi: can you show your Cauchy solution ? I've ended with nothing  :(

Here is the outstanding solution using Cauchy-Schwarz. It is useless to say it does not belong to me. Take $ S=\sum_{n\geq 1} {a_n^2}, T=\sum_{n\geq 1}{n^2 a_n^2}$. Take also $y=\frac{T}{S}$. Using Cauchy we have $ (\sum_{n\geq 1}{a_n})^2\leq\sum_{n\geq 1}{a_n^2(T+n^2S)}\cdot\sum_{n\geq 1}{\frac{1}{T+n^2S}=2TS\cdot \sum_{n\geq 1}{\frac{1}{T+n^2S}}}$. Now, a majorization of the sum with an impropoer integral finishes the proof. Impossible, isn't it?

Before Harazi post this solution, I found a similar solution.
Set $a=\frac{x_1^2+...+n^2x_n^2}{ x_1^2+...+x_n^2}$,   $1\leq a \leq n$.
The inequality becomes
$\pi^2[a(x_1^2+...+x_n^2)+( x_1^2+...+n^2x_n^2)]^2\geq4a(x_1+...+x_n)^4$,
$\pi[(a+1)x_1^2+...+(a+n^2)x_n^2)] \geq 2\sqrt{a}(x_1+...+x_n)^2$.
Applying Cauchy-Schwarz, it reduces to
$\pi\geq 2\sqrt{a}(\frac{1}{a+1}+...+\frac{1}{a+n^2})$.


Has anyone tried the book Geometry Revisited?  It looks interesting.  I have a copy of a book similar to it (I think), but its in Chinese  :? .

Yep.  One of my favorites.  What do you want to know about it?

Well, I just want to know what kind of stuff it covers.  Does it talk about ways to solve problems, such as what ways a midpoint can be used to solve a problem, or does it just give you problems and explanations, or a little of both or neither?

Whoops.  I never got back to you about this.  Sorry.

So let's see...

It presents lots of cool theorems (with awesome proofs), but it doesn't really have a huge emphasis on problem solving.  Of course, that doesn't mean you can't take what you learn and use it to solve problems, because you certainly can.  There are exercises at the end of each section, and there are hints and answers, but most of the problems don't have full solutions in the book.  Still, it's one of my favorites.


Let $(O)$ be a circle and $A,B,C$ be points lying inside the circle. Let $C_1, C_2$ be the intersections of $AB$ and $(O)$. Let $f(C) = (ABC_1C_2)+(ABC_2C_1)$ (cross ratio). Define $f(A), f(B)$ similarly. Prove that there is a unique positive real number $k$ such that $f(A) = f(B) = f(C)= k$.
   Can anyone find $k$ ?

Probably I don't understand something, but doesn't f(C) depend only on the circle (O) and on the points A and B, but not on the point C ? If this is right, then we could move the point C around, without changing f(C), but certainly f(A) and f(B) would change... So is the problem correct??

  Darij

I am really sorry. In fact, the previous problem is wrong. Let me write this problem again. (forget the old one  :oops: )
  Denote by $f(XY)$ the sum $(XYZ_1Z_2) + (XYZ_2Z_1)$ (cross ratio), where $X, Y$ are points inside $(O)$ and $Z_1, Z_2$ are two intersections of $XY$ and the circle.
  Let $B$ and $C$ be points inside the circle $(O)$. Prove that there are exactly two distinct points $A$ inside the circle $(O)$ which lie on different sides of $BC$ such that $f(AB) = f(AC) = f(BC)$.

It seems that this problem is not interested  :ninja:


[quote="GameBot"]Ohio entered the Union on March 1, 1803, which can be written as $ 3/1/03$. The date $ 3/1/03$ is a $ ^{\prime\prime}$friendly date$ ^{\prime\prime}$ because the product of the month and day is equal to value of the last two digits of the year. How many friendly dates occur during the year 2009?[/quote]

The last two digits are 09. Thus,

1/9/09
9/1/09
3/3/09.

So 3.

[hide="my carefully planned answer"]for this problem i will answer this in a series of steps.

1.a friendly date must be equal to 09

2.so for this all we need to do is just find the multiples of 09 (1,3,3,9)

3.so with this we get the dates

1/9/09
9/1/09
3/3/09.

4.[hide="answer"]3[/hide]

[/hide]


Players A and B alternate tossing a biased coin, with A going first. A wins the coin if A tosses a tail before B tosses a head; otherwise B wins. If the probability of a head is $p$, and it makes the game fair to both players, find $p^2+p$

The probability that A wins is the probability he wins on the first try plus the probability that A and B lose and then he wins plus the probability that A and B lose twice and then he wins, etc.

The probability that B wins is formulated similarly.

Let $A = \text{probability that A wins}$ and $B = \text{probability that B wins}$

$A = (1-p) + p(1-p)^{2} + ...+p^{k} (1-p)^{k+1}+...$
$B = p^{2} + p^{3}(1-p) +...+p^{k+2} (1-p)^{k}+...$

$A = \frac{1-p}{1-p(1-p)}$
$B = \frac{p^2}{1-p(1-p)}$

$A = B \rightarrow 1-p = p^2 \rightarrow p^2+p = 1$


Let $n$ is arbitrary positive integer. Prove that:

$\sqrt{2}\sqrt[4]{4}\sqrt[8]{8}\cdots\sqrt[2^n]{2^n}<n+1$

(I have a solution)

Are you sure the problem is right?

$\sqrt[2^n]{2^n} = 2^{ \frac{n}{2^n} }$

Thus the LHS can be written $2^{ \sum_{k=1} \frac{k}{2^k} } < 4$.  Simple casework for $n = 1, 2$.


When am I going to hear about which division I should belong?

I only got an email from Rob that says what problems I got right.

The new divisions should be announced sometime tomorrow, when appeals are finished and results become official.
By the way, what did you think of the problems? :)

Well, this was my first time taking a USACO contest, so I was foreign to some features. For example, I used the back button on my explorer several times. This resulted in the contest page showing me faulty time. When I only had 20 minutes, it said that I had 1 hour and 20 minutes... :(

I also did not know that you could submit more than 3 problems. I also did not know that the problems were not ordered by difficulty. I got number one and two easily, but got stuck for almost two hours on three (watering hole). This was when I realized I only had 20 minutes... I rushed to code pwalk, but did not complete it on time.

I am still proud of my quad (fences) solution. Instead of finding all good ways to cut the board, I subtracted the number of wrong ways to cut the board from total possible ways (board-1 choose 3). This gave me an O(n) solution that only looped 1250 times at most. My 250 point problem ran faster than my 70 point problems  :P

Overall, I thought that the contest was fun and interesting. I am looking forward to further contests this year.

Since this was your first contest, I think you shouldn't worry too much about your performance this month - you'll definitely get better at it as you participate more. By the way, if analyses are posted [url=http://ace.delos.com/OCT08]here[/url] if you haven't already seen them.
Another thing to note is that by using a bit of combinatorics, your idea on 'quad' can even be made into a constant time solution: :D
[code]int main ()
{
    long long N;
    fscanf (in, "%lld", &N);
    fprintf (out, "%lld\n", (N - 1) * (N - 2) * (N - 3) / 6 - 2 * (N / 2) * (N / 2 - 1) * (N / 2 - 2) / 3);
    return 0;
}[/code]
Anyway, it's good to hear that you're looking forward to upcoming contests, as they're certainly a good opportunity.


Can someone name a few famous mathematical problems/puzzles that aren't too hard to explain to a class?

How to find all the pythagorean triples. :)

[quote="aidan"]A set $\{a, b, c\}$ of three integers satisfying the equation $a^2 + b^2 = c^2$ is calle a [i]Pythagorean triple[/i]. The Pythagorean triple is [i]primitive[/i] if $a, b, c$ are positive and the greatest common divisor of $a$, $b$ and $c$ is $1$. If $\{a, b, c\}$ is a Pythagorean triple then so is $\{ka, kb, kc\}$ for any integer $k$. All primitive Pythagorean triples $\{a, b, c\}$ are given parametrically by \[ a = 2mn, b = m^2 - n^2, c = m^2 + n^2  \] where the integers $m$ and $n$ are coprime, one even, the other odd, and with $m > n$.

P.S. I don't know how to prove this though...  :blush:[/quote]

Just draw a right angle triangle ABC with $c$ as hypotenus and you should be able to get it .