SFFT
quadratics
conics
hyperbola
algebra
system of equations
number theory

1.xy=2
2.3x+4y=11
using SFFT,you'll get(x,y)=(1,2)
but  solving as a simultaneous equation you'll get 2 solutions each for x and y.so, is SFFT lacking something[quote]i want to be a genius[i]plz help me[/i][/quote]

[hide="Solution"]
$ xy\equal{}2$
$ 3x\plus{}4y\equal{}11$

Isolate $ y$:   $ y\equal{}\frac{2}{x}$

Substitute $ \frac{2}{x}$ for $ y$:   $ 3x\plus{}4(\frac{2}{x})\equal{}11\rightarrow 3x\plus{}\frac{8}{x}\equal{}11$

Multiply through by $ x$:   $ 3x^2\plus{}8\equal{}11x$

Rearrange:   $ 3x^2\minus{}11x\plus{}8\equal{}0$

Solve the quadratic:   $ x\equal{}\frac{8}{3}, 1$

Isolate $ x$:   $ x\equal{}\frac{2}{y}$

Substitute $ \frac{2}{y}$ for $ x$:   $ 3(\frac{2}{y})\plus{}4y\equal{}11\rightarrow \frac{6}{y}\plus{}4y\equal{}11$

Multiply through by $ y$:   $ 6\plus{}4y^2\equal{}11$

Rearrange:   $ 4y^2\minus{}11y\plus{}6\equal{}0$

Solve the quadratic:   $ y\equal{}2,\frac{3}{4}$

So the solutions are $ (1,2)$ and $ (\frac{8}{3},\frac{3}{4}$

[/hide]

I'm not sure what SFFT is, but it must not work for non-linear systems. As we can see, this system is non-linear because the equation $ xy\equal{}2$ is not linear.

SFFT = [url=http://www.artofproblemsolving.com/Wiki/index.php/SFFT]Simon's Favorite Factoring Trick[/url]

Adding the equations and using SFFT yields:
$ (x \plus{} 4)(y \plus{} 3) \equal{} 25$

However, adding the two equations and factoring this way would suggest that for every x value, there is a y value which satisfies the equation. I'm really not sure what to make of that. Not to mention that this equation suggests lots of other solutions which certainly don't work. Indeed, SFFT is lacking in something. It's interesting to notice that when we add the equations:

$ 3x \plus{} xy \plus{} 4x \equal{} 13$

Then any (x,y) determined from factoring (as shown above) works for that new equation. However, it does not work in either of the individual equations.

This is my theory.
The reason why the new equation (the equation formed after adding the two given equations and applying SFFT) has many solutions, only two of which work for the original system of equations, is because our steps aren't reversible.
Just like in proof-writing, you can't write a backwards proof and assume it is correct; you must write it forwards to make it rigorous.

Let's take a look at the new equation formed by adding the two given equations and applying SFFT:
$ (x + 4)(y + 3) = 25$

Now, let's expand it:
\begin{align}(x + 4)(y + 3) = 25\Rightarrow xy + 3x + 4y + 12 = 25 \Rightarrow (xy) + (3x + 4y) = (2) + (11) \\
xy = 2 \\
3x + 4y = 11\end{align}

You might be wondering why I put parentheses around the expressions in $ (1)$.  This is because I wanted to emphasize the fact that we can add $ (2)$ and $ (3)$ to yield $ (1)$.  Now we see why our problem occurs:
We can add $ (2)$ and $ (3)$ to yield $ (1)$, but we notice that we can NOT break apart $ (1)$ to yield $ (2)$ and $ (3)$, in the same way that we can't deduce that $ 1 = 2$ and $ 4 = 3$ just because $ 1 + 4 = 2 + 3$.
In other words, any solutions $ (x,y)$ to equation $ (1)$ are not necessarily solutions to equations $ (2)$ and $ (3)$ because we are not allowed to break up equation $ (1)$ into $ (2)$ and $ (3)$.

Notice that if we use the other method, which is substitution, then our steps are reversible and no extraneous solutions are introduced:
$ xy = 2\Rightarrow y = \frac {2}{x}$

From that first step which was performed by rearranging the first given equation, we can substitute for $ y$ in the second given equation.
This will give:
$ 3x + 4(\frac {2}{x}) = 11$

Now we see that we can form our first equation with this second equation.  If the second equation is true, then we can let some variable $ y$ be defined such that $ y = \frac {2}{x}$, or $ xy = 2$.
In other words, we can derive $ (1)$ from $ (2)$ and $ (2)$ from $ (1)$, so our steps are reversible, and only valid solutions should occur (although sometimes there are exceptions).

MathAndKnowledge, I'm not sure what you mean. The reason the equation has two solution sets, $ (1,2)$ and $ (\frac{8}{3},\frac{3}{4})$, is because the line $ 3x\plus{}4y\equal{}11$ passes through the rectangular hyperbola $ xy\equal{}2$ at two points. Its that simple.

When you combine the two equations into one equation (by SFFT etc), you indeed lose some information (as MathandKnowledge noted). However, SFFT is generally only helpful for solving equations in integers. You could still solve it from SFFT; for example $ (x\plus{}4)(y\plus{}3) \equal{} 25 \Longrightarrow x \equal{} \frac{25}{y\plus{}3} \minus{} 4 \Longrightarrow 3\left(\frac{25}{y\plus{}3} \minus{} 4 \right) \plus{} 4y \equal{} 11$ etc. So while SFFT is a legit technique, it doesn’t actually simplify the situation at hand.

[quote="JZambrano201"]The reason the equation has two solution sets, $ (1,2)$ and $ (\frac {8}{3},\frac {3}{4})$, is because the line $ 3x \plus{} 4y \equal{} 11$ passes through the rectangular hyperbola $ xy \equal{} 2$ at two points. Its that simple.[/quote]
I know what you mean.  But that wasn't what Lulze and the original poster were asking.  They were asking why SFFT seems to "fail" in this situation.  You gave a geometric reason that doesn't really answer the question because SFFT is an algebraic tactic.  So, that's why I gave the algebraic reason.
I showed that the algebraic reason for why SFFT "fails" in this case is because that the statement, "If there are two equations, $ 3x \plus{} 4y \equal{} 11$ and $ xy \equal{} 2$, then the simultaneous solutions of the two equations is given by $ (x \plus{} 4)(y \plus{} 3) \equal{} 25$," is not true, and I also showed why.

[quote="JZambrano201"]MathAndKnowledge, I'm not sure what you mean.[/quote]
Since you say you don't know what I mean, I'll try to give a different explanation (if you still don't understand after this, read my first post more carefully).  SFFT is known, especially for AoPSers (if not only for AoPSers) to be a useful and reliable algebraic tactic for solving equations (usually Diophantine equations, but not necessarily).  But when SFFT was used here for a non-Diophantine equation, it implied that there are infinitely many solutions to the original two equations because there are infinitely many solutions to the equation $ (x \plus{} 4)(y \plus{} 3) \equal{} 25$.  So, why did our trusty technique fail us?
And after that, as you can see by my first post, I explained why.
As I said, I know what you mean, but it's not that simple (you said, "Its that simple.").

I have two more algebraic arguments (I hope that they are valid) for why SFFT "failed" us.

First, SFFT was designed for when we [i]start[/i] with a single equation (such as $ xy\plus{}4y\equal{}3x\plus{}19$).  One can seldom use it legitimately for a single equation that is obtained [i]after combining two or more equations[/i], as shown by this problem.

Second, this problem was not a linear system of equations, though it did happen to have two solutions in two variables like other two-variable linear systems of equations.  We must be careful when we employ our tactics for linear systems of equations to solve systems of (pardon my bad vocabulary) un-linear systems of equations.  They will not necessarily hold.  Thus, we must always check that the solutions work in the end when we solve a problem in such a manner.

"MathAndKnowledge" is right at least twice.
1) It is always a good strategy to check your answers to make sure an error in logic or arithmetic did not introduce an extraneous solution.
2) If $ d(x)\equal{}e(x)$ and $ f(x)\equal{}g(x)$, we can safely conclude that $ d(x)\plus{}f(x)\equal{}e(x)\plus{}g(x)$ is necessarily true, but there is no way to prove that the first two equalities are necessarily true knowing only the third. That is why when we look for solutions to a system of equations (linear or not) we replace one equation with an equivalent one, keeping the same number of equations and being careful to check that all steps were reversible.

If the problem was worded differently, SFFT could have been very useful.
SFFT as used in this problem gave a necessary condition
$ (x \plus{} 4)(y \plus{} 3) \equal{} 25$
$ (1,2)$ is one solution of that equation. $ (21,\minus{}2)$ and $ (\minus{}9,\minus{}8)$ are some of the other few integer solutions.
SFFT would have been very useful if it was a multiple choice question, or if the problem asked for just one solution, or if the problem asked about all integer solutions only.
Unfortunately, there are also infinite real pairs of $ (x,y)$ solutions for the SFFT equation. Because not all of them will make the two original equations true at the same time, we need to put more constraints on x and y $ ($like $ xy\equal{}2)$, which brings us back to a system of equations just as difficult to solve as the first one.


calculus
MIT
college
Princeton
complex analysis

Umm, it's definetly not out of reach. There's no "late" for taking any class really. Everyone takes it when they're ready, it's not a race. And don't limit yourself to only MIT, there are some other schools out there that you should look into.

I like [url=http://epgy.stanford.edu]EPGY[/url] for online courses. The EPGY program was designed for students like you in school districts without advanced math courses.

No, it's not too late. In the 1970s, I graduated from a high school that offered only a very meager beginning calculus course as its highest math course, and MIT still traveled there to recruit students (none of us attended MIT, as I think none of us applied). Overcoming the limitations of your environment is more important than being in a great environment.

Do apply to a variety of colleges; some other college may fit you better than MIT.

another source for distance learning (cheaper, but less actual instruction than EPGY)
http://www.utexas.edu/cee/dec

Also check out MIT OpenCourseWare. Not for credit, but the price is right (free).
http://ocw.mit.edu/index.html

Books, libraries, and more recently the internet have more than you could read in a lifetime.
MIT and its equivalents are aware of the limitations of most high schools, such as the absence of math courses past calculus.

[quote]
Books, libraries, and more recently the internet have more than you could read in a lifetime. [/quote]

Totally agree. I didn't take calculus until university, and I'm doing fine in grad school.

[quote="lund"]I go to a rural public high school with limited oppurtunities in math.  Our community college only goes through Calc I and our high school through Pre-Calc.  I am currently taking Calc I as a sophmore and looking into what classes I should take next year.  I will probably do an independent study, any recommendations (courses or even text)?  Or, is there a good place to take on-line courses?  Also, I would like to go to MIT.  Is that out of reach, having taken Calculus relatively late?[/quote]

Relatively late?

[quote="yif man12"]
Relatively late?[/quote]
I was just referring to the belief that you would have to take Calc earlier than as a sophmore in order to be a math major at a prestigous school.

I took calculus as a sophomore and was accepted to Princeton (with the intention of majoring in math).  Of course, I could be the exception that proves the rule, but I doubt it.  Moreover, you can certainly take calculus later - even in your senior year (I know someone who fits this profile) - and be accepted into an Ivy League school with the intention of majoring in math.

[quote="lund"]
I was just referring to the belief that you would have to take Calc earlier than as a sophmore in order to be a math major at a prestigous school.[/quote]

I don't think this belief has any basis in fact.  (I'm saying that as a mathematics professor.)  The bulk of the students I know who have gone on to super-prestigious colleges took Calc (BC, without taking AB previously) as juniors.  I also know plenty of students who took Calc in their senior years and went on to major in math at prestigious schools.  There are also students, though they're somewhat rare these days because most high schools offer Calculus, who arrive at prestigious colleges not having taken Calculus and who successfully major in math.  At not-particularly-prestigious colleges, I'd estimate that as much as 1/3 of the graduating majors take Calc upon arrival in college.

[quote="smbelcas"]At not-particularly-prestigious colleges, I'd estimate that as much as 1/3 of the graduating majors take Calc upon arrival in college.[/quote]
Or more than 1/3. In our case, a significant number of our graduating majors transferred from community colleges, and those students frequently started out with courses before calculus. Among our own freshmen, AP Calculus AB is fairly common but BC is uncommon. We do send some students to Ph.D. programs, and in the last few years, we've had two students place among the top 500 in the Putnam. Both of them arrived with AP Calculus AB and started with our second semester calculus course.

Go look at some of the things rrusczyk has written about what he calls the "calculus trap." Now, most of the most impressive high school students I have known have taken a fair amount of college level math while still in high school - but that's much more a consequence of who they were than a cause. It isn't what made them impressive.

[quote="lund"]I go to a rural public high school with limited oppurtunities in math.  Our community college only goes through Calc I and our high school through Pre-Calc.  I am currently taking Calc I as a sophmore and looking into what classes I should take next year.  I will probably do an independent study, any recommendations (courses or even text)?  Or, is there a good place to take on-line courses?  Also, I would like to go to MIT.  Is that out of reach, having taken Calculus relatively late?[/quote]

You definitely shouldn't think of sophomore year as 'too late'.  In fact, from our experience at AoPS, I might even argue that for most people, earlier than that is 'too early', even for students who are getting 5's on the AP exam.  This doesn't mean they know how to think.  (And MIT knows this, too.)

53% of this year's entering freshman class at MIT started out with first semester calculus. Some of those folks had actually had some calculus in high school, but simply didn't have enough to get a 5 on the calc BC exam to place out of first semester calculus. But I am very sure that a significant portion of them had no calculus at all in high school. I don't think any of them are going to be handicapped in any way when it comes to majoring in whatever they want.

When I was a grad student at MIT, I used to grade the precalculus diagnostic test giving to incoming freshmen (to help place them into a freshman math course).  Many incoming MIT freshmen don't even have a good understanding of [i]precalculus[/i], let alone calculus, perhaps because many of them crammed for the AP Calculus exam rather than actually [i]learning[/i] calculus.

I took Calculus BC and got a five freshman year.  This is very surprising to me because I realize how little calculus I know, even now.  I am quite fortunate to have Complex Analysis and Differential Equations taught by an excellent professor who comes to my school.  He would often refer to proofs of things like the chain rule or make analogies to concepts we learned from single-variable calculus only to receive blank stares.  That was a whole class full of people who got fives on the AP exam.  After taking that class last year and a p-adic analysis course over the summer, I knew that I didn't really know calculus.  Point is, AP tests and "advanced math" in high school can often be aced without really knowing anything.  What you learn from the class is more important than the name of the class.


Hello I am new here and LOVE math.

I just created a website dedicated to helping people out with their math problems.

http://mathhelpfast.weebly.com/index.html

I just created the site so any comments and feedback will be GREATLY appreciated!

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Thanks and hope you enjoy it!

Coming from a background where a number of individuals [i]really[/i] struggle with the subject, and I mean struggling in having trouble finding the right directions or not working with math naturally as we do (or similar to an autistic individual struggling with reading comprehension), I believe it's good to help in these cases, primarily giving helpful directions and hints where they struggle so they can do them independently. However, there's a drawback.

A downside I commonly see, and occurs somewhat more frequently, is that some post a series of questions and only desire answers, not helpful hints or directions. This definitely seems the case where someone's not being specific where they struggle and just gives a response similar to, "I can't do them! I need solutions to all of them plz!," or, "I have the answer, but I want to double-check [but doesn't disclose his answer or solution]."

Just giving answers or solutions upfront is not helpful, and is easily subject to plagiarism (in fact there are a few examples of these around the forums).

For something like your site it would be highly recommended to help by giving hints or directions instead of solutions directly.

You're website's not bad, I like the implement of the live chat system but I think it needs to be LaTeX enabled.

Thanks for your all's suggestions!

@ Patterns: I understand exactly what you are saying. At school when I go around the class and help students with their math, I NEVER give them the answers. Instead, I give them the "road" to their answers. I let them do the work but I lead them towards it. (If that makes any sense).

@ 7h3.D3m0n.117: I was going to do that but I will have to wait until I can get the non free webhosting. I am making a little bit of money from the google ads, but not enough to keep a site up and running.

Thanks very much for your replies and I hope to see you on my site!

Paul

[quote="Nexmus"]To Whom it may concern,

every time you click on his websites and click on one of the ads that are sponsored by google, he get's \$5 or something like that. so, if you want to earn him some money, start clicking on the ads.[/quote]

Thanks for posting that!

I am not aloud to direct people to clicking on my ads, but other people can.

Also, ALL the money I make from the ads will be placed into the site so that I can get webhosting, unlimited bandwith, my own chatbox etc...

So thanks very much.

Also, I added a join section if you would like to be an official member of the site!

Thanks

Does anyone wanna help me get a forum up and running for my site?

I would do it alone but I am extremely busy just with this site and my school work and everything.

I would let you be on administration for whoever helped me out!

Thanks very much

[quote="1=2"][quote="7h3.D3m0n.117"][quote="mrmac"]Does anyone wanna help me get a forum up and running for my site?[/quote]

I can run an Invisionfree forum site with a short-URL, all for free. It might take me a while to set up since I'm busy with school, but I'll happily volunteer for that. Just send me specifics (color of backgrounds, forum name, etc.) in this thread and I'll try and create one by the end of this week.

The only bad thing is that it won't be directly integrated into the site so you'll have to post a link to it.[/quote]

UUhhh... I just became a member last night and we are beginning to set up the forum on freeforum, but it is a huge project... and I don't know if mrmac would like the work that I have done so far...[/quote]

Actually he has done a TERRIFIC job setting up the forums!

Once we get it all ready to go, there will be a forums section on the site!

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Thanks for all the help on clicking ads. I might have enough money to get webhosting soon!!

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Also, don't forget to check out http://mathhelpfast.weebly.com/

Thanks!

[quote="Cyrazeno"]Nice colors ...  :lol:[/quote]

Do you really like them?

Or was that sarcasm.

It's hard to match the colors without being able to use HTML on the site layout.

Thanks

[quote="alkjash"]Two suggestions:

The color on the forums is really bright. Make it darker please.

$ \text{\LaTeX}$ would really help on a math forum.[/quote]

I will do fix the color scheme.

I am just testing out a lot of different colors as I just made the forums.

I was going for a Mac type color scheme.

About the latex, I'm not sure if you could implement latex into a freeforums site.

If anyone can help me with that, I would greatly appreciate it.

Also, is latex in HTML?

Thanks,

Paul

No, I was not being sarcastic.
I do like the colors. =)
Change them if you like

[quote="Nexmus"]Personally, I like the colors and no problem...[/quote]

Thanks very much.

There aren't a lot of people joining the forums.

There are only 5 people there so far.

Please help out and join!

Thanks

[quote="sailingdog"]The colors are a little overwhelming...

But great site!

Please remember all of us when you become a millionaire.  :lol:[/quote]

I will remember you if you join my site!

j/k

:lol:

If you give some love, I will promise to give some back, some how, some way.

I just need to get this site bigger and make some money so I can get webhosting. I am getting paypal tomorrow so now I all need is the money!

Thanks very much!

[quote="Cyrazeno"]Awesome  :lol:  Very nice job, mrmac

How old are you anyways[/quote]

I am 16.

Thanks for the compliments guys!

Please join and help the site grow!

Just to update you all on the site a little bit, I added a donate button on the front page because the ad thing isn't getting me enough money for web hosting.

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So please consider donating or joining the forums!

Thanks a ton everyone for coming to my site, I TRULY appreciate it!

[quote="sailingdog"]Wait...
What is web hosting?
 :blush:[/quote]

It's where you get your site. Like....

http://en.wikipedia.org/wiki/Web_hosting

It will also be so I can buy my own chat (or make it) and implement LaTeX to make it a more pleasant stay while getting help.

Thanks,

Paul


JBL wrote:Given positive numbers w, x, y, and z such that
2(wx + wy + wz + xy + xz + yz) = 3(w:^2: + x:^2: + y:^2: + z:^2:), compute the value of (w - x)(x - y)(y - z)(z - w).

[hide]2(wx + wy + wz + xy + xz + yz) = 3(w:^2: + x:^2: + y:^2: + z:^2:) => (w-x):^2:+(w-y):^2:+(w-z):^2:+(x-y):^2:+(x-z):^2:+(y-z):^2:=0. Hmm, that means w=x=y=z. Thus that expression thing is 0.[/hide]

strange, but yes, i got the same thing, since a square can't be negative they must all be 0.

oh, well it said "positive" so i thought 0 was not a possibility

No, each of the numbers don't have to be zero -- however, their differences do.  For example, the quadruple (w, x, y, z) = (3, 3, 3, 3) works nicely.

Might I ask how someone would ever think up of a solution like that?

well . . . it's pretty much the only thing i could do. whenever i have squares and nothing else, i always say HM! i bet the sum of a bunch of squares is zero, which would give me a lot of information (because then all the squares must be zero). so i bring everything to one side and try to factor it into squares.

Because, well, you have x:^2:, and y:^2:, and xy, so what else are you going to do?  It's just something to look for, I guess -- you can test out AM-GM or Cauchy, but also just look for squares.  There was a problem on the USAMO whose final step was just that a couple or three years ago.  It's an important trick to know.  Here is that USAMO question -- I couldn't believe when I got to the end of it that its last step was so simple:

A2.  The triangle ABC satisfies the relation
cot:^2:(A/2) + 4cot:^2:(B/2) + 9cot:^2:(C/2) = 9(a+b+c):^2:/(49r:^2:), where r is the radius of the incircle (and a = |BC| etc, as usual). Show that ABC is similar to a triangle whose sides are integers and find the smallest set of such integers.

[hide]

u^2/r^2+4v^2/r^2+9w^2/r^2=36(u+v+w)^2/(49r^2)

u^2+4v^2+9w^2=36(u+v+w)^2/(49)

consider cauchy:

(u^2+4v^2+9w^2)(1+1/4+1/9)>=(u+v+w)^2

equality occurs iff u:1=2v:1/2=3w:1/3, ie u=4v=9w

notice the equality does occur

so the triangle must be similar to

40:45:13

[/hide]

they shouldnt give 4.5 hr on only three questions that are that easy!  APMO questions are much hader than this and its 4hr for 5 questions!

to get back on the perfect squares topic, cauchy can be proved using the fact that a perfect square is always non-negative

You really don't need Cauchy for that USAMO question, that was my point -- if you expand it out, you can just write it as sums of squares equal to zero.  Anyhow, APMO problems are not easier than USAMO questions . . . the APMO is more like the old USAMO.  And anyhow, this was the easiest question from that year, I thought.  There're always much harder questions on USAMO.

in that problem, how do you get from the equation with the cotangents to the equation bugzpodder solved?

Draw a triangle and its incircle.  Now, tangents to a circle are equal.  So give the six pieces of sides of the triangle, which have together 3 distinct lengths, names u, v, and w.  Then it follows.  Make sense?

yes, that makes a lot more sense now. :D  thanks


in quadrilateral ABC, the measure of ange A is 10 degrees larger than the measure of angle C; the measure of angle D is twice the measure of angle A; and the measure angle B is three times the measure of angle D. what is the number of degrees in the measure of angle D.

how would u do this problem? :rotfl:  :ninja:  :welcome:

what's up with the random smileys? :huh:
[hide]$\angle C=x$
$\angle A=x+10$
$\angle D=2x+20$
$\angle B=6x+60$
All of them added is $360^\circ$
$10x+90=360$
$x=27$
$\angle D=2x+20=74$[/hide]

[hide]Write everything in terms of angle A. So let the measure of angle A be $x$. The C measures $x-10$ degrees, D measures $2x$ degrees, and B measures $6x$ degrees. Since the sum of the interior angles must be 360, we have $10x-10=360$, and $x=37$, and angle D measures $2*37=74$ degrees.[/hide]


abstract algebra
group theory
vector
superior algebra
superior algebra unsolved

assume that $S_\infty$ is all permutations of $\mathbb{N}$.can we embed additive group of real number in $S_\infty$? what about $SO(3)$.

The additive group of the reals is a divisible group. So a necessary condition is that there are enough permutations in $S_\infty$ that have every n-th root, i.e. permutations $\sigma : \mathbb N \to \mathbb N$, such that for every $n \in \mathbb N$ there is a permutation $\sigma_{n}$ with $(\sigma_{n})^{n}= \sigma$. Do these permutations exist?

LiMa

I don't even know where to start the problem concerning $SO_{3}$, but the answer to the first question is affirmative: it is indeed possible to embed $\mathbb R$ in the infinite symmetric group $S(\Omega)$, where $\Omega$ is a countably infinite set. In fact, the following stronger assertion holds:

(*) Let $\Omega$ be an infinite set, and let $S(\Omega)$ be the group of permutations on $\Omega$. Then, every abelian group of cardinality $\le 2^{|\Omega|}$ can be embedded in $S(\Omega)$.

The idea is quite simple, but it may be a little difficult to explain properly.

[i]Proof of (*)[/i]

If $\Omega'$ is a subset of $\Omega$, I'll denote by $S(\Omega')$ the subgroup of $S(\Omega)$ which fixes everything outside $\Omega'$. Also, if $p$ is a prime number, I'll denote by $\mathbb Z_{p^{\infty}}$ the group of roots of unity having order equal to a prime power $p^{n}$ (for some $n$).

Let $G$ be a non-trivial group of cardinality $\le 2^{|\Omega|}$. Its divisible hull has the smallest infinite cardinality $\ge|G|$, so nothing in our problem changes if we assume $G$ to be divisible. It follows then that $G$ is a direct sum of groups isomorphic to $\mathbb Q$ (the torsion-free part $G_{\mathbb Q}$ of $G$) and groups isomorphic to $\mathbb Z_{p^{\infty}}$ for various primes $p$ (the torsion $TG$ of $G$). The isomorphism class of $G$ is uniquely determined by the dimension $\alpha_{\mathbb Q}(G)$ of $G_{\mathbb Q}$ over $\mathbb Q$ and by the cardinalities $\alpha_{p}(G)$ of the sets of summands isomorphic to $\mathbb Z_{p^{\infty}}$ for the various primes $p$. Clearly,  these $\alpha(G)$'s do not exceed $2^{|\Omega|}$, so the problem will be solved if we manage to embed in $S(\Omega)$ a divisible group $H$ with all $\alpha(H)$'s equal to $2^{|\Omega|}$.

Partition $\Omega$ into sets $\Omega_{\mathbb Q},\ \Omega_{p}$ for all primes $p$, each part having cardinality $|\Omega|$. Then partition $\Omega_{\mathbb Q}$ into $|\Omega|$ many parts $\Omega_{\mathbb Q}^{\omega},\ \omega\in\Omega$, each of cardinality $|\Omega|$; similarly, partition $\Omega_{p}$ into parts $\Omega_{p}^{\omega},\ \omega\in\Omega$. It is possible to embed any countable group into $S(\mathbb N)$ (just make the group act by left multiplication on itself; it becomes a permutation group on a countable set), so we can embed $\mathbb Q$ in each $S(\Omega_{\mathbb Q}^{\omega})$, and we can embed $\mathbb Z_{p^{\infty}}$ in each $S(\Omega_{p}^{\omega})$. Once we do this, we have embedded in $S(\Omega)$ a group which is isomorphic to the direct product of $|\Omega|$ copies of $\mathbb Q$ and $|\Omega|$ copies of $\mathbb Z_{p^{\infty}}$ for each prime $p$. It suffices now to prove that (a) the dimension of the direct product of $|\Omega|$ copies of $\mathbb Q$ over $\mathbb Q$ is $\ge 2^{|\Omega|}$, and (b) the torsion of the direct product of $|\Omega|$ copies of $\mathbb Z_{p^{\infty}}$ (which is a $p$-primary divisible group and hence a direct sum of $\mathbb Z_{p^{\infty}}$'s) is a direct sum of at least $2^{|\Omega|}$ copies of $\mathbb Z_{p^{\infty}}$.

(a) follows from the observation that the cardinality of an infinite-dimensional $\mathbb Q$-vector space is equal to its dimension, and the obvious fact that a product of $|\Omega|$ copies of $\mathbb Q$ has cardinality $2^{|\Omega|}$. For (b) we do something similar, but this time we work over the field $\mathbb F_{p}$ with $p$ elements instead of $\mathbb Q$. Just notice that if $H$ is a $p$-primary divisible group, then $\alpha_{p}(H)$ is equal to the dimension of $H/pH$ over $\mathbb F_{p}$, and that the torsion of the product of $|\Omega|$ copies of $\mathbb Z_{p^{\infty}}$, reduced modulo $p$, has cardinality $2^{|\Omega|}$.


LaTeX
inequalities
function
number theory unsolved
number theory

Find all triplets $(x,y,n)$ of positive integers which satisfy:
\[ (x-y)^n=xy  \]

We see that both x and y must be even, so write x = k*2^r, y = s*2^t, where k and s are odd. Assume that t >= r (we also have x > y). We then get
2^(rn)*(k - 2^(t-r)*s)^n = 2^(r+t)*ks. Because k - 2^(t-r)*s and ks are odd we must have r+t = rn => r(n-1) = t =>
(k - 2^(t-r)*s)^n = k*s. Assume (k,s) = 1. Let p be a prime dividing k - 2^(t-r)*s. Then p also divides k*s => it divides k or s. If it divides s then p must divide k too, contradiction. So assume p divides k => p divides 2^(t-r)*s => p divides 2^(t-r) => p = 2. Hence k - 2^(t-r)*s = 2^t for some t => 2^tn = k*s, clearly a contradiction because both k and s are odd. If k - 2^(t-r)*s = 1 we get k = s = 1, but then the former is not true.

So we must have (k,s) = d > 1. Write k = d*a, s = d*b, where (a,b) = 1. Then
d^n*(a - 2^(t-r)*b)^n = d^2*ab => d^(n-2)*(a - 2^(t-r)*b)^n = ab. Now if a - 2^(t-r)*b > 1, by the same reasoning as before we get p = 2 and we get a contradiction. So we must have a - 2^(t-r)*b = 1. Then
ab = (1 + 2^(t-r)*b)*b = d^(n-2) => (1 + 2^(r*(n-2))*b)*b = d^(n-2).

Assume n = 3. Then d = (1 + 2^(r)*b)*b and we get a solution: x = (2^r)*b*(1+b*2^r)^2, y = (2^t)*b^2*(1 + b*2^r)

Because (b, 1 + 2^(r*(n-2))*b) = 1, we get b = f^(n-2), 1 + 2^(r*(n-2))*b = g^(n-2) for some g and f that are odd.
Assume now n is even => n - 2 = 2t for some t. Then
g^(n-2) - 2^(r*(n-2))*f^(n-2) = 1 => g^(2t) - 2^(2rt)*f^(2t) = (g^t - 2^(rt)*f^t)*(g^t + 2^(rt)*f^t) = 1, which would imply that f is zero and g is one => b = 0 => s = 0 => y = 0 => x = 0, and n can be every even number.

Now assume n > 3 is odd. Then we get, by setting n-2 = t, where t is now odd:
1 = g^t - (f*2^r)^t = (g - f*2^r)*(g^(t-1) + 2^r*f*g^(t-2) + ... + (f*2^r)^t-1)
That implies f = 0, g = 1 => b = 0 => s = 0 => y = 0 => x = 0, and n can be every odd number.

Probably one can do same calculations for t < r.

Please, edit your post to make a use of LaTeX - it'll make clearer what you wrote

Yeah, though I don't know how to write in Latex :S
I'll try do it soon, kind of busy for the moment.

Also I have discovered some more solutions, which can be seen clearly when doing the proof there, as for example when r > t we get x = 4, y = 2, n = 3 as a solution. Maybe there are more solutions hidden there in that investigation, but hopefully they can all be discovered in those steps.

Denote $x=dx'$ and $y=dy'$ with $(x',y')=1$. Now we have:

$x'y'=d^{n-2}(x'-y')^n$

But if $(x',y')=1$ then $x'-y'|x'y'$ implies $|x'-y'|=1$ (in fact if we have $(x'y',x'-y')|(x',x'-y')(y',x'-y')=1$). Substitutig we get:

$x'y'= d^{n-2}$

Now we have that $x'$ and $y'$ are consecutive positive integers so $x'y'\geq2$ and so $n>2$. but if $n>3$ we have that two consecutive numbers are both $m$-th powers with $m>1$ and so it is impossible. Thus we have $n=3$ and so the general solution is (remembering that $x'>y'$ else we have $RHS<0$):

$(a(a+1)^2,a^2(a+1),3)$   $a \in N_0$

OK just explain me please why
"x'-y'|x'y' implies |x'-y'|=1"
and
''if n>3 we have that two consecutive numbers are both m-th powers with m>1 and so it is impossible.''

??

OH come on!
I`ll be very thankful if somebody helps me

[quote="Beat"]
"x'-y'|x'y' implies |x'-y'|=1"
and
''if n>3 we have that two consecutive numbers are both m-th powers with m>1 and so it is impossible.''
[/quote]

1) because of $(x',y')=1$ - If you cannot see this straight away suppose that |x'-y'|>1 so that there's a prime $p| x'-y' |x'y'$ and think why this is a contradiction

2) Suppose to the contrary that $y^m-1=x^m$ but $y\geq x+1$ so $y^m \geq (x+1)^m > x+1$ contradiction

''2) Suppose to the contrary that y^m-1=x^m but y\geq x+1 so y^m \geq (x+1)^m > x+1 contradiction''
Where did you get that from??
$y^m-1=x^m$
And where`s the contradiction

And secondly how will we solve the inequality for n =3?
how do we get a(a+1)^2 and a^2(a+1)

If you are asking yourself something about me.. then the answer is yes!
Maybe i`m blind or stupid.. doesn`t matter. I want to understand the problem. Thank you..

I've just done the following:

$a=x^m$
$a+1=y^m$

so from this you get the equation. And the inequality contradicts this equation

My sulotion:

1.    easy to see no way to get a triple if n=1 or n=2;
2.    let Z=X-Y, the original function can transform to Z^n = (Z+Y)*Y;
3.    because n>2, to any Z's prime divisor p, if p^(k+1) is not Z's divisor but p^k|Z, we must have p^k|Y.
4.    till now, we get Z|Y. let Y=wZ. So Z^(n-2)=w*(w+1)
5.    easy to see the solution is (n, Z, w) = (3, w(w+1), w) for any positive integer w.

For $n=1$ we get $x-y = xy \iff (x+1)(y-1) = -1$ which has no solutions in positive integers.

For $n=2$ we get $x^2 + y^2 = 3xy$. Setting $d = \gcd(x,y)$ implies $x = dx_1,\ y = dy_1,\ \gcd(x_1, y_1) = 1$ and our equation becomes:
$$x_1^2 + y_1^2 = 3x_1y_1$$
Since $x_1 | y_1^2$ and $y_1 | x_1^2$ we get $x_1 = y_1 = 1$ which is not a solution so this case has no solutions.

For $n \geq 3$, setting $m = n - 2,\ d = \gcd(x,y)$ implies $m \geq 1,\ x = dx_1,\ y = dy_1,\ \gcd(x_1, y_1) = 1$ and our equation becomes:
$$d^m (x_1 - y_1)^{m+2} = x_1y_1$$
Firstly, assume $|x_1-y_1| > 1$. If a prime $p$ divides $x_1 - y_1$, then it divides at least one of $x_1$ or $y_1$ and since it divides their difference, it must divide both of them which is a contradiction because $\gcd(x_1, y_1) = 1$. Since $|x_1 - y_1|$ can't be zero because that would imply $x_1y_1 = 0$, we must have $|x_1 - y_1| = 1$. The expression $(x_1 - y_1)^{m+2}$ can't be equal to $-1$ because that would leave left side negative while the right side is positive. Letting $t = \min(x_1, y_1)$ and combining this with $|x_1 - y_1| = 1$ gives us $x_1y_1 = t(t+1)$. Thus, our equation becomes:
$$d^m = t(t+1)$$
For $m \geq 2$ we get that if a prime $p$ divides $t$, $p | t \implies p | d \implies p^m | d^m \implies p^m | t$ and this holds for all primes that divide $t$, which means $t = t_1^m$ for some positive integer $t_1$. Similarly, we get that $t+1 = t_2^m$ for some positive integer $t_2$. This would mean that $t$ and $t+1$ are perfect $m$-th powers where $m \geq 2$ which differ by $1$ and that is impossible for positive integers.
So we're left with the case $m = 1$ which gives us $x_1 > y_1 = t,\ d = t(t+1),\ n = 3,\ x = t(t+1)^2,\ y = t^2(t+1) \ \forall t \geq 1$ and those are indeed the solutions to the starting problem.


calculus
integration
number theory
greatest common divisor
Ring Theory
algebra
function

Good afternoon =) I have to prove that every maximal ideal in a commutative ring is prime (without using the characterization of prime and maximal ideals for quotient rings (fields, integral domains).

My solution I feel is not very strong. Could I get some help please ? :) ($ X \equal{} <x>$, similarly for $ y.$

Let $ R$ be a commutative ring, and $ M$ a maximal ideal. Let $ x,y \notin M$. Thus $ M \subset ( X \plus{} M)$ and $ M \subset (Y \plus{} M)$ since M is maximal and $ 0 \in X,Y$.
Thus $ X \plus{} M \equal{} Y \plus{} M  \equal{} R$
Now $ R \equal{} (X\plus{}M)(Y\plus{}M) \equal{} XY \plus{}XM\plus{}YM\plus{}MM \subset XY\plus{}M \subset R$.
Thus $ R \equal{} XY\plus{}M \supset M$ so $ xy \notin M$. Therefore $ M$ is a prime ideal

Thanks

Let M be a maximal ideal of R. Choose $ a, b \in R$ and suppose that $ ab \in M$ with $ b \notin M$.
Since M is maximal we must have $ M \plus{} (b) \equal{} R \equal{} (1)$, which implies that there exist $ m \in M$ and $ c \in R$ such that $ m \plus{} cb \equal{} 1$. Now write $ a \equal{} a\cdot 1 \equal{} a(m \plus{} cb) \equal{} am \plus{} cab$. Since $ m, ab \in M$, we must have $ a \in M$.

Could you please explain why mine is wrong (if it is?) =)

Oh, wicked =)
Mhmm jmerry I believe you're a professor. As such which would you prefer a student to hand in? I am going to use my own solution but just to understand how other's think about the differences.

While I have some teaching experience, most of it is as a TA. I'm not a professor yet.

I would take either of these arguments, pretty much equally. They both use the same idea; it's just that they use contrapositive forms of the definition of a prime ideal.

REMARK When learning new abstractions such as ideals it is essential to compare them to their better-known concrete instances. Here the identity is merely an ideal-theoretic analog of a well-known identity for GCDs, namely (AB,C) = (A(B,C),C). Thus  (B,C) = 1 => (AB,C) = (A,C), as in the 2nd proof above. If, further, (A,C) = 1 then (AB,C) = 1, as in the 1st proof. It's an ideal-theoretic analog of Euclid's Lemma. Here (B,C) = B+C is the ideal sum, which behaves like a GCD in many ways. The proof is a simple application of basic laws of ideal arithmetic, namely

AB + C  =  AB + AC + C . . . via AC < C , Absorption, Associativity

. . . . . .    =  A (B+C) + C . . . via Distributive Law

In Z or any PID, i.e. a domain where all ideals are principal, the ideal/gcd view coincides since (a,b) = (c) for c = gcd(a,b);  also contains = divides: (a) > (b) <=> a|b, etc. When you go on to study ideal/divisor theory you will find that one defines GCD, LCM for ideals to exploit this analogy.  For another similar simple analogy see my post [1] on the Freshman's Dream (A + B)^n = A^n + B^n, which mentions connections to Dedekind/Prufer domains, CRT (Chinese Remainder), Gauss' Lemma, etc.

--Bill Dubuque

[1] sci.math, 19 Nov 2008, (A + B)^n = A^n + B^n for GCDs/Ideals, if  A+B cancellative,
http://google.com/groups?selm=y8zej18x733.fsf%40nestle.csail.mit.edu
http://google.com/group/sci.math/msg/ba5ac69dddfc4a67


How would someone be able to thoroughly prepare for an Exam like the Putnam without any prior Olympiad Experience.  That is, how can one maximize their chances of scoring high, relative to how they would've done w/o preparation, using various practice problems from previous years, olympiad books, specific preparation techniques, a studying regimen, etc.  What kind of specifics are needed in the aforementioned ways to prepare to improve one's ability to solve Putnam-type questions?  Just give me some of your examples of how you guys did this sort of thing in high school (and college as well, possibly), as I never prepared for Olympiads in high school, and now that I am in college I feel as if I can distinguish myself by doing well on the Putnam Competition.  Thank you for your input.

There's already a topic opened for the same reasons, but I can't find it now. I think it was created by Peter VDD ... :?

Well, at least here (GATech) 90-95% of the people that participate in the Putnam in general does not have any prior experience in Math Olympiads at all, and some of them got good scores (last year I think someone got around 60, not sure, and I know that is quite lower than the big scores, but if you count that the mean is 2 points or less). The technique is simple: A professor has organize for around 15 years the competition, and the "team" meets once a week, there is a small lecture (sometimes given by me, but most of the time given by Prof. Wang), and we assign a fair number of problems a week for the students to try (and we actually review them). That's basically it. I know that the guys that are actually quite interested meet at least three or four extra hours a week by themselves. If you personally want to do something, start just playing around with the problems, and really make some compromise (kind of I AM GOING TO SOLVE THIS ONE BY MYSELF) and if it takes you forever, it doesn't matter, research if you need about what kind of theory you need to solve it.

Now, if you are a freshman, well, without any prior experience in problem solving, it may be tough the first year, but if you make a compromize with yourself and continue solving problems all the time you can, the second year will be easier (not easy but easier) and by the fourth year you will have a lot of experience!

It was just a bunch of ideas.


A man is running down the railroad tracks in a tunnel just large enough for the train to get through. When he is $\frac{2}{5}$ of the wa through, he hears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or runs back, he will reach safety at the end of a tunnel at the same time that the train reaches that end. At what rate, in mph, is he running?

Billy

[color=green][hide]When the man runs the 2/5 of the tunnel back the train will reach the tunnel at the same time.
If he runs the 3/5 left, the train will reah the tunnel and go through it.
That means he runs 1/5 of the tunnel while the train goes through it.
So he goes 1/5 as fast of the train, so he goes 12 mph[/color][/hide]

[hide]t=time to reach tunnel
T=time to exit tunnel from time of entry
t=2/5
T+t=3/5
T=1/5
60*1/5=12
>>12?<<[/hide]


function
algebra
linear equation
algebra unsolved

Let $n \geq 2$ be a positive integer. Find all functions $f: \mathbb R \to \mathbb R$ such that
\[f(x^n \plus{} f(y)) \equal{} f^n(x) \plus{} y\]
for any real $x$ and $y$.

$ f(f(0))\equal{}f^n(0)$ for all $ n\ge 2$ and since the LHS is constant we have $ f(0) \equal{} 1$ or $ 0 \qquad(1)$

Similarly, $ f(1\plus{}f(0))\equal{}f^n(1) \Rightarrow f(1) \equal{} 1$ or $ 0 \qquad(2)$

Also $ f(f(1))\equal{}f^n(0) \plus{} 1$ so from (1) and (2) it follows that $ f(0)\equal{}0, f(1)\equal{}1$
=========================================================

Let $ y\equal{}0$, $ f(x^n)\equal{}f^n(x)$, let $ y\equal{}1$, $ f(x^n\plus{}1) \equal{} f^n(x) \plus{} 1$

Subtracting the two equations gives $ f(x^n\plus{}1) \equal{} f(x^n)\plus{}1$

so let $ x^n\equal{}a$ and define $ g$ such that $ f(x)\equal{}x\minus{}g(x)$

Plugging into the equation gives $ f(a\plus{}1)\equal{}f(a)\plus{}1$ and $ g(a\plus{}1)\equal{}g(a) (*)$
=========================================================

Now let $ x\equal{}0$ and we find $ f(f(y))\equal{}y$

From (*) $ y\minus{}g(y) \minus{} g(y\minus{}g(y)) \equal{} y \Longrightarrow g(y) \equal{} \minus{}g(y\minus{}g(y)) \Longrightarrow 0\equal{}g(0)\equal{}g(\minus{}g(y))$

so $ g(y)\equal{}0$

Hence $ f(x)\equal{}x\minus{}g(x) \equal{} x$

[quote="ocha"]$ f(f(0)) \equal{} f^n(0)$ for all $ n\ge 2$ and ...[/quote]

I think you misread the statement : $ n$ is fixed and considered as a parameter. So you cant write $ f(f(0)) \equal{} f^n(0)$ [b][u]for all [/u][/b]$ n\ge 2$

[quote="moriarti"]Let $ 2 \le n \in N$. Find all functions f:R -> R such that

$ f(x^n \plus{} f(y)) \equal{} f^n(x) \plus{} y$

for any x, y.[/quote]
Let $ P(x,y)$ be the assertion $ f(x^n\plus{}f(y))\equal{}f(x)^n\plus{}y$

If $ f(y_1)\equal{}f(y_2)$, and comparing $ P(x,y_1)$ and $ P(x,y_2)$ we get $ y_1\equal{}y_2$ and $ f(x)$ is injective.
$ P(x,u\minus{}f(x)^n)$ $ \implies$ $ f(\text{something})\equal{}u$ and $ f(x)$ is surjective, so is a bijection.

let then $ f(0)\equal{}a$ and $ b$ such $ f(b)\equal{}0$ and $ c$ such that $ f(c)\equal{}b$

$ P(0,x^n)$ $ \implies$ $ f(f(x^n))\equal{}x^n\plus{}a^n$
$ P(x,b)$ $ \implies$ $ f(x^n)\equal{}f(x)^n\plus{}b$ and so, using line above : $ f(f(x)^n\plus{}b)\equal{}x^n\plus{}a^n$
$ P(f(x),c)$ $ \implies$ $ f(f(x)^n\plus{}b)\equal{}f(f(x))^n\plus{}c$ and so, using line above : $ f(f(x))^n\plus{}c\equal{}x^n\plus{}a^n$
$ P(0,x)$ $ \implies$ $ f(f(x))\equal{}x\plus{}a^n$ and so, using line above : $ (x\plus{}a^n)^n\plus{}c\equal{}x^n\plus{}a^n$ $ \forall x$ and so $ a\equal{}b\equal{}c\equal{}0$

$ P(0,x)$ $ \implies$ $ f(f(x))\equal{}x$
$ P(x,0)$ $ \implies$ $ f(x^n)\equal{}f(x)^n$
$ P(x,f(y))$ $ \implies$ $ f(x^n\plus{}y)\equal{}f(x^n)\plus{}f(y)$

1) $ n$ is even
==============
$ f(x^n\plus{}y)\equal{}f(x^n)\plus{}f(y)$ $ \implies$ $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x\geq 0$, $ \forall y$
Setting $ y\equal{}\minus{}x$ in the above equation, we get $ f(\minus{}x)\equal{}\minus{}f(x)$ $ \forall x\ge 0$ and so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y$
This is a Cauchy equation and we also have, using $ f(x^n)\equal{}f(x)^n$ and $ n$ even : $ f(x)\ge 0$ $ \forall x\ge 0$
So $ f(x)$ has a lower bound on $ [0,\plus{}\infty)$ and so $ f(x)\equal{}ax$ for some real $ a$
Plugging this back in the original equation, we get $ a\equal{}1$

And the solution $ f(x)\equal{}x$

2) $ n$ is odd
=============
$ f(x^n\plus{}y)\equal{}f(x^n)\plus{}f(y)$ $ \implies$ $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y$
$ P(1,0)$ $ \implies$  $ f(1)\equal{}f(1)^n$ and, since $ f(1)\ne 0$ ($ f(0)\equal{}0$ and $ f(x)$ is injective) and $ n$ odd : $ f(1)\equal{}1$

So we get $ f(p)\equal{}p$ and $ f(px)\equal{}pf(x)$  $ \forall p\in\mathbb Z$

Consider then the $ n\minus{}1$ quantities $ y_k\equal{}f(x^k)$, $ k\in[2,n]$
$ f((x\plus{}p)^n)\equal{}f(x\plus{}p)^n$ is a linear equation of $ y_k$
Using as many different values of $ p$ as we need, it's easy to show that $ y_k\equal{}f(x)^k$ $ \forall k\in[2,n]$

So $ f(x^2)\equal{}f(x)^2$ and we can then use the demonstration in $ 1)$ above to conclude $ f(x)\equal{}x$

Hence the unique solution $ \boxed{f(x)\equal{}x}$

oh sorry, how about this

First show that $ f$ is a bijection
$ f(a) \equal{} f(b) \Rightarrow f(a^n \plus{} f(a)) \equal{} f(a^n \plus{} f(b)) \Rightarrow f^n(a) \plus{} a \equal{} f^n(a) \plus{} b \Rightarrow a \equal{} b$, so $ f$ is injective

Also for any real number $ c$, we can fix $ x$ and let $ y \equal{} c \minus{} f^n(x)$, which gives $ f(x^n \plus{} f(c \minus{} f^n(x))) \equal{} c$ so $ f$ is surjective.
=========================

Now show that $ f(0) \equal{} 0$
let $ x \equal{} 0 \Longrightarrow f(f(y)) \equal{} y \plus{} f^n(0)$ Using this result

$ f(x^n \plus{} f(f(y))) \equal{} f^n(x) \plus{} f(y)$
$ \therefore f(x^n \plus{} y \plus{} f^n(0)) \equal{} f^n(x) \plus{} f(y)$
$ \therefore f(f(x^n \plus{} y \plus{} f^n(0)) \equal{} f(f^n(x) \plus{} f(y))$
$ \therefore x^n \plus{} y \plus{} 2f^n(0) \equal{} f^n(f(x)) \plus{} y \equal{} x \plus{} f^n(0) \plus{} y$
$ \therefore f^n(0) \equal{} 0 \Longrightarrow f(0) \equal{} 0$
==========================

Now we have $ f(f(x)) \equal{} x$ and $ f(x^n \plus{} f( \minus{} f^n(x))) \equal{} 0 \Longrightarrow x^n \plus{} f(f^n( \minus{} x)) \equal{} 0$
$ \therefore f( \minus{} x^n) \equal{} \minus{} f(x^n)$

[b]For $ n$ even:[/b]
If $ a > b$ then there exists an $ x > 0$ such that $ a \equal{} x^n \plus{} b$
$ \therefore f(a) \equal{} f(x^n \plus{} b) \equal{} f(x^n \plus{} f(f(b))) \equal{} f^n(x) \plus{} f(b) > f(b)$
So $ f$ is increasing, but since $ f(f(x)) \equal{} x$ we must have $ f(x) \equal{} x$

[b]For $ n$ odd:[/b]
$ f(x \minus{} f(x)) \equal{} f((x^\frac {1}{n})^n \plus{} f(\minus{}x)) \equal{} f(x) \minus{} x$

However if we have an $ a$ such that $ f(a) \equal{} \minus{} a$ then
$ f(a^n \minus{} a) \equal{} f(a^n \plus{} f(a)) \equal{} a \minus{} a \equal{} 0$
So $ a^n \minus{} a \equal{} 0$ and $ a \equal{} 0$ or $ 1$, so $ f(x) \equal{} x$ or $ x \plus{} 1$, the later leading to an obvious contradiction

So we have $ f(x) \equal{} x$ as the only solution

But when $ n$ is odd, function $ f(x)\equal{}\minus{}x$ also satisfies original equation. I think you missed something.

[quote="turbo taskytojas"]But when $ n$ is odd, function $ f(x) \equal{} \minus{} x$ also satisfies original equation. I think you missed something.[/quote]

Ooops,  :blush: , you're obvously right.

Here is the error in my own proof :
[quote="pco"]2) $ n$ is odd
=============
$ f(x^n \plus{} y) \equal{} f(x^n) \plus{} f(y)$ $ \implies$ $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ $ \forall x,y$
$ P(1,0)$ $ \implies$  $ f(1) \equal{} f(1)^n$ and, since $ f(1)\ne 0$ ($ f(0) \equal{} 0$ and $ f(x)$ is injective) and $ n$ odd : $ f(1) \equal{} 1$[/quote]

$ f(1) \equal{} f(1)^n$ and $ f(1)\ne 0$ implies $ f(1)^{n\minus{}1}\equal{}1$ and since $ n$ odd, $ n\minus{}1$ is [u][b]even [/b][/u]and then $ f(1)\equal{}1$ or $ f(1)\equal{}\minus{}1$ (I missed this).

Thanks for your remark.

Still I am not sure I completely understood the way you showed $ y_{k}\equal{}f(x)^{k}$ in your solution. Changing $ p$ we can get a system of $ n\minus{}1$ nonequivalent linear equations in $ n\minus{}1$ variables $ y_{k}$. Am I right, claiming that this system has at most one unique solution and then just checking and confirming that solution $ y_{k}\equal{}f(x)^{k}$ satisfies equations?

@pco: thanks to give me this link, your solution is so nice, if compare with my solution I had read (it used more in analysis). :oops:
and I think you should put $ y_k\equal{}f(x^k)\minus{}f(x)^k$ so we have a linear system equation with variable is $ y_k$, so $ y_k\equal{}0$, it will be clearer. :wink:


\[ \begin{array}{l} x_1 x_2 .....x_n \equal{} a^n \\
ais\_constant, \\
for\_some\_positive\_constant\_k,find\_least\_value\_of \\
\prod\limits_{i \equal{} 1}^n {(x_i \plus{} k)} \\
\end{array}
\]
[hide]ans is
[hide]\[ (a \plus{} k)^n
\][/hide]
[/hide]

[quote="apratimdefermat"]ur answer is correct. i got the same ans . induction solves it. :D[/quote]
hmm it's false in some case where x_i are negative , when x_i are all positive , it follows directly  Holder's inequality


What can be said about where h''(x) = g''(x), if h(x) = -x^3-3x^2-3x-1 and g(x)= (x+1)^3?

a. This might be a point of inflection
b. h''(x) is concave down with g''(x) is concave up
c. never, w(q) is always concave down
d. none of the above

[quote="HilbertThm90"]Can I ask a silly question? What is w(q)?[/quote]

To be honest, it was confusing to me as well.  This is a study guide to a test upcoming, I'm not sure what w(q) had to do with that question....

The question looks a bit garbled. For the parts that make sense, note that $ h(x)\equal{}\minus{}g(x)$. Also, you can easily calculate second derivatives.

The title typo is entertaining- I'm not sure what to make of an "infliction point", other than that it sounds dangerous.

[quote="jmerry"]The title typo is entertaining- I'm not sure what to make of an "infliction point", other than that it sounds dangerous.[/quote]
Well, painful at the very least. I let the typo stand precisely because it was entertaining.

I think inflection point was meant. http://en.wikipedia.org/wiki/Inflection_point


What is the probability that an integer in the set {1,2,3,...,100} is divisible by 2 and not divisible by 3?

$\left(A\right) \frac{1}{6}$
$\left(B\right) \frac{33}{100}$
$\left(C\right) \frac{17}{50}$
$\left(D\right) \frac{1}{2}$
$\left(E\right) \dfrac{18}{25}$

[hide]C. There are 50 multiples of 2. If we subtract the multiples of two that are also multiples of 3, we would have the answer. The multiples of both 2 and 3 are multiples of 6. There are 16 of these. $\frac{50-16}{100}=\frac{17}{50}$[/hide]


Hi !
For the moment I learn both Maths and Physics in a Belgian University (second year). Next year I will graduate. I'd like to study abroad in a famous university... So I wonder which ones I should apply to. How much is the financial investiment needed, the atmosphere (sport, parties) etc...
I'm looking for courses that concern mathematical physics. If someone foolows such a degree course please tell me :)

Thanks A LoT !

no one  :(  ?

Does someone know something about CAMS in Cambridge UK ? It seems possible to take also physics courses.
Is there someone here who did this ?

Hi me too, I'm a student here in Belgium in BXL, soon I should go to Mons Polytech. but I'd really like to go to a US university too, are you at ULB ? And I understand from this forum that to get into a top US university you got to have some Intl.Math.Olympiad. experience, have you got any ?

At Cambridge (and indeed practically every good university) you can take physics courses  on top of your maths ones. If I remember correctly you don't have much choice in first year - you've [i]got[/i] to take a few physics courses. If you want even more physics whilst still doing maths you can take "maths with physics"; you are then able to choose what to specialise in later on. I'm not at university yet and when I go I will try to minimise the amount of physics I have to do so I can't tell you much more.
(P.S. It's a suprise so few forum members apply to Cambridge - it's a lovely location, the tutorial (called supervisions) system is excellent and you don't have to do all those non-maths courses American universities seem to demand (unless you want to)).

Oh,I'd be happy to apply to Oxbridge but isn't the tuition there even more astronomical than in US institutions ?

Appolonius - I don't think that's true but I'm not too sure. Presumably tuition varies wildly among U.S colleges. For more information on the relative expense google "Cambridge" and look around the forum here for info. on the better U.S universities - there's already a ton of facts since most posts ask about getting into Harvard, Princeton etc... Plus Cambridge is closer to Belgium :lol:

Yes,yes but closer to Belgium doesn't mean easier for a student in Europe to get into, the anglo-saxon system is quite original why can't they have for all their schools simply an admission exam let's say in maths, physics...Wouldn't that be fair and give everybody who can succeed there equal chances ? This is what most continental Europe countries do, why they don't ?I simply don't get it, eerm could anybody check my topic :blush: ???

There is such a large variety of universities in the United States that there isn't a general answer to the questions in the original post in this thread. Some of those universities would be very happy to have more students from continental Europe, I am sure, but many are very competitive for international applicants. Costs also vary quite a lot.

@ Appolonius : now I study at the UCL. [quote]have you got any ?[/quote] Yes I went to Mexico two years ago.
@ tokenadult : Yes I agree with you, but I think UK universities are in the same case however it is often cheaper.
I gave up the idea to study in the US because it seems that universities like Cambridge, Oxford, Durham, Warwick are as good as US ones. Moreover it is closer for belgian people. Nevertheless I'm afraid it is also very very difficult for applicants in my situation to get in.
The best for me is to go to Cambridge and follow the Certificate of Advanced Study in Mathematics ( CASM and not CAMS, sorry for the typo in my second post ). There is a lot of theorical physics courses available while it is still possible to keep mathematical ones. I just regret that it is only one year. Not enough to taste the nice atmosphere there (another good reason to try to study at Cambridge).
I wonder if someone knows what kind of studying openings are conceivable after this Certificate, I mean something else than a Phd (I'd like to wait one more year before thinking about it).
Thanks for your answers


Harvard college geometry AMC USA(J)MO USAMO AIME

[hide="yay i got 36!"][url]http://www.mathleague.com/reports/report0405_6.htm[/url][/hide]

[hide="36"]
Olga Mandelshtam, University HS, CA
[b]Shrenik Shah, Monta Vista HS, CA[/b]
[b]Conner Rogers, D'Evelyn Jr/Sr HS, CO[/b]
[b]Jeffrey Rosen, Choate Rosemary Hall, CT[/b]
[b]Eric Stansifer, Westshore Jr/Sr HS, FL[/b]
Joshua Weiss, Stoneman Douglas HS, FL
Richard Ragland, Southside Comprehensive HS, GA
Natache Roger, Southside Comprehensive HS, GA
[b]Qi Gan, Ames High School, IA[/b]
Jonathan Swenson, Barrington HS, IL
[b]David Benjamin, South Vigo HS, IN[/b]
[b]Rahul Banerjee, St. John's HS, MA[/b]
Daniel Engel, Framingham HS, MA
[b]Nat Pitiwan, Brooks School, MA[/b]
[b]Pawat Technapougtada, Middlesex School, MA[/b]
[b]Erbo Xu, Lexington HS, MA[/b]
[b]Tim Zheng, Algonquin Regional HS, MA[/b]
Matt McCutchen, Montgomery Blair HS, MD
Sam Bhagwat, Livonia Math/Sci/Comp., MI
Jeremy Diepenbrock, Enloe HS, NC
[b]Tom Belulovich, AAST Mu, NJ[/b]
Jeong Wook Choi, AAST Mu, NJ
[b]Caroline Drucker, AAST (9-10), NJ[/b]
[b]Zhou Fan, Parsippany Hills HS, NJ[/b]
Jiashuo Feng, AAST (9-12), NJ
Chris Kennedy, AAST Mu, NJ
[b]HyunSoo Kim, AAST Mu, NJ[/b]
[b]Brian Lee, AAST (9-10), NJ[/b]
Illan Levy, Freehold HS, NJ
[b]Aaron Potechin, Millburn HS, NJ[/b]
Jacob Sanders, AAST (11-12), NJ
Michelle Ton, AAST (11-12), NJ
YoungBin Won, AAST (11-12), NJ
[b]Bohua Zhan, W. Windsor-Plainsboro HS South, NJ[/b]
Fan Zhang, AAST Mu, NJ
Jay Gupta, Ward Melville HS, NY
[b]Geehoon Hong, Benjamin Cardozo HS, NY[/b]
Albert Ko, Ward Melville HS, NY
[b]Neal Wadhwa, Ward Melville HS, NY[/b]
Jason Chen, Oklahoma School of Science & Math, OK
Michael Liang, Oklahoma School of Science & Math, OK
[b]Conghua Ye, Jenks HS, OK[/b]
Daniel Droz, Veritas Academy, PA
Adam Hallowell, Emmaus HS, PA
Tim Heath, Cumberland Valley HS, PA
John Lesieutre, State College HS, PA
Zack Pierpoint, Freedom HS, PA
[b]Jason Trigg, Unionville HS, PA[/b]
[b]Xander Marcus, Wheeler School, RI[/b]
Josh Lim, The Oakridge School, TX
[b]Jeff Nanney, Plano East Sr. HS, TX[/b]
[b]Jesse Geneson, T. Jefferson HS-Sci/Tec, VA[/b]
[b]Thomas Mildorf, T. Jefferson HS-Sci/Tec, VA[/b]
Andrew Bolanowski, Rufus King HS, WI
[/hide]

[hide="35"]
John Gipson, Randolph School, AL
[b]David Harris, Vestavia Hills HS, AL[/b]
Don Ding, Little Rock Central HS, AR
Charles Wei, Highland HS, AZ
[b]Tao Chen, Cupertino HS, CA[/b]
[b]Yingkun (Roger) Li, Mark Keppel High School, CA[/b]
Justin Ng, San Marino HS, CA
[b]David Philipson, Harvard-Westlake Sch, CA[/b]
[b]Yi Sun, The Harker School, CA[/b]
[b]Andy Scacco, Smoky Hill HS, CO[/b]
Bill Magin, Brookfield HS, CT
Seth Meltzer, Brookfield HS, CT
Paul Salcuni, Brookfield HS, CT
Jeff Birnbaum, Stoneman Douglas HS, FL
Danny Puller, Stoneman Douglas HS, FL
[b]Nimish Ramanlal, Seminole HS, FL[/b]
Andrew Westerdale, Stoneman Douglas HS, FL
Aaron Wysocki, Buchholz HS, FL
Joshua Moloney, Ames High School, IA
[b]Scott Abrams, Naperville North HS, IL[/b]
Carolyn Auchter, Barrington HS, IL
[b]Cliff Chang, New Trier HS, IL[/b]
Patrick Eibl, Barrington HS, IL
Jennifer Iglesias, West Aurora HS, IL
Aviad Rubinstein, Bloomington HS South, IN
[b]Sway Chen, Lexington HS, MA[/b]
[b]Boping Lai, Roxbury Latin School, MA[/b]
Daniel Suo, Lexington HS, MA
Miles Kaufmann, Traverse City Central HS, MI
Stephanie Cheng, Mounds View HS, MN
[b]Daniel Gibson, Highland Sr. HS, MN[/b]
Patrick Miles, Mounds View HS, MN
Mikhail Lavrov, Enloe HS, NC
[b]Richard Walsh, Manchester West HS, NH[/b]
Ben Hsieh, AAST (11-12), NJ
[b]Kevin Lee, AAST (9-10), NJ[/b]
Rajesh Ramakrishman, AAST Mu, NJ
Anna Shneidman, AAST (11-12), NJ
Michael Ty, AAST (9-12), NJ
Alex Yang, W. Windsor-Plainsboro HS North, NJ
[b]Hooyoung Chung, Stuyvesant HS, NY[/b]
Teddy Singer, Scarsdale HS, NY
Andrew Violino, Holy Trinity HS, NY
Barry Weng, Bronx HS of Science, NY
[b]Danny Zhu, Stuyvesant HS, NY[/b]
Greg Ebersole, Strongsville HS, OH
Aaron Friedman, Sycamore HS, OH
Jon Krause, Strongsville HS, OH
John Bauman, Boyertown Area HS, PA
Garrett Goldfinger, East Stroudsburg HS North, PA
[b]Vincent Roscioli, Freedom HS, PA[/b]
[b]Ameya Velingker, Parkland Sr. HS, PA[/b]
[b]Jeffrey Chen, A & M Consolidated HS, TX[/b]
Yuhong Liu, Pearland HS, TX
[b]Alex Schwendner, Austin Area Homeschoolers, TX[/b]
Kevin Welch, Austin Area Homeschoolers, TX
[b]Tim Pollio, Blacksburg HS, VA[/b]
[b]Daniel Schafer, T. Jefferson HS-Sci/Tec, VA[/b]
Daniel Graves, Nicolet HS, WI
[/hide]

[hide="34"]
Lawrence Watts, Little Rock Central HS, AR
Peiran Gao, University HS, CA
Albert Liu, Fremont HS, CA
Macky Mackisumi, Polytechnic School, CA
Neel Shah, Chadwick School, CA
Jason Shu, Mission San Jose HS, CA
Elizabeth Tang, William S. Hart HS, CA
[b]Henry Tung, Torrey Pines HS, CA[/b]
Timmy Wong, Fremont HS, CA
Seth O'Brien, Monarch HS, CO
Leonid Ovanesyan, Smoky Hill HS, CO
Edwin Chen, Choate Rosemary Hall, CT
[b]Khoa DoBa, Taft School, CT[/b]
[b]Arthur Huang, The Charter School of Wilmington, DE[/b]
Andrew Brannon, H. B. Plant HS, FL
Daniel Levy, Stoneman Douglas HS, FL
Miorel Palii, Eastside HS, FL
Andy Davenport, Westminster Sch, GA
William Drobny, Parkview HS, GA
Robert Gaines, Southside Comprehensive HS, GA
Shawndria Miller, Southside Comprehensive HS, GA
Josh Wilkinson, Grayson HS, GA
Livia Zarnescu, Alan Pope HS, GA
Chris Chang, Naperville Central HS, IL
Sam Dittmer, Zionsville Comm. HS, IN
Fern Sha, Indiana Acad-Sci/Math, IN
Fuhe Xu, Harrison HS, IN
[b]Nick Fortino, Phillips Academy, MA[/b]
Tom Goldstein, Hudson HS, MA
Sean Kessler, Xaverian Brothers HS, MA
Pranau Nanda, Roxbury Latin School, MA
[b]Jacob Sharpe, Canton HS, MA[/b]
Nate Sharpe, Canton HS, MA
David Wen, Lexington HS, MA
Andrew Drechsler, Hereford HS, MD
Jason Kleban, Sidwell Friends Sch, MD
Lidang Li, River Hill HS, MD
Jessica Yuan, River Hill HS, MD
Lou Zellinger, Maine Sch. of Sci./Math, ME
David Budde, Catholic Central HS, MI
Dawson Zhou, Troy HS, MI
[b]Sam Beck, Jackson Preparatory Sch, MS[/b]
Vivek Bhattacharya, Enloe HS, NC
Don Lackman, St. Paul's Sch, NH
Christine Chang, AAST (9-12), NJ
Hyun Soo Cho, AAST (9-12), NJ
Chang Hoon Hahn, AAST (9-10), NJ
Edward Hong, AAST (9-10), NJ
Edward Hong, AAST (9-12), NJ
Ben Huan, AAST (9-10), NJ
Joe Huan, AAST (9-10), NJ
LeeYun Kim, AAST (9-12), NJ
[b]Dan Le, W. Windsor-Plainsboro HS South, NJ[/b]
Eunice Nam, AAST (9-12), NJ
Chad Nathanson, Voorhees HS, NJ
David Puthens, AAST (11-12), NJ
Arthur Safira, AAST (9-12), NJ
Ben Yang, AAST (9-10), NJ
Bixiao Zhao, W. Windsor-Plainsboro HS South, NJ
Adrian Zielenski, AAST (9-12), NJ
Yang Bai, Benjamin Cardozo HS, NY
Alex Dubbs, Horace Mann School, NY
Maria Guirguis, Brooklyn Technical HS, NY
[b]Sunny Kam, Great Neck South HS, NY[/b]
Stan Kats, Brooklyn Technical HS, NY
Michael Kruskal, Harrison HS, NY
Christopher Lim, Trinity School, NY
Steven Margolin, Scarsdale HS, NY
Aaron Bendix-Balgley, Seven Hills School, OH
[b]Eric Mecklenberg, Hawken School, OH[/b]
[b]Aaron Pollack, Liberty HS, OH[/b]
[b]Mitchell Wang, Lakota West HS, OH[/b]
Piotr Fidkowski, Emmaus HS, PA
Marshall Grinstead, Strath Haven HS, PA
[b]Maria Monks, Hazleton Area High School, PA[/b]
Kurt Schneider, Wissahickon Sr. HS, PA
[b]CJ Smith, Hempfield HS, PA[/b]
[b]Zachary Abel, Greenhill School, TX[/b]
Jonathan Einav, Lyndon B. Johnson HS, TX
Erik Feng, Plano East Sr. HS, TX
Drew Fisher, Texas Academy of Math & Science, TX
Jessica Wang, A & M Consolidated HS, TX
Samson Zhou, A & M Consolidated HS, TX
[b]Peter Diao, T. Jefferson HS-Sci/Tec, VA[/b]
[b]Menyoung Lee, T. Jefferson HS-Sci/Tec, VA[/b]
Frederick Mako, T. Jefferson HS-Sci/Tec, VA
Palmer Mebane, Maggie L. Walker Governor's Sch., VA
Winston Yan, T. Jefferson HS-Sci/Tec, VA
Rick Eyers, Homestead HS, WI
[/hide]

Would it be too easy to ask what the bolded names mean?

Congrats! But I don't think we need two topics on the same thing.

Did somebody else post the same thing?

or you are talking about what i wrote on the AMC forum?
hmm.. maybe youre right..

but after I wrote this here, somebody congratulated all people who got 34, 35 or 36
and i thought it was reasonable, since one or two mistakes frequently happen -
so i thought it would be nice to see all people who got 34, 35 or 36.

but after that, i was curious about how many people among them were USAMO qualifiers, and I bolded their names -
so I would like to people who goes on the AMC forum to see this post also.
because it seems like many more people goes on the AMC forum than here.
plus, i also noticed that this post is not going to be shown in "last post" on the main page of Forum.

that's why i wrote that on AMC forum also.
I agree its redundant, but i dont think its a big deal.

I was actually referring to Joml88's topic on this forum, a little further down. But it's nice that you're congradulating everyone.

I think there might be some errors in the results. For [url=http://www.mathleague.com/reports/2004_05/MD6.HTM]this[/url] one at least.  I mean look at Montgomery Blair.  They have 30's and one 29...and only one person on the high scoring list?

Well, it is possible.
In fact,

[u]6th one[/u]
John Kim, Montgomery Blair HS
Brian Lawrence, Montgomery Blair HS
Randy Li, Montgomery Blair HS
Matt McCutchen, Montgomery Blair HS
Max Wasserman, Montgomery Blair HS

[u]5th one[/u]
John Kim, Montgomery Blair HS
Randy Li, Montgomery Blair HS
Richard McCutchen, Montgomery Blair HS
Alex Mont, Montgomery Blair HS
Jordan Turner, Montgomery Blair HS

[u]4th one[/u]
John Kim, Montgomery Blair HS
Brian Lawrence, Montgomery Blair HS
Matt McCutchen, Montgomery Blair HS
Henry Scher, Montgomery Blair HS
Chealsea Zhang, Montgomery Blair HS

[u]3rd one - they got 29 here [/u]
Matt Jordan, Montgomery Blair HS
Mitchell Katz, Montgomery Blair HS
Brian Lawrence, Montgomery Blair HS
Matt McCutchen, Montgomery Blair HS

[u]2nd one[/u]
Brian Lawrence, Montgomery Blair HS
Eric Ma, Montgomery Blair HSl
Matt McCutchen, Montgomery Blair HS
Alex Mont, Montgomery Blair HS
Renee Park, Montgomery Blair HS

[u]1st one[/u]
John Kim, Montgomery Blair HS
Brian Lawrence, Montgomery Blair HS
Benjamin Lee, Montgomery Blair HS
Randy Li, Montgomery Blair HS
Matt McCutchen, Montgomery Blair HS

However, suspicion comes from different place.
Since both Brian Lawrence and Matt McCutchen were on the list who got 6 five times each, they had to have 30~35.
However,
1) Matt McCutchen got 36
2) Brian Lawrence didnt get 30 - there is a person who got 30 on the final high scorer list, but Brian Lawerence wasnt.

It is really obvious that there is an error if you look at [url=http://www.mathleague.com/reports/2004_05/MD5.HTM]here[/url]

by the way, it's amazing that everytime they scored 30, they had exactly five perfect scorers.

One of the 6 Math league tests was on the same day as the AIME I. Quite a few people taking AIME that day skipped that math league test. Some of the USAMO qualifiers got 29 out of 5 tests. Other USAMO qualifiers did not do well for that particular math league, since their minds did not work well after 3 hours of AIME.

Why have I not heard of this contest before? :huh: I always seem to miss the important competitions... I had never even heard of the AMC until about a year ago.

[quote="Kent Merryfield"]I see a lot of names I recognize on the California list. Congatulations to all of you, and especially to Olga Mandelshtam, Roger Li, Justin Ng, Robbie Gao, and Macky Makisumi, all of whom scored 34, 35, or 36.[/quote]

Thank you. They misspelled my last name on the website: "Macky Mackisumi."  :(

By the way, can someone post (or point to a linke with) a nice solution for round 2 problem #2?

What was the problem?

[quote]
Why have I not heard of this contest before? Huh? I always seem to miss the important competitions... I had never even heard of the AMC until about a year ago.[/quote]

I don't really classify Math League as one of the more important tests...nowhere near AMC at least.  It's a nice little fun test that gets you (me, at least) out of class for one period 6 times a year :D .

That's funny... my name is up there but I don't remember ever taking this competition.

When was it? Is it just called "math league"?

The name you saw it under would have included the state as well. It's the same questions nationally, but the results and names are broken down by state.

wow...im there as  34......but im AAST so weve got a lot of freshmen with a 34, and i think one 35, but i only glanced at the people

Why do they always take so long to post the scores... though I seem to remember they were really far behind last year.

I wonder if they'll notice that 3 of the Indiana scores mentioned aren't exactly highschoolers :)

w/e, they said we were allowed to take it.

do you get any prizes(preferably math books) for getting a perfect score?

i believe you get a plack for highest score and that's about it :(

Yep, you get a nice looking plaque that you can hang on your wall.

you got 107/108 from 00-01 to 02-03!!

great job! :clap2:
im not surprised tho. :)
(I'm wondering if you remember which mistake did you make?)

Haha, no, I don't.  Not only that, but the first 5 contests of the year I missed a question were so easy (>135, >145, >125, >120, >75 perfect scores in New York State) that I can't find out which exam had the question I missed.  All I know is that I was one of the 5 people who got a perfect on the last one :)

Also, there is a middle school version of the test for which I won a plaque twice -- the first time, as a fifth grader in the sixth grade test.

Okay, enough massaging of my ego.  Going back through the middle school history is fun, though -- I can see a number of my friends then and friends now who did very well on those exams.


I came across this interesting little puzzle that I can't seem to figure out:

The code used is similar to ROT13

The person who made the code calls it "Leap Year" offset substitution code.

A=L while Z=Z.

Use the code to decrypt the following coordinates:

JKOSCSVKJDJYSCOYYQDWMKDJSSVKYDBCSZYOKVYQSZYOKJDJYRDUYSCOYYZYOKMKDJSRDUYKJYZYOK

I can only assume that due to the fact that these are coordinates, that the encrypted phrase may also contain numbers, but not sure of that. Can anyone here help decypher this for me?

I think its somewhere in Texas and that the N(orth) coordinate might be close to 29 Degrees.

Thanks.

i know the answer, but where does the puzzle come from? it looks like it's from some sort of contest.

It's from http://www.geocaching.com.

If you'd rather not post the answer, could you give me a clue in decyphering it?

Thanks.

It's just a simple substitution code, with keyword LEAP YEAR.

In other words, the ciphertext corresponding to the plaintext alphabet is LEAPYRBCDFGHIJKMNOQSTUVWXZ.

Thanks, I got the answer with your help. I'm going to have to figure out an easier way to do this tuff.

Thanks again.


Let a,b,c >0; Find Max and Min of: $ P\equal{}\frac{a}{a\plus{}b}\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}a}$

$ 1 < P < 2$ One bound implies the other.
note that $ P\geq \frac {a \plus{} b \plus{} c}{max(a \plus{} b,b \plus{} c,c \plus{} a)}\geq 1$
to see why it can not be improved let $ a \equal{} b^2$ and see $ b\rightarrow 0$

[quote="cdsppy"]Let a,b,c >0; Find Max and Min of: $ P \equal{} \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a}$[/quote]

Let a,b,c >0, n>1.
Can Find Max and Min of: $ P \equal{} (\frac {a}{a \plus{} b})^{n} \plus{} (\frac {b}{b \plus{} c})^{n} \plus{} (\frac {c}{c \plus{} a})^{n}$


Using only the digits 7, 8 and 9, how many positive seven-digit integers can be made that are palindromes?

Can you use only one or two of the numbers as well as using all three to create the seven-digit numbers, or do all three numbers have to be used to create the seven-digit numbers? I'm just trying get a feel for the intent of the problem.

example using only one number: 7777777
example using only two numbers: 8989898
example using all three numbers: 7899987

I think it's 1, 2, or 3, because if you use 1 or 2, you are just using the unused digits 0 times.

The palindrome is abcdcba (each representing a digit). Each digit has 3 choices, so 3^4=81


That $ \angle{FED} = 90^o$ is equivalent to $ \angle{APC} = 90 + B$ (Just making some computations using that: $ AFPE$ & $ EPDC$ are cyclic. Making the correct computations, we get: $ \angle{AQC} = 90^o$. $ AQ$ is the isogonal conjugate of $ AP$ (Diameter of $ \triangle{APE}$ $ \Rightarrow$ $ AQ\perp FE$ analogously: $ CP\perp ED$, by Carnot, with points $ ABC$ and $ \triangle{FED}$, (and $ Q$) we get $ BQ\perp FD$ (it can also be seen because the isogonal conjugates of $ AP,BP,CP$ must concur, so $ BQ$ must be the isogonal conjugate of $ BP$). $ \triangle{AQC}\equiv \triangle{PDC}$ $ \Rightarrow$ $ \frac {QC}{AC} = \frac {DC}{PC}$ but $ \angle{QCD} = \angle{PCA}$ $ \Rightarrow$ $ \triangle{QCD}\equiv \triangle{ACP}$ then $ \angle{DQC} = \angle{PAC}$ analogously: $ \angle{AQF} = \angle{PCA}$ then $ \angle{AQD} = 180^o - B$ so $ Q$ is the orthocenter of $ \triangle{BFD}$.

[quote="Mathpro"]Let $ P$ and $ Q$ be interior points of triangle $ ABC$ such that
$ \angle ACP \equal{} \angle BCQ$and $ \angle CAP \equal{} \angle BAQ$. Denote by $ D;E$ and
$ F$the feet of the perpendiculars from$ P$ to the lines$ BC, CA$
and $ AB,$ respectively. Prove that if$ \angle DEF \equal{} 90^o$ then $ Q$ is the
orthocenter of triangle$ BDF$[/quote]
[hide="Solution"]
Notice that\[ \angle ECD \equal{} 180 \minus{} \angle DEC \minus{} \angle EDC \equal{} 90 \minus{} \angle DEC \equal{} \angle PEC \minus{} \angle DEC \equal{} \angle PED\]and since $ \angle PEC \plus{} \angle PDC \equal{} 90 \plus{} 90 \equal{} 180$, we have that $ PECD$ is cyclic.  Therefore, $ \angle ECD \equal{} \angle PED \equal{} \angle PCD$, so $ P$ lies on $ CE$, which means that $ P$ lies on $ AC$.  Then, $ P \equal{} E$.  Now, $ \angle FPD \equal{} 90 \equal{} \angle PFB \equal{} \angle PDB$, so\[ \angle ABC \equal{} \angle FBD \equal{} 360 \minus{} \angle PFB \minus{} \angle BDP \minus{} \angle FPD \equal{} 90\]Since $ \angle BAQ \equal{} \angle CAP \equal{} 0$ and $ \angle QCB \equal{} \angle PCA \equal{} 0$, we have that $ Q$ is the intersection between $ AB$ and $ BC$, which is $ B$.  Therefore, $ Q \equal{} B$, and $ B$ is the orthocenter of $ \triangle ABC$ since $ \angle ABC \equal{} 90$. [/hide]


geometry trapezoid parallelogram ratio invariant rhombus symmetry

A circle is inscribed in the trapezoid [i]ABCD[/i]. Let [i]K, L, M, N[/i] be the points of tangency of this circle with the diagonals [i]AC[/i] and [i]BD[/i], respectively ([i]K[/i] is between [i]A[/i] and [i]L[/i], and [i]M[/i] is between [i]B[/i] and [i]N[/i]). Given that $AK\cdot LC=16$ and $BM\cdot ND=\frac94$, find the radius of the circle.

[color=red][Moderator edit: A solution of this problem can be found on http://www.ajorza.org/math/mathfiles/scans/belarus.pdf , page 20 (the statement of the problem is on page 6). The author of the problem is I. Voronovich.][/color]

Let $P, Q, R, S$ be the tangency points of the sides $AB, BC, CD, DA$ with the trapezium incircle $(I)$ and $r = IP = IR$ the incircle radius. Denote $a = AP = AS$, $b = BP = BQ$, $c = CR = CQ$, $d = DR = DS$. According to the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23514]Trapezium[/url] in the Olympiad Section, Geometry, Unsolved Problems, the products $k = AK \cdot CL$, $l = AL \cdot CK$, $m = BM \cdot DN$, $n = BN \cdot DM$ are given by

$l, k = ac + 2r^2 \pm 2r \sqrt{ac + r^2}$

$n, m = bd + 2r^2 \pm 2r \sqrt{bd + r^2}$

Let $Z$ be the intersection of the lines $BC, DA$ and $z$ a line through the point $Z$ parallel to the lines $AB \parallel CD$. There exists a central projection, which carries the circle $(I)$ into itself and the line $z$ to infinity. Since no finite point of the lines $AB, CD$ is carried to infinity, the projected lines $A'B' \parallel C'D'$ remain parallel. Since the point $Z$ is carried to infinity, the projected lines $B'C' \parallel D'A'$ become parallel and the trapezim $ABCD$ is projected to a parallelogram $A'B'C'D'$. At the same time, this parellelogram has the incircle $(I)$. Hence, it is a rhombus.

The cross-ratios $\frac{\frac{AK}{AL}}{\frac{CK}{CL}} = \frac{\frac{A'K'}{A'L'}}{\frac{C'K'}{C'L'}}$ and $\frac{\frac{BM}{BN}}{\frac{DM}{DN}} = \frac{\frac{B'M'}{B'N'}}{\frac{D'M'}{D'N'}}$ are preserved in this central projection, i.e.,

$\frac{AK \cdot CL}{AL \cdot CK} = \frac{A'K' \cdot C'L'}{A'L' \cdot C'K'}$

$\frac{BM \cdot DN}{BN \cdot CM} = \frac{B'M' \cdot D'N'}{B'N' \cdot D'M'}$

Denoting similarly $a' = A'P'$, $b' = B'P'$, $c' = C'R'$, $d' = D'R'$ and $k' = A'K' \cdot C'L'$, $l' = A'L' \cdot C'K'$, $m' = B'M' \cdot D'N'$, $n' = B'N' \cdot D'M'$, we have

$\frac k l = \frac{ac + 2r^2 - 2r \sqrt{ac + r^2}}{ac + 2r^2 + 2r \sqrt{ac + r^2}} = \frac{a'c' + 2r^2 - 2r \sqrt{a'c' + r^2}}{a'c' + 2r^2 + 2r \sqrt{a'c' + r^2}} = \frac{k'}{l'}$

$\frac m n = \frac{bd + 2r^2 - 2r \sqrt{bd + r^2}}{bd + 2r^2 + 2r \sqrt{bd + r^2}} = \frac{b'd' + 2r^2 - 2r \sqrt{b'd' + r^2}}{b'd' + 2r^2 + 2r \sqrt{b'd' + r^2}} = \frac{m'}{n'}$

This is possible only if $ac = a'c'$ and $bd = b'd'$. This implies that the products $k, l, m, n$ are also invariant under this central projection, i.e., $k = k', l = l', m = m', n = n'$. Hence, we can find the radius $r$ of the incircle $(I)$ just for the rhombus $A'B'C'D'$. From now on, all primes in the notation are dropped and we consider the trapezium $ABCD$ to be a rhombus. From symmetry of this rhombus, $a = AP = CR = c$, $b = BP = DR = d$ and $AK = CL$, $AL = CK$, $BM = CN$, $BN = CM$.

$l, k = AK^2, AL^2 = a^2 + 2r^2 \pm 2r \sqrt{a^2 + r^2}$

$n, m = BM^2, BN^2 = b^2 + 2r^2 \pm 2r \sqrt{b^2 + r^2}$

In addition, the diagonals $AC, BD$ of the rhombus intersect at its incenter $I$ and $AL = AK + 2r$, $BN = BM + 2r$. Consequently,

$k = AK^2 = a^2 + 2r^2 - 2r \sqrt{a^2 + r^2}$

$l = (AK + 2r)^2 = AK^2 + 4r AK + 4r^2 = a^2 + 2r^2 + 2r \sqrt{a^2 + r^2}$

$a^2 + 2r^2 - 2r \sqrt{a^2 + r^2} + 4r \sqrt k + 4r^2 = a^2 + 2r^2 + 2r \sqrt{a^2 + r^2}$

$4r \sqrt k + 4r^2 = 4r \sqrt{a^2 + r^2}$

$\sqrt k + r = \sqrt{a^2 + r^2} = e$

and similarly,

$\sqrt m + r = \sqrt{b^2 + r^2} = f$

where we denoted $e = AI = \sqrt{a^2 + r^2}$, $f = BI = \sqrt{b^2 + r^2}$ the half-diagonals of the rhombus $ABCD$. Since the diagonals fo a rhombus are perpendicular to each other, the angle $\measuredangle AIB = 90^o$ is right and the inradius $r = IP$ its altitude from the vertex $I$ to its hypotenuse $AB$. From similarity of the right angle triangles $\triangle API \sim \triangle IBP \sim \triangle AIB$,

$\frac{AP}{AI} = \frac{\sqrt{e^2 - r^2}}{e} = \frac{IP}{BI} = \frac r f$

$e^2 f^2 = r^2 (e^2 + f^2)$

Substituting the expressions for $e, f$ into the last equation, we get an equation for the inradius $r$ in terms of the products $k, m$:

$(k + 2r\sqrt k + r^2)(m + 2r\sqrt m + r^2) = r^2(k + 2r\sqrt k + r^2 + m + 2r\sqrt m + r^2)$

${r^4 - 4r^2 \sqrt{km} - 2r \sqrt{km}(\sqrt k + \sqrt m}) - km = 0$

Substituting the values $k = 16$, $m = \frac 9 4$, this becomes

$r^4 - 24r^2 - 12r \left(4 + \frac 3 2\right) - 36 = r^4 - 24r^2 - 66r - 36 = 0$

We could look for rational roots of this quartic equation with the help of the rational root theorem, but there are just too many possibilities $(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36)$. It is easier to find the roots numerically and factor out any rational roots we spot. Numerical solution gives 2 real roots $r = 6$ or $r \doteq -0.74$. The rational root can be factored out:

$r^4 - 24r^2 - 66r - 36 = (r - 6)(r^3 + 6r^2 + 12r + 6) = 0$

If the remaining cubic factor is zero, the cubic equation can be written as

$(r + 2)^3 - 2 = 0$

which has a single real root equal to $r =\ ^3\sqrt 2 - 2 \doteq -0.74$. Since the negative root is not acceptable, the inradius of the original trapezium $ABCD$ is $r = 6$.

Thank you for your detailed solution Yetti but since I've no knowledges about projective geometry yet :? , it's not hard to conclude that I didn't understand it at all.If you've got another idea, maybe simpler don't hesitate to post it(...I wanted to say don't laze to :D  ;) )
Best regards
Artiona  :)

Since I love geometric transformations, I try to insert them, whenever I can. I used the following 2 theorems of projective geometry (see, for example, Geometric Transformations III by I. M. Yaglom):

[b]1.[/b] For any circle and a line not having a common point, a central projection exists, which carries the given circle into a circle and the given line to infinity.

[b]2.[/b] The cross ratio of 4 points on a line is preserved in a central projection.

However, there is a far simpler way to arrive to the same quartic equation for the radius $r$ of the incircle $(I)$. The previous solution was based on the formulas for the products $k = AK \cdot CL$,  $l = AL \cdot CK$,  $m = BM \cdot DN$,  $n = BN \cdot DM$ in terms of the inradius $r$ and the tangent lengths $a = AP$, $b = BP$, $c = CR$, $d = DR$ from the vertices $A, B, C, D$ to the incircle $(I)$, obtained in the thread [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23514]Trapezium[/url]. In the current notation, the products $k, l, m, n$ were obtained as the roots of the quadratic equations

$k^2 - 2(ac + 2r^2)k + (ac)^2 = 0$ [color=white].[/color] (with the $\pm$ roots $l, k$)

$m^2 - 2(bd + 2r^2)m + (bd)^2 = 0$ [color=white].[/color] (with the $\pm$ roots $n, m$)

We actually do not have to solve these quadratic equations for $k, m$, because these quantities are given. Instead, we denote $x = ac, y = bd$ and solve these quadratic equations for $x, y$:

$x^2 - 2kx + k^2 - 4kr^2 = 0$

$y^2 - 2my + m^2 - 4mr^2 = 0$

$x = ac = \frac{2k \pm \sqrt{4k^2 - 4(k^2 - 4kr^2)}}{2} = k \pm 2r \sqrt k$

$y = bd = \frac{2m \pm \sqrt{4m^2 - 4(m^2 - 4mr^2)}}{2} = m \pm 2r \sqrt m$

It is clear that we are interested at the roots with the plus signs, because the products $k, m$ can become arbitrarily small, when moving the points $C, D$ on the rays $RC, RD$ and keeping the lines $BC, DA$ tangent to the incircle $(I)$. Thus the roots with the minus signs can become negative, while the products $ac, bd$ are always positive. Hence

$ac = k + 2r \sqrt k$

$bd = m + 2r \sqrt m$

Multiplying these 2 equations, we get

$abcd = (k + 2r \sqrt k)(m + 2r \sqrt m) = km + 2r \sqrt{km} (\sqrt k + \sqrt m) + 4r^2 \sqrt{km}$

But according to the thread [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28864]Looks So Easy[/url] (my last reply, please), the inradius $r$ of a tangential trapezium is given by

$r^4 = abcd$

Combining these 2 results leads directly to the quartic equation for $r$ obtained previously:

$r^4 - 4r^2 \sqrt{km} - 2r \sqrt{km} (\sqrt k + \sqrt m) - km = 0$

After substituting $k = 16$ and $m = \frac 9 4$, this quartic equation has 2 real roots $r = 6 > 0$ and $r =\ ^3\sqrt 2 - 2 < 0$ and only the positive one is acceptable.

Regards, Yetti

A circle is inscribed in the trapezoid $ABCD$. Let $K, L, M, N$ be the points of tangency of this circle with the diagonals $AC$and $BD$ respectively ($K$ is between $A$and $L$ and $M$ is between $B$ and $N$). Given that $AK\cdot LC=16$ and $BM\cdot ND=\frac94$, find the radius of the circle.


function
algebra
polynomial
modular arithmetic
absolute value
algebra unsolved

Let f be a polynomial with integer coeffcients. Let Zn be the set of remainders modulo n
for any integer n >= 2. Note that f can be thought of in a well dened way as a function
f : Zn to Zn.
Find all such polynomials, f, such that for each n > 1 the function f : Zn to Zn is surjective.

Bomb

This is problem 2 from 12'th grade, Romanian NMO 2004 (I was in that contest).

Let $f=a_mx^m+\ldots+a_1x+a_0,\ m\ge 1$. Then, given $p,q$, we have $f(p)-f(q)=(p-q)[a_m(p^{m-1}+\ldots+q^{m-1})+\ldots+a_2(p+q)+a_1]\ (*)$. Since $\mathbb Z_n$ is finite, $f$ being surjective is equivalent to it being injective, so all we need is that whenever $p-q\ne 0\pmod n$, we also have $f(p)-f(q)\ne 0\pmod n$.

Assume $m>1$. Take $p\ne q$ large enough that the second factor of the RHS in $(*)$ is large in absolute value ($>1$ or $<-1$ will do :)), and take $n>p,q$ divisible by the RHS in $(*)$. Then $f$ will not be injective when regarded as a function $:\mathbb Z_n\to\mathbb Z_n$. This means that $m=1$, so $f=ax+b$. Now, for $|a|>1$ we'll find that $f$ does not satisfy the requirements, by writing $(*)$ again (with $m=1$) and doing the same thing as above, so $|a|=1$.

Conversely, it's clear that $f=x+b$ or $-x+b$ are Ok, so these are the solutions.

I might have made some mistakes because I'm a bit in a hurry, but this is the main idea (I know very few people who went to that contest and did not get maximum score for this problem :)).


Solve $ \sqrt {4x \minus{} x^2} > 3 \minus{} x$.

hello, because of the square root we have $ 0\le x\le 4$. If we have  $ 3 \le x$ then the inequality is fulfilled for all real $ x$ with $ 3\le x\le 4$. For $ x < 3$ we can square both sides of the inequality and we get
$ 0 > 2x^2 \minus{} 10x \plus{} 9$ which is for fulfilled for all real $ x$ with $ \frac{5}{2}\minus{}\frac{1}{2}\sqrt{7}<x<\frac{5}{2}\plus{}\frac{1}{2}\sqrt{7}$ so we have as the solution the interval $ \frac{5}{2}\minus{}\frac{1}{2}\sqrt{7}\le x \le 4$ and $ x \in \mathbb{R}$.
Sonnhard.

[hide]The LHS only makes sense for $ x \in [0,4]$

Within these values, if $ x > 3$, then the RHS is nonpositive, so we automatically have that the interval $ (3,4]$ works.

If both are positive, then we can square both sides and compare:

$ 4x \minus{} x^2 > x^2 \minus{} 6 x \plus{} 9$
$ 0 > 2 x^2 \minus{} 10 x \plus{} 9$
$ \frac {5 \minus{} \sqrt {7}}{2} < x < \frac {5 \plus{} \sqrt {7}}{2}$

This interval overlaps with $ (3,4]$, so we take their union as the solution set:

$ \boxed{\frac {5 \minus{} \sqrt {7}}{2} < x \leq 4 }$[/hide]

[b]Edit: @ Sonnhard:[/b] Your inequality for $ x<3$ should be $ 2x^2\minus{}10x\plus{}9$, which does have solutions.

That's correct. I solved the inequality by graghs.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=256529]See here[/url]

hello TZF, thanks for your hint, i have corrected it.
Sonnhard.


A small body A starts  sliding from the height h down an inclined groove passing into a half-circle of radius h/2. Assuming the friction to be negligible, find the velocity of the body at the highest point of its trajectory(after breaking off the grove)
I'm confused about the part of a solution, which is:At point O, the N normal reaction$ \equal{}0$. Moreover, I don't know how to decompose the normal reaction into xy-plane since if the object is on the different position, then the normal reaction will exist and is still $ N$(no $ N.sin{alpha}$ or $ N.cos{alpha}$ as I think it is) :( . You will see what I meant when you see the following figure as well as the solution:

[quote="ffao"][quote="ghjk"]
I'm confused about the part of a solution, which is:At point O, the N normal reaction$ \equal{} 0$.[/quote]

How can there be normal reaction if the object lost contact with the surface? (this happens at point O).[/quote]
Wait! Does the ball come to point  O first and then lose the contact? Then I think you're right :oops: . How about the case when the ball doesn't lose the contact? I meant how can we decompose O into x and y components like we did with gravitational force?

Exactly as we did with gravitational force -- in terms of some angle between the normal force and the x (or y) direction.

The angle $ \theta$ in the solution is a good call ;)


geometry
incenter
parallelogram
geometry unsolved

Let ABC be a triangle with incentre I, and let M be the midpoint of AB. The points P and
Q lie on the sides CA and CB respectively, so that PQ is parallel to AB and tangent at a
point D to the incircle of ABC. Let MI intersect PQ at E. Prove that PD = QE if and
only if AB = 2PQ.

See same problem at this link:
http://www.mathlinks.ro/viewtopic.php?t=184582
http://www.mathlinks.ro/Forum/viewtopic.php?t=5811
http://www.mathlinks.ro/Forum/viewtopic.php?t=112509
http://www.mathlinks.ro/Forum/viewtopic.php?t=106337
http://www.mathlinks.ro/Forum/viewtopic.php?t=15932

 :rotfl:  :rotfl:  :rotfl:

[quote="CHunter"][color=darkred]Let $ ABC$ be a triangle and let $ M$ be the midpoint of the side $ [BC]$ . Let $ P\in (AB)$ , $ Q\in (AC)$

be two points so that $ PQ\parallel BC$ and the line $ PQ$ is tangent at a point $ D$ to the incircle $ w \equal{} C(I)$ .

Denote  the intersection $ E\in MI\cap PQ$ . Prove that $ PD \equal{} QE\ \Longleftrightarrow\ BC \equal{} 2\cdot PQ$ .[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. that $ b > c$ . Denote $ S\in (BC)$ where the $ A$ - exincircle touches the side $ [BC]$ . Prove easily

that $ S\in AD$ and $ \overline {MIE}\parallel\overline {ADS}$ . Thus, $ \frac {DP}{p \minus{} c} \equal{}$ $ \frac {DQ}{p \minus{} b} \equal{}$ $ \frac {PQ}{a} \equal{}$ $ \frac {p \minus{} a}{p} \equal{}$ $ \frac {DP \minus{} DQ}{b \minus{} c}$ and $ ED \equal{} MS \equal{} \frac {b \minus{} c}{2}$ .

$ 1.\blacktriangleright$ $ BC \equal{} 2\cdot PQ\ \Longleftrightarrow\ a \equal{} \frac {2a(p \minus{} a)}{p}\ \Longleftrightarrow\ \boxed {\ 2(p \minus{} a) \equal{} p\ }$ .

$ 2.\blacktriangleright$ $ PD \equal{} QE\ \Longleftrightarrow\ DP \minus{} DQ \equal{} DE$ $ \Longleftrightarrow$ $ \frac {(b \minus{} c)(p \minus{} a)}{p} \equal{} \frac {b \minus{} c}{2}$ $ \Longleftrightarrow$ $ \boxed {\ 2(p \minus{} a) \equal{} p\ }$ .

[b]Remark I.[/b] $ 2(p \minus{} a) \equal{} p\ \Longleftrightarrow\ b \plus{} c \equal{} 3a\ \Longleftrightarrow\ IG\perp BC$ , where $ G$ is the centroid of $ \triangle ABC$ .

Thus, $ BC \equal{} 2\cdot PQ\Longleftrightarrow PD \equal{} QE\Longleftrightarrow IG\perp BC\Longleftrightarrow b \plus{} c \equal{} 3a$ .

[b]Remark II.[/b] Denote the point $ T$ where the $ A$ - exincircle touches the sideline $ AB$ . Then

$ \boxed {P\in MI\ \Longleftrightarrow \ IG\perp AB\ \Longleftrightarrow\ SA\perp ST}\ \Longleftrightarrow\ a \plus{} b \equal{} 3c$ . See [/color][url=http://www.mathlinks.ro/viewtopic.php?t=198162][size=117][color=red][b]here[/b][/color][/size][/url] [color=darkblue]this property.[/color]

[hide="Solution"]
Let the incircle of $ \triangle ABC$ meet $ AB$ at $ F$.  Extend $ CE$ to meet $ AB$ at $ Y$.  Now, from this link, $ MI$ passes through the midpoint of $ CF$: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184582.  Since $ I$ the the midpoint of $ EF$ and let $ N$ be the midpoint of $ CF$, $ NI$ is a midline of $ \triangle FEC$, so $ NI\parallel CE\implies MD\parallel EY$.  Also, $ DE\parallel MY$, so $ DEYM$ is a parallelogram, from which it follows that $ DE = MY$.  Also, since $ DE\parallel FM$ and $ IE = IF$, we see that $ \triangle DEI\cong \triangle MFI\implies FM = DE = MY$.

Now, if $ PQ = \frac {AB}{2}$, then since $ \triangle CPQ\sim \triangle CAB$, we have that $ P$ and $ Q$ are the midpoints of $ AC$ and $ BC$ respectively.  Since $ PE\parallel AY\implies \triangle CPE\sim \triangle CAY$, which means that $ EP$ is a midline of $ \triangle ACY$, so $ E$ is the midpoint of $ CY$.  Hence, $ \frac {PE}{AY} = \frac {CE}{CY} = \frac {1}{2}\implies \frac {PD + DE}{AF + 2DE} = \frac {1}{2}\implies 2PD = AF$.  Also, $ \triangle CEQ\sim \triangle CYB$, so
\[ \frac {EQ}{YB} = \frac {1}{2}\implies 2EQ = YB = BM - MY = AM - MF = AF = 2PD
\]
which means that $ PD = EQ$, as desired.

If $ PD = EQ$, then we have that, as before, $ \triangle CEQ\sim \triangle CYB$ and $ \triangle CEP\sim \triangle CYA$.  Also, $ AF = AM - FM = MB - MY = YB$.  Now,
\[ \frac {EQ}{YB} = \frac{CE}{CY}={\frac {PE}{AY} = \frac {QE + DE}{2DE + YB}\implies EQ\cdot YB + 2DE\cdot EQ = QE\cdot YB + YB\cdot DE\implies 2EQ = YB
}\]
Thus, $ \frac {QC}{BC} = \frac {EQ}{YB} = \frac {1}{2}$, so $ Q$ is the midpoint of $ AB$.  Similarly, $ P$ is the midpoint of $ AC$, and so $ 2PQ = AB$, as desired. [/hide]


factorial
algebra
polynomial
binomial coefficients
combinatorics proposed
combinatorics

This might sound dumb, but it serves as a somewhat nice clearification:

I was using Vandermonde's Identity to prove another identity, and accidentally came across this identity... :D

(n C 2)=n(n-1)/2=(1-n)(-n)/2=(1-n C 2). But isn't this suggesting like 2 C 2 = -1 C 2? Since when were binomial coefficients (n C k) defined to negative n? (Or is it just that I remembered it wrong...cuz I thought you can't find n! for negative n :? )

When you define binomial coefficients, you need not stick on natural numbers. You can define binomial coefficients by the formula

$\dbinom{n}{k} = \frac{n \cdot \left(n-1\right) \cdot \left(n-2\right) \cdot ... \cdot \left(n-k+1\right) }{k \cdot \left(k-1\right) \cdot \left(k-2\right) \cdot ... \cdot 1}$.

And it is clear that the right hand side of this formula is defined for any real number $n$. Hence [i]we can define the binomial coefficient $\displaystyle \dbinom{n}{k}$ for any real number $n$[/i]. Here, however, [i]the number $k$ must still be a natural number[/i] (otherwise, the expressions $n \cdot \left(n-1\right) \cdot \left(n-2\right) \cdot ... \cdot \left(n-k+1\right)$ and $k \cdot \left(k-1\right) \cdot \left(k-2\right) \cdot ... \cdot 1$ would be nonsense).

And actually, with the above definition of $\displaystyle \dbinom{n}{k}$ for any real number $n$ and any natural number $k$, you do have $\displaystyle \dbinom{n}{2} = \dbinom{1-n}{2}$, and in particular, $\displaystyle \dbinom{2}{2} = \dbinom{-1}{2}$.

Above I said that we can extend the number $n$ to all real numbers, but the number $k$ must remain natural. Actually, we can even define the binomial coefficient $\displaystyle \dbinom{n}{k}$ for non-natural $n$ and non-natural $k$ ! But in order to do this, we must leave elementary mathematics and use the Gamma function. See

  http://mathworld.wolfram.com/BinomialCoefficient.html

  Darij

I'm not sure I understood your question but $C_n^p$ is not defined for $n<0$ if you consider it to have a combinatorial meaning (or define it to be $0$).

In another hand, for notations convenience, for real $x$, the number $\frac {x(x-1) \cdots (x-p+1)} {p!}$ is sometimes denoted by $C_x^p$.

Pierre.

If you try to extend the definition of factorial to negative integers, the values are infinite. This gives an inf/inf expression for the binomial coefficient, which doesn't rule out the possibility of a meaningful value. For positive integers k, we can define (n choose k) = n(n-1)...(n-k+1)/k!; this is a polynomial which has meaning for all n.

I'm not sure what combinatorial meaning you give to these, but they make perfect sense in terms of algebra.

Ok. Since we all wrote in the same time, I think we all agree  :D .

Pierre.

Thanks to all of you...now it's crystal clear...(I wasn't looking for any combinatorial meaning, so algebraic manipulations suit me just well.) :lol: