AMC USA(J)MO USAMO trigonometry induction geometry AIME

Hello! I've been lurking here for a while, but this is my first post on the AoPS forums. First, I must just thank this organization for revitalizing my interest in math. Before I came here (and ordered the books), I didn't realize how fun and interesting math could be. The problems I've been exposed to here that require thinking and creativity are much better than the "Formula X for Problem Type Y" problems I've been mostly exposed to in my mathematics education.

Which brings me to my next question. After discovering the AMC, I have a strong desire to make the USAMO next year. I have covered all of the typical high school math curriculum with a little extra - I know trigonometry, induction, basic combinatorics, etc. But I have had little experience with problem solving as such (except for part of a year of geometry which exposed me to proofs - and which has been my favorite part of math so far).

Am I being too ambitious in my goal of making USAMO? Should I set my sights lower? I have set out a rigorous study plan and begun working through the first AoPS book, but I don't know if it will be possible. I know the AIME is pretty difficult. What do you think?

I guess my first question is, how old are you? And my second would be, what math competition experience can you get locally?

you can never be too ambitious.. its always better than not being ambitious enough..
i've heard a quote... it goes like "aim for the moon, even if you miss, you'll get the stars" or something like that... so.. aim high..

"Aim for the moon, for if you miss, you'll still land among the stars."

Well yes, it really depends on your age.  Age is a major factor in determining who gets into USAMO.  If you are freshman and below, it's definately not too late to start prepping seriously for the USAMO.  If you are a sophomore it might be a little more difficult, but still attainable, and if you are a junior and above, then it's frankly very difficult to get into USAMO, unless you work 24/7 for the next eight months.  I've seen many people that set their goal as USAMO only after one year of exposure to contest math during their freshman/sophomore year, and they still made their goal.  So definately don't think that you are too "ambitious", though there is quite a fair amount of stuff you have to prep for USAMO.  If you don't make USAMO at first, be encouraged that many people don't even make AIME on their first try at the AMC system.

Good luck!

I think you would still have a good chance even if you're a junior or senior.

-interesting_move

I am heading into 11th grade this coming school year. I'm homeschooled, so I can tailor my math studies to focus on the math needed for the AMC/AIME and I have the flexability to spend time on it. I plan to take some AoPS online courses in the fall and spring.

I have had "some" exposure to math competitions before, if you count my competing in Mathcounts when I was in 8th grade. However, I was poorly prepared and only made a mediocre showing in the regional competition. I hope to prepare better for the AMC. As far as current opportunities for math competitions, I know of one local math competition that I may be able to compete in that I will investigate further and certainly I will be doing the USAMTS over the next year. Unfortunately, most math contests seem to be team contests and I don't know any fellow homeschoolers in my area who are interested in math. There must be some, but I am not aware of them. I believe there is also an ARML team in my locality, but I don't think they would be interested in someone who is inexperienced in the ways of contest math.

I know it will not be easy, but I am willing to put in the hard work. 24/7 is obviously impossible, but I am going to put in as much time and effort as I can. I don't know if I will be able to make it next year - probably not - but I just have this burning desire to try.

pascal012,

If you are planning to take courses with us and gear your math curriculum toward AMC material then I would suggest looking into our independent study program.

In particular I would suggest taking a look at our Intermediate level Independent Study.  The program includes two classes per semester, reading assignments in problem solving mathematics, many problems from AMC level to early Olympiad level including many like those on the USAMTS, ARML power questions, and Mandelbrot team problems.  The curriculum is based largely on the curriculum covered by the AMC and AIME exams and a bit of that from the USAMO.  We will also provide office hours to Independent Study students to discuss each week's material.  We in fact design the curriculum to focus on everything on the AMC/AIME that is not well-covered in ordinary textbooks.  Plus a few other topics that teach general mathematical insight and exploration.

If you have any questions, feel free to ask.

[quote="pascal012"]Unfortunately, most math contests seem to be team contests and I don't know any fellow homeschoolers in my area who are interested in math. There must be some, but I am not aware of them.[/quote]

Where are you? I might be able to make some suggestions. I sponsor a very active homeschool math team in Central Texas.

texas137:

I can assure you that I do not live within easy driving distance of Central Texas. In fact, I live in Eastern Pennsylvania.

[quote="tokenadult"]I too would be glad to advise on how to stir up interest for a homeschool math team. Is your family part of a homeschooling support network now?[/quote]

Yes, we are part of our local homeschool support network. Most Mondays during the school year there is an enrichment meeting in which various classes - such as literature, art, foreign languages, and last year a prealgebra class - take place. The main problem I would see in sparking interest in getting people to see that math can be interesting. A sample of comments I have gotten from homeschooling friends about math:

"Math is boring."

"I just can't get math."

"We just want to scrape by with the bare minimum of math." [Said by a parent when I offered to show her the AoPS books as possible math curriculum for her daughter.]

[quote="pascal012"]
A sample of comments I have gotten from homeschooling friends about math:

"Math is boring."

"I just can't get math."

"We just want to scrape by with the bare minimum of math." [Said by a parent when I offered to show her the AoPS books as possible math curriculum for her daughter.][/quote]

I hate those kind of comments.

If they're just going to go by a bare minimum, how in the world are they going to get a decent job. Math is pretty much important in nearly everything you do.

[quote="Oasis"]"Aim for the moon, for if you miss, you'll still land among the stars."
[/quote]

uh isnt the moon closer than the closest star :D

[i]edit[/i]  oo never mind .. u might overshoot
but if u relate this analogy to the mathworld..
then wouldnt it mean he becomes greater in math than he originally wanted to

..which would be a good thing

pascal012,

might i recommend the American High School Internet Mathematics Competition?  (http://www.ahsimc.com or http://www.americanmathcompetition.com)

completely free to enter, and we already have several homeschoolers signed up.  from the sounds of it, the AHSIMC could be a useful to you as a great way to benchmark yourself at the start of the year.

it will be highly interesting to see how homeschoolers fare against public and private high schools - one of many underlying storylines to the inaugural AHSIMC Exam.

certainly, we would welcome your participation...if you aren't already one of the homeschoolers signed up!

best of luck in your continuing math education.


inequalities
real analysis
real analysis unsolved

Study the convergence of $\sum_{n\geq 1}\frac{\sqrt{p_{n+1}}-\sqrt{p_n}}{n^b}$

where $b$ a real number

$p_n$ is n-ème  prime number

Do they ask for an elementary solution? I mean, are we allowed to use the prime number theorem? Something tells me the answer is negative. Edit: actually, we do not need such strong tools, just use Abel summation and some weak bpunds for $p_n$ (the ones that follow from Chebyshev works)/


hello, can you prove this inequality ?

\[ (a\plus{}b\plus{}c)(\sqrt{1\minus{}a^2}\plus{}\sqrt{4\minus{}b^2}\plus{}\sqrt{9\minus{}c^2}) \leq 18\]

I suppose is positive .. but im not sure cuz i don't actually know.
But can you solve them in appropriate interval that $ a,b,c$ would probably be

thnx

Ok I think it holds for all reals.

Notice that $ (1 \minus{} a^2) \equal{} (1 \minus{} a)(1 \plus{} a)$ and $ (4 \minus{} b^2) \equal{} (2 \minus{} b)(2 \plus{} b)$ and $ 9 \minus{} c^2 \equal{} (3 \minus{} c)(3 \plus{} c)$. Then by Cauchy Shwartz, we have

$ \sqrt {1 \minus{} a^2} \plus{} \sqrt {4 \minus{} b^2} \plus{} \sqrt {9 \minus{} c^2} \leq \sqrt {[(1 \minus{} a) \plus{} (2 \minus{} b) \plus{} (3 \minus{} c)][(1 \plus{} a) \plus{} (2 \plus{} b) \plus{} (3 \plus{} c)]} \equal{} \sqrt {(6 \minus{} \sum a)( 6 \plus{} \sum a)} \equal{} \sqrt {36 \minus{} (\sum a)^2}$

Let $ t \equal{} \sum a$, then it suffices to prove that

$ t \sqrt {36 \minus{} t^2} \leq 18$

$ t^2 (36 \minus{} t^2) \leq 18^2$

$ 0 \leq t^4 \minus{} 36t^2 \plus{} 18^2$

$ 0 \leq (t^2 \minus{} 18)^2$

which is clearly true by the Trivial Inequality.

QED


geometry
analytic geometry
graphing lines
slope

Hi guys i got these 2 equation of the line question which i'm not getting.  I'm not good at math so any help would be really apericated.  I'm not sure if this is the correct area to post this question so i apologize, i am new to the forums thanks

Question:

Find the equation of the line through the following pairs of points:

1 A) (3, -2) and (7, -3)

  B) (-2, -2) and (1, 4)

Thank you very much.

No offense, but this probably belongs in a lower forum (somewhere in the Middle School section), but I'm gonna do it anyways.

All you have to know is $y=mx+b$ and that $m=\frac{y_1-y_2} {x_1-x_2}$

[hide="Click for solution to #1"] You plug the points into $m$ (the slope). $m=\frac{-2--3} {3-7}=-\frac{1} {4}$ So now that you know the slope is $-\frac{1} {4}$, plug that back into $y=mx+b$ to get $y=-\frac{1} {4}x+b$. Then you plug any set of x and y in, such as $(3,-2)$ to get $-2=-\frac{3} {4}+b$. Solve that to get $b=-1.25$. Your final line answer will be $\boxed {y=-\frac{1} {4}x-1.25}$[/hide]

I'm purposely not doing the other one because then you won't learn anything. Try to mimic what I did with the first one. Check your answer though.

[hide="Answer to #2"]$y=2x+2$[/hide]

no offence take, but thank you very much for your help! I really apericate it! thank you

I got 1 question though, i'm not too sure where you got y = - 1/4x +b after you found that the slope was 2.

if coorindates are (3, -2) and (7, -3)

isn't that

x1 = 3
x2 = 7

y1 = -2
y2 - -3?

you have it as
x1 = 3
x2 = -2

y1 = 7
y2 = -3

Am i wrong? i'm so confused :(

More generally:
If we have two distinct points: $A=(x_1;y_1) \quad B=(x_2;y_2)$ Then a line through this two points have the equation:

$(x_2-x_1)(y-y_1)=(y_2-y_1)(x-x_1)$

thanks guys, i got the answers and all the work.  Thanks for the help! :)

I'll move it to Getting Started.  The moderators there can move it further down if they wish.


geometry
geometric transformation
reflection
rotation
combinatorics unsolved
combinatorics

How can I find the number of (completed) [url=http://en.wikipedia.org/wiki/Sudoku]Sudokus[/url] of size n^2 X n^2.?

for instance let n=3 .

the answer is :

For the 9 X 9 case the allowed equivalences are: relabeling entries; reflection; rotation;

permutation of blocks of columns 1-3, 4-6 and 7-9;

permutation of blocks of rows 1-3, 4-6 and 7-9;

permutation of columns 1-3; permutation of rows 1-3;

permutation of columns 4-6; permutation of rows 4-6;

permutation of columns 7-9; permutation of rows 7-9.

How could I do the calculation ?   :?:

I think it's an open question for $ n>3$.
See how hard it was to compute this number for $ n\equal{}3$: http://www.afjarvis.staff.shef.ac.uk/sudoku/sudgroup.html


Prove that if $ a,b,c,d\geq0$, then
\[ a^6\plus{}b^6\plus{}c^6\plus{}d^6\plus{}44(abcd)^{3/2}\geq 3(abc\plus{}bcd\plus{}cda\plus{}dab)^2.\]

[quote="pvthuan"]Prove that if $ a,b,c,d\geq0$, then
\[ a^6 \plus{} b^6 \plus{} c^6 \plus{} d^6 \plus{} 44(abcd)^{3/2}\geq 3(abc \plus{} bcd \plus{} cda \plus{} dab)^2.
\]
[/quote]
Try $ a\equal{}1$ and $ b\equal{}c\equal{}d\equal{}2.$ :wink:

Thanks. Is the following correct, Arqady?
\[ a^6\plus{}b^6\plus{}c^6\plus{}d^6\plus{}12(abcd)^{3/2}\geq(abc\plus{}bcd\plus{}cda\plus{}dab)^2.\]

[quote="pvthuan"] Is the following correct, Arqady?
\[ a^6 \plus{} b^6 \plus{} c^6 \plus{} d^6 \plus{} 12(abcd)^{3/2}\geq(abc \plus{} bcd \plus{} cda \plus{} dab)^2.
\]
[/quote]
I think it's true, pvthuan. The Mixing Variables  kills it and there is a stronger inequality. :wink:

The following inequality is true too:
For non-negative $ a,$ $ b,$ $ c$ and $ d$ prove that:
\[ a^6\plus{}b^6\plus{}c^6\plus{}d^6\plus{}28\sqrt{a^3b^3c^3d^3}\geq2(abc\plus{}abd\plus{}acd\plus{}bcd)^2\]


Hello. I was working on a complicated RL circuit problem, and after several algebraic steps, I am supposed to calculate the values of t which make this true:

e^[(-t)/(1.295E-4)] - 2.5406e^[(-t)/(2.289E-4)] + 1.5406 = 0.

Clearly, t = 0 seconds satisfies this equation. However, there should be another value for t which makes this equation true, and I am having trouble determining what it is. I have done all of the physics steps for this question, and it whould be a shame if I got caught up in the algebra and was unable to solve the problem. [b]If anyone could please show me what to do in order to find the second t value I would really really appreciate it.[/b] Thank you!

PS: If it helps, here is the question which yielded the above equation.

1) In Fig. 30-76a, switch S has been closed on A long enough to establish a steady current in the inductor of inductance L1 = 5.45 mH and the resistor of resistance R1 = 23.8 . Similarly, in Fig. 30-76b, switch S has been closed on A long enough to establish a steady current in the inductor of inductance L2 = 3.73 mH and the resistor of resistance R2 = 28.8 . The ratio 02/01 of the magnetic flux through a turn in inductor 2 to that in inductor 1 is 1.44. At time t = 0, the two switches are closed on B. At what time (in s) is the flux through a turn in the two inductors equal?

t = 0 is the only solution of your equation. Rewriting it as

e^[(-t)/(1.295E-4)] + 1.5406 = 2.5406 e^[(-t)/(2.289E-4)]

e^(-7721 t) + 1.5406 = 2.5406 e^(-4367 t)

we have a decreasing function through (0, 2.5406) with limit 1.5406 for t -> infinity on the left and a faster decreasing function through (0, 2.5406) with limit 0 for t -> infinity on the right. Their graphs intersect only at t = 0. But t = 0 is not a solution of the problem, because at t = 0, the 2 magnetic fluxes are different, their ratio being 1.44. It follows that your equation cannot be correct.

$\Phi$ is the magnetic flux and N the number of coil turns. At $t \le 0$,

$\Phi_{10} = \frac{L_1}{N_1} I_{10},\ \Phi_{20} = \frac{L_2}{N_2} I_{20},\ \frac{\Phi_{20}}{\Phi_{10}} = 1.44$

At time $t > 0$ for both circuits,

$IR = U_R = U_L = -N \frac{\text d\Phi}{\text dt} = -L \frac{\text dI}{\text dt}$

$\frac{\text dI}{\text dt} + \frac{R}{L}\ I = 0$

$I = I_0 \exp\left(-\frac R L\ t\right)$

$\Phi = \frac L N\ I = \frac L N\ I_0 \exp \left(-\frac R L\ t\right) = \Phi_0 \exp \left(-\frac R L\ t\right)$

$1 = \frac{\Phi_2}{\Phi_1} = \frac{\Phi_{20}}{\Phi_{10}}\ \exp \left[-\left(\frac{R_2}{L_2} - \frac{R_1}{L_1}\right) t \right]$

$\left(\frac{R_2}{L_2} - \frac{R_1}{L_1}\right) t = \ln \frac{\Phi_{20}}{\Phi_{10}}$

etc.

Thank you for your help. I just have a question about the algebra you used.

In the second to the last line, why did you set the ratios of the flux's equal to 1?

Can you substitute my values into your equation and tell me what answer you get? I substituted them in and I got an answer around 9000 seconds and that seems a little odd. I could be wrong though, but if you tell me what answer you get I would really appreciate it.

Thanks!

I set the ratio of fluxes to 1, because the fluxes are supposed to be equal at the time to be determined. Substituting the given values,

$t = \frac{\ln \frac{\Phi_{02}}{\Phi_{01}}}{\frac{R_2}{L_2} - \frac{R_1}{L_1}} = \frac{\ln 1.44}{\left(\frac{28.8}{3.73} - \frac{23.8}{5.45}\right) \times 10^3} =$

$= \frac{0.365 \times 10^{-3}}{7.72 - 4.37} = \frac{0.365 \times 10^{-3}}{3.35} = 0.109 \times 10^{-3}\ \text{s} = 0.109\ \text{ms}$

Oh I see. Coincidentally, that was the original mistake that I made as well. Thank you very much for your help. I really appreciate it yetti.


A solid box is $ 15$ cm by $ 10$ cm by $ 8$ cm. A new solid is formed by removing a cube $ 3$ cm on a side from each corner of this box. What percent of the original volume is removed?

$ \textbf{(A)}\ 4.5 \qquad
\textbf{(B)}\ 9 \qquad
\textbf{(C)}\ 12 \qquad
\textbf{(D)}\ 18 \qquad
\textbf{(E)}\ 24$

The easiest way to do this is by working with the numbers you are given
[hide]The big box's volume 15*10*8 or 1200 cubic centimeters. Since none of the smaller removed cubes overlap with each other (because all the side lengths are greater than six), their combined volume is $ 8(3*3*3)$, or 216 cubic centimeters. This is $ \frac{216}{1200}$ of the cube, or [u]18%[/u], [b]D[/b][/hide]

A box has 8 corners, each with a volume of $27$. The box also has a total volume of $1200$, so therefore the box without the corners has a volume of $984$. $1200 - 984 = 216$, and since $1\%$ of $1200$ is twelve, the increase is $\frac{216}{12} = \boxed{18}$.


How many liters of pure water must be evaporated from 60 liters of a 3% salt water solution to produce a solution that will be a 5% salt water solution?

We have 3% x 60 = 1.8 liters of salt.
We then have 1.8 = 5%x. (where x is the amount of water remaining)
Solving for x gives us 36.
Then we have 60-36 = $ \boxed {24}$

yay


A truck is traveling due north at 30 km/hr approaching a crossroad. On a perpendicular road a police car is traveling west toward the intersection at 40 km/hr. Both vehicles will reach the crossroad in exactly one hour. Find the vector currently representing the displacement of the truck with respect to the police car.


[quote="shyong"][quote="prototypex24"]What kind of function is $tg$ ?[/quote]

I guess he meant $tangent$ :D[/quote]

Yes tg means tangent

[quote="felipesa"][quote="white_horse_king88"]After looking over several of your problems felipesa, please read the difficulty level which is appropriate for this forum. For example, this problem should belong in Intermediate.[/quote]

What forum is this ?? Intermediate ?So I think i am right.[/quote]

When he said this it was in another forum and got moved to intermediate.

[quote="felipesa"]BE
$3tg3x=[3(lnk)^2-4lnk+2]tgx$

 for what  k interval the equation has solution[/quote]
Is $lnk$ natural log of $k$.  Lets make it prettier:
$3\tan{3x}=[3\ln{k}-4\ln{k}+2]\tan{x}$
try setting $\ln{k}=y$ and $\tan{x}=z$

[quote="felipesa"][quote="santosh_blr"][quote="felipesa"]Could someone solve?thanks[/quote]

Is it   ( 0, e-1 ) U (e 1/3 , e) U (e 7/3,infinity)[/quote]

can you show how?[/quote]

Well, using the identity tan3x = (3tanx - tan3x)/(1-3tan2x),

we can transform the eqn to

 3(3-tan2x)/(1-3tan2x) = 3y2 - 4y + 2 where y= ln(k)

Now substituting tan2x by A and rearranging the above eqn,

we arrive at the following

              A =  (3y-7)(y+1)/(3y-1)(y-1)

But A = tan2x > 0

So (3y-7)(y+1)/(3y-1)(y-1)  > 0

=> y i.e. ln(k) belongs to (-infinity , -1) U (1/3,1) U (7/3, infinity)

=> k belongs to ( 0, e-1 ) U (e 1/3 , e) U (e 7/3,infinity)


what's the limit of f(x) when x approaches positive infinity and negative infinity?
f(x)=2xe^(2x)

Some things a student should just [i]know[/i]:

1. You should know in which direction an exponential tends to infinity and in which direction it tends to zero.

2. You should know that the either the rate of growth or the rate of decay to zero of an exponential function is sufficiently rapid to overwhelm the growth or decay of any mere power of the variable.

In other words, it should take you about five seconds to answer this question. Do it in your head and save your scratch paper for something else that (unlike this) requires a computation.


can someone do this problem, or is it that noone wants to do this problem?

i need some help on this one....

[quote="IntrepidMath"]Albatross is "decking the halls" with ivy. He is changing the oreientation of the ivy. The first one is |, the second \. The pattern is:
| \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ /...
Albatross drops evry perfect square numbered branch.
What is the sum of the the branches he drops from 1 to 4242?
What is the 4242th orientation?[/quote]
If you mean there are 4242 branches, then:
[hide="ANSWER"]
$1^2+2^2+3^2$...$+65^2 = 93655$
After he drops the first, the 4242 will be a | next, if he drops another one, / ... after 65 drops the $4242^{nd}$ one will be a |.[/hide]

EDIT: Do you mean the sum of the branch#??


function
functional equation
number theory
IMO 1996
IMO
algebra
IMO Shortlist

Let $ \mathbb{N}_0$ denote the set of nonnegative integers. Find all  functions $ f$ from $ \mathbb{N}_0$ to itself such that
\[ f(m \plus{} f(n)) \equal{} f(f(m)) \plus{} f(n)\qquad \text{for all} \; m, n \in \mathbb{N}_0.
\]

Its not hard to guess that

$f(x)=x$

works well. Actually, I just don't know if it's the only possible solution for that. It's quite reasonable to observe that an aplication of f sends 0 into itself. And by taking m=0, we opberve that

$f(f(n))=f(n), \forall n$

So i can garantee that there are more fixed points...don't know if that's a nice observation, because I can't conclude much about...but i wanted to discuss about this problem

There are definitely more solutions. I don't have the time to write a detailed solution, and it's not that hard anyway, but I believe the solutions can be described like this:

Take some $r\in\mathbb N_0$, let $f$ map $1,2,\ldots,r-1$ onto some multiples of $r$ arbitrarily, put $f(kr)=kr,\ \forall k\in\mathbb N_0$, and extend the function to the other naturals by setting $f(u+r)=f(u)+r,\ \forall u\in\mathbb N_0$.

Kalva Solution:

Setting $m = n = 0$, the given relation becomes: $f(f(0)) = f(f(0)) + f(0)$. Hence f(0) = 0. Hence also $f(f(0)) = 0$. Setting $m = 0$, now gives $f(f(n)) = f(n)$, so we may write the original relation as $f(m + f(n)) = f(m) + f(n)$.

So $f(n)$ is a fixed point. Let $k$ be the smallest non-zero fixed point. If $k$ does not exist, then $f(n)$ is zero for all $n$, which is a possible solution. If $k$ does exist, then an easy induction shows that $f(qk) = qk$ for all non-negative integers $q$. Now if $n$ is another fixed point, write $n = kq + r$, with $0$ ≤ $r$ < $k$.
Then $f(n) = f(r + f(kq)) = f(r) + f(kq) = kq + f(r)$. Hence $f(r) = r$, so $r$ must be zero. Hence the fixed points are precisely the multiples of $k$.

But $f(n)$ is a fixed point for any $n$, so $f(n)$ is a multiple of $k$ for any $n$.
Let us take $n_1, n_2, ... , n_{k-1}$ to be arbitrary non-negative integers and set $n_0 = 0$.
Then the most general function satisfying the conditions we have established so far is:

      $f(qk + r) = qk + n_rk$ for $0$ ≤$r$ < $k$.

We can check that this satisfies the functional equation.
Let $m = ak + r$, $n = bk + s$, with $0$ ≤ $r, s$ < $k$.
Then $f(f(m)) = f(m) = ak + n_rk$, and $f(n) = bk + n_sk$,
so $f(m + f(n)) = ak + bk + n_rk + n_sk$, and $f(f(m)) + f(n) = ak + bk + n_rk + n_sk$.
So this is a solution and hence the most general solution.

[quote="Philip_Leszczynski"]I'm not sure about this... but it seems right.

[hide]

Put $m,n=0: f(f(0)) = f(f(0)) + f(0)$, so $f(0) = 0$.
Put $m=0: f(f(n)) = f(f(0)) + f(n)$, so $f(f(n)) = f(n)$.
So the original becomes $f(m+f(n)) = f(m) + f(n)$.
If we plug in $f(m)$ for $m$, we have $f(f(m)+f(n)) = f(f(m)) + f(n) = f(m) + f(n)$.
So if we have $m=n$, then $f(2f(m)) = 2f(m)$.
Since $2f(m)$ must be in the range of $f$ for all $m$, we can use a simple induction on $k$ to show that for all $k,m \in \mathbb{N}_0$, $f(kf(m)) = kf(m)$.
Let $f(1)=c$. Then $f(kf(1)) = f(kc) = kf(1) = kc$.
So $f(x)=cx$, where $c=f(1)$.
But then $f(f(x)) = cf(x) = c^2x > cx$. Contradiction if $c \ge 2$ (since $f(f(x)) = f(x)$)
So $c=0,1$.
But if $c=1$, then $f(x)=x$, and we obviously have a contradiction in the original. So $c=0$, and $f(x)=0$ for all x.

[/hide][/quote]

wrong solution.  ;)

[quote="Philip_Leszczynski"]
But then $f(f(x)) = cf(x) = c^2x > cx$. Contradiction if $c \ge 2$ (since $f(f(x)) = f(x)$)
So $c=0,1$.
But if $c=1$, then $f(x)=x$, and we obviously have a contradiction in the original. So $c=0$, and $f(x)=0$ for all x.
[/quote]

This part here seems very strange to me... :huh:
First because you weren't so clear about your contradiction [even though you didn't assume anything to contradict!]...second because $f(x)=x, \forall x\in \mathbb{N}_0$ works very well =]

[quote="e.lopes"]
We can check that this satisfies the functional equation.
Let $m = ak + r$, $n = bk + s$, with $0$ ≤ $r, s$ < $k$.
Then $f(f(m)) = f(m) = ak + n_rk$, and $f(n) = bk + n_sk$,
so $f(m + f(n)) = ak + bk + n_rk + n_sk$, and $f(f(m)) + f(n) = ak + bk + n_rk + n_sk$.
So this is a solution and hence the most general solution.[/quote]
When you are verifying, I think you should not make the assumption $f(f(m))=f(m)$.
If $f(qk+r)=qk + n_r k$, how do you prove that $f(f(m))=f(m)$ ?
Suppose otherwise that $f(f(m))=f(m)$. Then $qk+n_r k = f(qk+r) = f(f(qk+r))=f(qk+n_r k)=qk + n_{n_r k}k$, so  $n_r = n_{n_r k}$ : but why should they be equal ?

[quote="quantumbyte"]Can't we conclude from f(f(n))=f(n), that f(n)=n.[/quote]
Actually we can, but first we must show that $f$ is injective.

Can you explain to me how we could do that?

I don't think so. I have written this statement for the general case.

I hope I am not reviving anything.

[hide="My Solution"]
First, set $m=n=0$. Thus, we have

$f(f(0))=f(f(0))+f(0)$.

Thus $f(0)=0$

Then, set $m=0$.

$f(f(n))=f(n)$

Next, let $m=n=f(b)$

$f(2f(b))=f(f(b))+f(b)$

$f(2f(b))=2f(b)$

Substituting $n=2f(b)$ and $m=f(b)$, we have

$f(3f(b))=f(2f(b))+f(f(b))=2f(b)+f(b)=3f(b)$.

By induction, we realize that

$f(cf(b))=cf(b)=f(cn)=cf(n)$

Thus, if $n$ is in the range of $f(x)$, all integers of the form $cn$ for a positive integer $c$ is also in the range of $f(x)$. Thus, the range of $f(x)$ encompasses all integers.

From this, we get

$f(f(n))=f(n)$

Letting $f(n)=k$,

$\fbox{f(k)=k}$.

This works when $f(b)\neq 0$, so the other solution is $\fbox{f(b)=0}$.[/hide]

Here's a new solution(i guess) by me :

suppose function is not constant(or else f=0) :

it is not hard to get f(0)=0, f(kf(n))=kf(n) (*),f(f(n))=f(n) (**), so if there exists some natural (or zero) i so that f(i)=1,then we will get
f(k)=k by (*), suppose the least [b]natural [/b]number which the function could produce is p, then by (*) we have f(kp)=kp, where k is a natural number, it's not hard to find that f(m+f(n))=f(f(m)) + f(n) is equal to f(m+f(n))=f(m)+f(n) by (**),plugging m=i<p,n=kp , we will get f(kp+i)=kp +f(i) ,so :

set n=kp+i,i<p,
f(f(n)=f(n) so f(kp+f(i))=kp+f(i),by the assumption of the minimality of function, we get f(i)=sp+j ,0<j<p
so f(kp+sp+j)=kp+sp+f(j)=kp+sp+j
so f(j)=j, contradiction with minimality,so in this case we get the least minimality is equal to 1 and so f(k)=k for all naturals,and it's Done.
.....
if j=0,then f(i)=sp for some arbitrary natural(or zero) s....the rest is Done.

[i]typos fixed.[/i]

Is it true?

Let $m=n=0$. Then $f(0+f(0))=f(f(0))=f(f(0))+f(0)$. Then $f(0)=0$. Then put in $m=0$. We get $f(f(n))=f(f(0))+f(n)=f(n). Then $f(m+f(n))=f(m)+f(n). Assume that $a, b, c$ are arbitrary integer nonnegative numbers. Then consider $f(a+b+f(c))$. Using condition we get:
1) $f(a+b+f(c))=f(a)+f(b+f(c))=f(a)+f(b)+f(c)$
2) $f(a+b+f(c))=f(a+b)+f(c)$.
So $f(a+b)=f(a)+f(b)$. It's well-known Cauchy's functional equation. As we have integer numbers, $f(x)=kx$ (k is an arbitrary nonngative number).
As $f(f(n))=f(n)$, $k^2n=kn$, so $k=0$ or $k=1$. So the only solvings are $f(x)=x$ and $f(x)=0$

@tenniskidperson3 : would you please check the end part of my solution? where i get in the last line : "f(i)=sp"?

i think i have a pretty idea why don't we generalize the exercise to be more free with it thing like working in R
in fact it will be more easy to deal with the equation
if we prove that f(1)=1 we re done

[quote="MBGO"]we will get f(kp+i)=kp +i[/quote]

False, we will get $f(kp+i)=kp+f(i)$.

[quote="MBGO"]we get f(i)=sp+j ,0<j<p[/quote]

False again, we can have $j=0$.

[quote="MBGO"]so f(j)=j, contradiction with minimality[/quote]

... unless $j=0$ for all integers.

first one was just a typo,EDITED, second one i just devide the solution in two cases,the case where 0<j and j=0...i the only thing i doubt on it is when f(i)=sp,where i<p

OK, I didn't understand what you were writing.  It looks OK to me now.

[quote="tenniskidperson3"][quote="MBGO"]you can only apply Cauchy's function if and only if f is continouse.
so you're not done.[/quote]

Actually, here it's ok because it's on the integers.  But you're right, McItran is not done.  His issue is here:

[quote="McItran"]1) $f(a+b+f(c))=f(a)+f(b+f(c))$[/quote]

You cannot just take $a$ outside into its own function.  You need justification for that, and since you have none, you are wrong.  You can also check that you are wrong because $f(x)=x$ if $x$ is even and $f(x)=x-1$ if $x$ is odd also works.[/quote]

Oh, I'm sorry  :blush:

[hide="Solution"]Suppose $P(m,n)\implies f(m+f(n))=f(f(m))+f(n)$
$P(0,0)\implies f(0)=0$ and $P(m,0)\implies f(m)=f(f(m))$. So the original equation turns into $f(m+f(n))=f(m)+f(n)$. From now we define this new equation to be $P(x,y)$.

$P(f(m),m)\implies f(2f(m))=2f(m)$ and by induction $f(nf(m))=nf(m)\forall n\in \mathbb N_0\quad (1)$. Suppose $a$ is the gcd of the all elements in the image of $f$. Now pick $c,d$ such that $\gcd (f(c),f(d))=a$. Now the diophantine equation $f(c)x-f(d)y=a$ has solution in positive integers. So using (1) we get, $P(xf(c)-yf(d),yf(d))\implies f(xf(c))=f(a)+f(yf(d))\implies f(a)=a$.

Now we can write the general solution: since $a|f(i)$, so define $f(i)=ac_i$ for $1\leq i<a$, $c_i\in \mathbb N_0$, $f(a)=a$ and for $i=ak+b>a$, define $f(i)=ak+f(b)$ with $b<a$. Very easy to prove this function works. (Can be done as other people did in the previous posts.) [/hide]
[hide="Comment:"]I too fell for $f(n)=n$. After proving $f(a)=a$ I tried too desperately to prove $f(1)=1$, but was in vain. Finally I understood even if $f(1)=0$, we still get functions satisfying the equation. After that, boom! I killed it :D[/hide]

Let $\Omega = \text{Im} (f) $. Throughout my solution, $a,b$ will denote elements of this set.

Notice that $f(m+a)=f(f(m))+a$, this is the statement. Then $f(a+b)=f(a)+b=f(b)+a$ and therefore $f(a)-a$ is constant $\forall a \in \Omega$. Also $a+c=f(a)=f(0+a)=f(f(0))+a=f(0)+a+c$ and so $f(0)=0$. Then if $a=0, c=0$ and so $f(a)=a \forall a \in \Omega$. Notice also $f(m+a)=f(m)+a$.

Notice that if $a,b \in \Omega$ then $f(a+b)=a+b$ so $\Omega$ is additive and if $b > a$ then $b=f(b)=f((b-a)+a)=f(b-a)+a$ then $\Omega$ is "subtractive". We obtain that $\Omega=\{0\}$ or the set of multiples of a positive integer $d$. In the first case, $f=0$ doesn't work. In the other case, we obtain $f(m+kd)=f(m)+kd$ and so the solution is:

Let $d$ be any positive integer and choose any $a_1,...,a_{d-1} \in \mathbb{N}_0$. Then $f(kd+i) = kd+a_i$ for $k \in \mathbb{N}_0$ and $1 \le i \le d-1$ and $f(kd)=kd$.

Let $g$ be an arbitrary positive integer and let $a_1,a_2,\cdots a_{g-1}$ be an arbitrary sequence of nonnegative integers, with $a_0=0$. Any solution in the form
\[
f(pg+r)=pg+a_r
\]
for $0\le r<g$ can easily be shown to be valid. Now we show that any possible function is in this form. Indeed, $P(0,0)$ implies $f(0)=0$, and $P(0,n)$ implies $f(f(n))=f(n)$. Then
\[
f(m+f(n))=f(m)+f(n)
\]
so $f$ is periodic modulo any integer in its range. Let $g>0$ be the gcd of the range of $f$ (noting that the solution $f(x)=0$ identically has been accounted for in the solution set, and so we may assume otherwise). Then some linear combination of elements of the range of $f$ form any multiple of $g$, so in particular
\[
f(m+kg)=f(m)+kg
\]
for any $m,k$. This implies that $f$ is in the above form.

Hmm... quite a nice yet nasty problem- the amount of time I tried to prove $f(n)=n$ is ridiculous.
First, we get over with the standard stuff- $P(0,0): f(f(0))=f(0)+f(f(0))=>f(0)=0$.
$P(0,n): f(f(n))=f(n)$.
Now, it becomes quite hard to play with the function too much, and after you get $f(m+f(n))=f(m)+f(n)$, the only way I found to move forward is noticing that $P(n,n):f(2f(n))=2f(n)$, $P(2f(n),n):f(3f(n))=3f(n)$, and so on- from here we can easily prove by induction $f(tf(n))=tf(n)$ for any non-negative integer $t$. Also notice that $P(m,tf(n)): f(m+tf(n))=f(m)+tf(n)$.
Funnily enough, we're almost done... just choose the smallest non-negative fixed point(if there exists one) $n_0$ such that $f(n_0)=n_0$ - thus $f(tn_0 +m)=f(m)+tn_0$.
Now let's consider other values of $f(n)$ - let one of them be $p => f(p)=p$. Say $kn \leq p <(k+1)n$ for some positive integer $k$- thus $P(p-kn_0,kn_0)$ gives us $f(p)=f(p-kn_0)+kn_0=>f(p-kn_0)=p-kn_0$-thus $ p-kn_0$ is a fixed point, but we assumed $n_0$ to be the smallest such number! Hence $p=kn_0$ for all $p$, or $f(n)=kn_0$ for all $n$ for some integer $k$... letting $n=an_0+b$, $0 \leq b \leq n_0-1$, we get $n_0 \mid an_0 +f(b)=>n_0 \mid f(b)$.

Here I couldn't find any other restriction on $f(b)$, and I spent more time than I'd like to admit trying to prove $n_0=1$ :D ... anyways it turns out you don't need any more restrictions on $f(b)$ [i]or[/i] $n_0$ - hence our general solution becomes $f(an_0+b)=an_0 + f(b), 1 \leq b \leq n_0 -1, n_0 \mid f(b)$, and it's easy to verify this works.

It really is quite a nice problem, and it could be quite deceiving as such a general solution is usually... unlikely (and very frustrating if you're on the wrong track), but still- an IMO P3? Most of the steps are actually quite natural, and the substitutions and even the fixed point method are also quite well known- maybe because it's a pre-2000 question, so functional equations weren't that common, I suppose.

The answer is all functions $f$ such that if $a$ is the greatest common factor of the range of $f$, then $f(aq+r)=aq+f(r)$ whenever $r<a$. It is easy to check that this works: if $m=aq_1+r_1$ and $n=aq_2+r_2$, where $r_1<a$ and $r_2<a$, then
\begin{align*}
    f(m+f(n))&=f(aq_1+r_1+f(aq_2+r_2))\\
    &=aq_1+f(r_1+aq_2+f(r_2))\\
    &=aq_1+aq_2+f(r_1)+f(r_2)\\
    &=f(aq_1+f(r_1))+f(aq_2+r_2)\\
    &=f(f(m))+f(n).
\end{align*}
Now we prove this describes all solutions to the functional equation. Let $P(m,n)$ denote the assertion.

First, $P(0,0)$ gives $f(f(0))=f(f(0))+f(0)$, so $f(0)=0$. Then $P(m,0)$ gives $f(m)=f(f(m))$. Hence the functional equation rewrites as \[f(m+f(n))=f(m)+f(n),\]
so $f$ is periodic modulo every element of the range of $f$. The greatest common factor $a$ of the range of $f$ is expressible as a linear combination of the range of $f$, so $f$ is also periodic modulo $a$. This completes the proof.

[hide=Solution]
By substituting $m=n=0$, we get $f(0)=0$, so $f(f(0))=0$. Now, substituting $m=0$ gives
$$f(f(m))=f(n).$$ Therefore, $f(x)=x$ if $x$ is in the range of $f$. This also means that the original equation is equivalent to $$f(m+f(n))=f(m)+f(n).$$ This implies that if $a$ and $b$ are in the range of $f$, then their sum is also in the range of $f$. We claim that the range of $f$ is all nonnegative multiples of a certain number $d$. If $f(x)\geq f(n)$, we can substitute $m=f(x)-f(n)$ to get $f(x)=f(f(x)-f(n))+f(n)$, or $f(f(x)-f(n))=f(x)-f(n)$. This means that the nonnegative difference of any two numbers in the range of $f$ is also in the range of $f$. Therefore, the range of $f$ must be all nonnegative multiples of $d$.

Now, let $n$ be an integer such that $f(n)=d$. Then, $f(m+d)=f(m)+d$. Therefore, if $x=qd+r$ where $0\leq r\leq d-1$, then $f(x)=qd+f(r)$. Since $d\mid f(x)$, we must have $d\mid f(r)$. Let $x_i$ be nonnegative constants with $0\leq i<d$ such that $x_0=0$ and $x_i\in\mathbb N_0$. Then, every possible function $f$ must be in the form $f(qd+r)=d(q+x_r)$. We claim that all functions of this form work. By definition, $f(m)$ is a multiple of $d$, and $f(qd)=d(q+x_0)=qd$, so $f(f(m))=f(m)$. Let $f(n)=da$ and let $m=qd+r$. Then, $f(m+f(n))=f((q+a)d+r)=qd+ad+x_r=f(m)+f(n)=f(f(m))+f(n)$.

Therefore, the only functions $f$ are $\boxed{f(qd+r)=d(q+x_r)}$ with $x_0=0$, $0\leq r\leq d-1$, and $q,d,r,x_r\in\mathbb N_0$, where $d$ and $x_r$ are fixed.

The answer is $f(x)=a\left\lfloor\frac xa\right\rfloor+g\left(x-a\left\lfloor\frac xa\right\rfloor\right)$ for some $a\in S$, where $g:\{0,1,\ldots,a-1\}\to\mathbb N$ satisfies $g(0)=0$. It's easy (but not trivial) to see that these work.

To see that these are all of the solutions:
$P(0,0)\Rightarrow f(0)=0$
$P(x,0)\Rightarrow f(f(x))=f(x)$
$P(m,n)\Rightarrow f(m+f(n))=f(m)+f(n)$

[hide = Solution]
The answer is $f(n) = 0$ for all $n$, and all functions of the following form: let $a$ be a positive integer, and let $c_0$, $c_1$, \ldots, $c_{a - 1}$ be integers such that $c_0 = 0$, and $i + c_i$ is nonnegative and $0$ mod $a$ for all $1 \leq i \leq a - 1$. Then for any $n$ with remainder $r$ mod $a$, \[f(n) = n + c_r.\] To show such functions work, note that $f(n) \equiv r + c_r \equiv 0 \pmod{a}$ for all $n$, so if $r$ is the remainder of $m$ mod $a$, then it is also the remainder of $m + f(n)$ mod $a$. So then \[f(m + f(n)) = m + f(n) + c_r = f(f(m)) + f(n),\] since $f(x) = x$ for all multiples of $a$ and $f(m) = m + c_r$ is a multiple of $a$.

Now we show that any $f$ is of this form.

Let the given assertion be $P(m, n)$. By $P(0, 0)$ we get \[f(f(0)) = f(f(0)) + f(0) \implies f(0) = 0.\] Then by $P(m, 0)$ we get \[f(m + f(0)) = f(f(m)) + f(0) \implies f(m) = f(f(m)).\]

Let $T$ be the image of $f$. Then $T$ is exactly the set of fixed points -- we showed above that every $t \in T$ is a fixed point, while every fixed point is in $T$ because if $f(x) = x$, then $x$ is in the image of $f$.

[color=#0B5394][b]Claim:[/b][/color] $T$ is closed under addition.

[i]Proof.[/i] Consider $m$ and $f(n)$ in $T$. Then by $P(m, n)$ we have \[f(m + f(n)) = m + f(n)\] (using the fact that $f(m)$ is a fixed point), so $m + f(n)$ is a fixed point. $\blacksquare$

If $T = \{0\}$ we must have $f(n) = 0$ for all $n$. Otherwise, let $a$ be the smallest positive integer in $T$. Then $a = f(b)$ for some $b$, so by $P(m, b)$ we have \[f(m + a) = f(m) + a\] for all nonnegative integers $m$.

Since $f(0) = 0$, this means all multiples of $a$ must be in $T$. Meanwhile, if $m \in T$, then $f(r) = r$ where $r$ is the remainder of $m$ mod $a$, so we must have $r = 0$ (or else it is a smaller positive integer). So then $T$ is exactly the set of multiples of $a$, and $f(m + a) = f(m) + a$, so taking $c_r = f(r) - r$ gives that $f$ must be of the form described.
[/hide]

The only functions are of the form $$f(qd+r) = qd + a_rd,$$ where $q$ ranges over all nonnegative integers $0 \leq r<d$, $a_i$ is an arbitrary sequence of nonnegative integers for $1 \leq i < d$ and $a_0=0$, and $d$ is a fixed nonnegative integer constant. First, we verify that these functions do indeed work. Let $m = q_0d+r_0$ and $n = q_1d+r_1$, such that $$f(m+f(n)) = f(q_0d+r_0+q_1d+a_{r_1}d) = (a_{r_1}+q_1+q_0 + a_{r_0})d,$$ and $$f(f(m)) + f(n) = f(q_0d+a_{r_0}d)+q_1d+a_{r_1}d = (a_{r_1}+q_1+q_0+a_{r_0})d,$$ so both sides are indeed equal.

Now we show these are the only such functions.

[b][color=#f00]Claim.[/b][/color] $f(0) = 0$.

Simply letting $m=n=0$ suffices. $\blacksquare$

[b][color=#f00]Claim.[/b][/color] $f(f(n)) = f(n)$ for all nonnegative $n$.

Let $m=0$ and use the first claim. $\blacksquare$

In particular, let $d$ be the smallest number in the range of $f$. Then by the second claim, $f(d) = d$. It follows that $$f(m+d) = f(m+f(d)) = f(f(m)) + f(d) = f(m) + d.$$ Thus, the values of $f(x)$ for $x \equiv i \pmod d$ for every $0 \leq i < d$ form an arithmetic sequence, and in particular $f(kd) = kd$ for any positive integer $k$. This yields the desired class of functions.

$P(0,0)\implies f(0)=0\implies f(f(n))=f(n),$ which is a fixed point.
$P(x,y)\implies f(m+f(n))=f(m)+f(n).$ It's easy to see $\boxed{f\equiv 0},$ is a working solution.
Suppose $f(x)\neq 0$ for some $x\in \mathbb{N}.$ Let $p$ be the least fixed point of $f.$ It follows, $2f(p)=f(p+f(p))=f(2p).$ By induction, $f(np)=np \forall n\in \mathbb{N}_0.$
Case 1: $p=1.$
Then $\boxed{f(x)=x}$ for some $x$, which is indeed a fitting solution.
Case 2: $p>1.$
Let $q$ be any fixed point of $f.$ Then, $q=cp+a,$ where $c\in \mathbb{N}_0, 0\leq a<p \implies cp+a=cp+f(a) \implies a=0.$ Thus $a=cp.$
Notice $f(f(u))=f(u) \implies f(u)=pn_u \forall u, 0\leq u <b,$ where $n_u$ is positive integer and $n_0=0.$
If $n=cp+u,$ then $f(u)=(c+n_u)p.$
It can also be noticed that for any $p>1, n_0=0$ and $n_1,n_2,\dots n_p-1\in \mathbb{N}_0$ the function $\boxed{f(n)=([\frac{n}{p}] +n_u)p},$ is yet another working solution. And we are done!

First, letting $m=n=0$, we obtain that $f(0)=0$. Letting $n=0$ only, we have that $f(f(n))=f(n)$. Let $P(x,y)$ denote the reduced assertion that $f(m+f(n)) = f(m)+f(n)$.

Note that, for any $m,n$ with $f(m)\geq f(n)$,

\begin{align*}P(f(m)-f(n),n)&\implies f(f(m)-f(n)+f(n))=f(f(m)-f(n))+f(n)\\
&\implies f(f(m)-f(n)) = f(m)-f(n) = f(f(m))-f(f(n)).\qquad (\clubsuit)
\end{align*}
For $P(f(m), n)$, we obtain $f(f(m)+f(n)) = f(f(m))+f(n) = f(m)+f(n) \quad (\spadesuit)$.

Let $k$ be the smallest nonzero value in the image of $f$, and suppose that $f(N)=k$. Note that $f(f(N))=f(N) = k\implies f(k)=k$.

We'll first show that $f(nk)=nk$ by induction: it suffices to show that $f(\alpha_n)=nk$ for some $\alpha_n$, since computing $nk=f(\alpha_n)=f(f(\alpha_n))=f(nk)$ would finish. We already have $f(k)=k$. Also, we have that $$(\spadesuit)\implies f(nk+k)=f(f(\alpha_n)+f(k))=f(\alpha_n)+f(k)=nk+k=(n+1)k,$$which shows what we want.

[b]Claim.[/b] $k$ divides every value $f(n)$.

[i]Proof.[/i] Suppose that there is a value $n'$ such that $k\nmid f(n')$. Then, use relation $(\clubsuit)$ repeatedly in the way: $k\nmid f(n')\implies k\nmid f(n')-ka$ and
$$f(n')-ka= f(n')-f(ka)=f(f(n')-f(ka))=f(f(n')-ka)\quad\forall a \text{ s.t. }ka\leq f(n')$$
Taking the largest possible $a$, we would get a nonzero value $f(n')-ka$ that is mapped to itself, but $f(n')-ka<k$ by construction, contradicting the minimality of $k$.

-----
Note that we have $f(m+ak) = f(m+f(ak))=f(m)+f(ak) = f(m)+ak$ for all $m$ and $a$, so defining $f$ on the integers $1$ to $k-1$ uniquely determines $f$. Indeed, we can show that such $f$ work as long as $k$ divides each value:

$$f(m+f(n)) = f\left(m+k\cdot \frac{f(n)}{k}\right) = f(m)+k\cdot \frac{f(n)}{k} = f\left(k\cdot \frac{f(m)}{k}\right)+f(n) = f(f(m))+f(n).$$


function
algebra
polynomial
absolute value
algebra unsolved

Find all functions $f : N \Rightarrow N$ such that
$[f(n)]^(19) + 97f(n) = 98n + 232$; $n \in N$  ;)

[quote="mps"]Find all functions $f : N \to N$ such that
$[f(n)]^{19} + 97f(n) = 98n + 232$; $n \in N$  ;)[/quote]

Should it be that? :)

Maybe similar technique as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84638
should work...
the characteristic polynomial has 18 ugly roots (the other two are 1), but the complex ones are all greater than 1 (in absolute value) so I suppose we can ignore them. Then the sequence n, f(n), f(f(n)),... is linear, by pluging into given equation we get the diference is 2, so the function we're looking for is
f(n)=n+2.


geometry
angle bisector
geometry proposed

$ PC'$ is the angular bisector of $ \angle{APB}$ and so the fact that $ Q$ is the intersection of the bisector of $ AB$ and the external angular bisector of $ \angle{APB}$ would imply $ AQPB$ is incriptible. This shows the first equality.

Let $ M$ be the intersection of $ PC'$ and the circle passing through $ A, Q, P, B$. Then $ \angle{PBQ} \equal{} \angle{PAQ} \equal{} \angle{PMQ}$. But we also have $ \angle{PMQ} \equal{} \angle{PCC'}$ (notice that $ MQ // CC'$). This completes the proof.

You forgot to prove this

[quote="mathangel"]$ PC'$ is the angular bisector of $ \angle{APB}$ and so the fact that $ Q$ is the intersection of the bisector of $ AB$ and the external angular bisector of $ \angle{APB}$ would imply $ AQPB$ is incriptible[/quote]
It follows from $ (A,B,C',D)=-1,$ where $ D\equiv{}A'B' \cap AB.$

Let [i]QD[/i] be the bisector of the triangle [i]AQB[/i]. Let [i]H[/i] be the orthocentre of triangle [i]ABC[/i]. Let [i]AB[/i] intersect [i]A'B'[/i] in [i]M[/i].

<[i]A'C'B[/i]=<[i]ACB[/i], because [i]AA' [/i]and [i]CC'[/i] are alptitudes.

Let [i]BB'[/i] intersect [i]DQ[/i] at [i]E[/i]. Then [i]B'ADE[/i] is circumcsribed(because <[i]AB'E[/i]=<[i]ADE[/i]=90 degrees), so <[i]EB'D[/i]=<[i]EAD[/i], but [i]DQ[/i] is bisector of [i]AB[/i],
so <[i]EAD[/i]=<[i]EBD[/i]=[i]90degrees[/i]-<[i]BAC[/i]. So <[i]DB'B[/i]=[i]90degrees[/i]-<[i]BAC[/i].
[i]A'HB'C[/i] is circumscribed, so <[i]BB'A[/i]'=<[i]C'CB[/i]=[i]90degrees[/i]-<[i]ABC[/i].
<[i]DB'A'[/i]=<[i]DB'B[/i]+<[i]BB'A'[/i]=[i]90degrees[/i]-<[i]BAC[/i]+[i]90degrees[/i]-<[i]ABC[/i]=<[i]ACB.[/i]

Then <[i]A'C'B[/i]=<[i]ACB[/i]=<[i]DB'A'[/i] and [i]A'B'DC'[/i] is circumscribed. So [i]MA*MB[/i]=[i]MD*MC'[/i]=[i]MA'*MB*.[/i]
<[i]QDM[/i]=<[i]C'PM[/i]=[i]90 degrees[/i], then triangles [i]MPC'[/i] and [i]MDQ[/i] are similar. So [i]MD*MC'[/i]=[i]MP*MQ[/i]. Then [i]MA*MB[/i]=[i]MP*MQ [/i]and [i]ABPQ[/i] is circumscribed. Then <[i]APB'[/i]=<[i]ABQ[/i]=<[i]BAQ[/i]=<[i]BPA'[/i]. Then <[i]APC'[/i]=<[i]C'PB[/i].

[i]PC'[/i] is bisector of <[i]APB[/i]. [i]QD[/i] is bisector of <[i]AQB[/i]. Let [i]PC'[/i] intersect [i]QD[/i] in [i]X[/i]. Then [i]AXBQP[/i] is circumscrtibed.

<[i]CC'B[/i]=<[i]QDB[/i], so[i] CC'[/i]||[i]QX[/i]. So <[i]PC'C[/i]=<[i]PXQ[/i]=<[i]PBQ[/i]=<[i]PAQ[/i], then [u][i]<PC'C=<PBQ=<PAQ[/i][/u].

Using nine point circle and given 2 circles,
if use power of some point, it is easy to show that

[quote="khashi70"]$ ABC$ is a triangle and $ AA'$ , $ BB'$ and $ CC'$ are three altitudes of this triangle . Let $ P$ be the feet of perpendicular from $ C'$ to $ A'B'$ , and $ Q$ is a point on $ A'B'$ such that $ QA \equal{} QB$ . Prove that :
$ \angle PBQ \equal{} \angle PAQ \equal{} \angle PC'C$[/quote]

Let I be the intersection of A'B' and AB.
We have $ (ABC'I)\equal{}\minus{}1$ and $ \angle C'PI\equal{}90^o$ so $ PC'$ is the bisector of $ \angle APB$
Let $ (ABP) \cap A'B'\equal{}\{Q'\}$
$ \Rightarrow \angle BAQ'\equal{}\angle BPA'\equal{}\angle APQ'\equal{}\angle ABQ'$ so $ Q'A\equal{}Q'B\Rightarrow Q'\equiv Q$
$ \Rightarrow \angle PBQ\equal{}\angle PAQ$
On the other hand, $ \angle PC'C\equal{}\angle C'IP\equal{}\angle ABP\minus{}\angle BPI\equal{}\angle ABP\minus{}\angle ABQ\equal{}\angle QBP$

[quote]the fact that  is the intersection of the bisector of  and the external angular bisector of  would imply  is incriptible.[/quote]

why is this the case? :?:

Let $J$ be the intersection point of $A'B'$ and $AB$ and drop $QX \perp AB$ with $X$ on $AB$.Then note that $X$ is the midpoint of $AB$ and since $AA',BB',CC'$ are concurrent we have $(J,C',A,B)=-1$.So $X$ being the midpoint of $AB$ and noting that $PC'XQ$ is cyclic( $\angle{PC'X}=\angle{PQX}=90^{\circ}$) we have $JA \cdot JB=JC' \cdot JX=JP \cdot JQ$ so $APQB$ is cyclic.So $\angle{PAQ}=\angle{PBQ}$.Next note that if $K=AQ \cap CC'$ then $\angle{AKC}=\angle{AQX}=\frac{1}{2}\angle{AQB}$.Also since $(J,C';A,B)=-1$ we see that $PJ,PC',PA,PB$ form a harmonic pencil.Now since $PC' \perp PJ$,$PC'$ must be the internal angle bisector of $\angle{APB}$.Thus $\angle{APC'}=\frac{1}{2}\angle{APB}=\frac{1}{2}\angle{AQC}$.So $\angle{APC'}=\angle{AKC'}=\frac{1}{2}\angle{AQC} \implies APKC'$ is cyclic.Thus $\angle{PAQ}=\angle{PAK}=\angle{PC'K}=\angle{PC'C}$ as desired.

 Nice and elegant.

Here's my solution:
Let $A'B'$ meet $AB$ at the point $T$.(working in the projective plane we wont bother about $T$ being the point at infinity.)
Now, observe that $C',P,Q,M$ are con-cyclic because of the perpendicularity conditions and that $(B,A;C',T)=-1$.
This along with the fact that $M$ is the midpoint of $AB$ gives that $TC'.TM=TA.TB=TP.TQ$ which in turn proves that points $B,A,P,Q$ are con-cyclic.
Therefore, $\angle PBQ=\angle PAQ$ and since, $Q$ lies on the perpendicular bisector of $AB$, $PQ$ bisects $\angle APB$.
Now, $\angle PC'C=\frac{(\angle A' -\angle B')}{2}$ and $\angle PAQ=\frac{(\angle PBA-\angle PAB)}{2}$. Now, the rest is trivial angle-chase.

[hide=Solution][quote=Iran TST#1 2009 P5]$ ABC$ is a triangle and $ AA'$ , $ BB'$ and $ CC'$ are three altitudes of this triangle . Let $ P$ be the feet of perpendicular from $ C'$ to $ A'B'$ , and $ Q$ is a point on $ A'B'$ such that $ QA \equal{} QB$ . Prove that :
$$\angle PBQ \equal{} \angle PAQ \equal{} \angle PC'C$$[/quote]
[b][color=#000]Solution:[/color][/b] Assume $P$ lies inside $\Delta A'B'C'$. Let $\odot (ABQ)$ $\cap$ $A'B'$ $=$ $P'$ and $BP' \cap CC'$ $=$ $R$. Thus,
\begin{align*} \angle AP'Q=\angle ABQ=\angle BAQ=\angle QP'R \end{align*}
hence, $P'Q$ bisects $\angle AP'R$ and,
\begin{align*} -1= (A,B; C', A'B' \cap AB) \stackrel{P'}{=} (AP' \cap CC' , R ; C' , P'B' \cap CC') \end{align*}
So, $\angle C'P'Q =90^{\circ}$ $\implies$ $P=P'$. Hence, $AQPB$ is cyclic $\implies$ $PC'$ bisects $\angle APB$. Let $AQ \cap CC'$ $=$ $S$
\begin{align*} \angle C'SA=90^{\circ}-\angle BAQ=\angle C'PA \end{align*}
$\implies AC'PS$ is cyclic $\implies$ $\angle PBQ=\angle PAQ=\angle PC'C \qquad \blacksquare$[/hide]

[hide=hint 1] Let $D$ and $E$ be the points where the lines perpendicular to line $AB$ passing through $A$ and $B$ respectively meet line $A'B'$, and try to prove that $\triangle C'AD \sim \triangle C'BE$ with SAS similarity.[/hide]
[hide=hint 2] Let lines $A'B'$ and $AB$ meet at $X$, and use the Ceva-Menelaus harmonic bundle. [/hide]
[hide=solution]
We will first prove that $PC'$ is the angle bisector of $\angle APB$.

Let $D$ and $E$ be the points where the lines perpendicular to line $AB$ passing through $A$ and $B$ respectively meet line $A'B'$. Also, let $X$ be the intersection point
of line $A'B'$ and line $AB$. By the Ceva-Menelaus harmonic bundle, $\frac{XA}{XB} = \frac{AC'}{C'B}$, but $\frac{XA}{XB} = \frac{AD}{BE}$, so
by SAS similarity $\triangle C'AD \sim \triangle C'EB$, so $\angle DC'A = \angle BC'E$ and $C'C$ is the angle bisector of $\angle ECD$. Let $Y$ be the intersection of $CC'$ and $A'B'$, and note that $\angle PC'A = C + 90 - (180 - 2B) = C + 2B - 90$. Meanwhile, $\angle DYC' = 180 - (90 - C + 180 - 2B) = C + 2B - 90$. Therefore, since $\angle C'DY = \angle C'AP$,
$\angle YC'D = \angle APC'$. Similarly, $\angle EC'Y = \angle C'PB$, so $PC'$ is the angle bisector of $\angle APB$, as required.

Now, by the incenter-excenter lemma, since $PQ$ is the external angle bisector of $\angle APB$, $APQB$ is cyclic, so $\angle QAP = \angle QBP$. Also note that
$\angle QAP = \angle BAP - (90 - \frac{\angle AQB}{2}) = \angle BAP + \frac{\angle APB}{2} - 90$. Then, since $\angle CC'P = 90 - (180 - \frac{\angle APB}{2} - \angle BAP) =
\angle BAP + \frac{\angle APB}{2} - 90$, $\angle CC'P = \angle QAP = \angle QBP$, as required. [/hide]


I have some questions to ask about the USA Mathematical Talent Search...

First, how old do you have to be to enter?

How do you enter?

How hard are the questions?

Is it worthwhile?

Thanks, guys!

To enter, go to [url]http://www.usamts.org/HowJoin/U_Join.php[/url].  I don't think there's a minimum age, but if you're under 13 you need a special form.  As for difficulty...take a look at the first round of problems...

[quote="buzzer11"]First, how old do you have to be to enter?[/quote]There is no age limit.  As long as you are a U.S. citizen or resident, and are not a high-school graduate, you can participate.  If you are under 13, you will need to send a parental permission form.

[quote="buzzer11"]How do you enter?[/quote]Go to [url]http://www.usamts.org[/url] and click on "Register".

[quote="buzzer11"]How hard are the questions?[/quote]They vary quite a bit in difficulty.  Almost everyone will find some of the questions challenging!

[quote="buzzer11"]Is it worthwhile?[/quote]It depends how you define "worthwhile".   :)  The big difference between the USAMTS and the other major contests is that you have to show full written justifications for your answers, and you will receive written feedback on those answers in addition to your score.  So in that sense it is very worthwhile!

Download the questions and try them and see!  You can download and try some of the Round 1 problems -- there's no obligation to actually send in solutions if you decide that it's not for you.


I think that it would be good to have physics olympiad resources also at
http://www.mathlinks.ro/resources.php

What do you say??  :roll:  :)

great.it would help us a lot in preparing for IPHO

@Aadil
How are resources added?
I've no clue whatsoever.

But I do think it'd be good to have not only math resources but also other subjects :)

Unfortunately, for Physics Olympiad stuff, they've only released tests from 4 years, which doesn't really help to prepare much. And they keep on changing the format every year, so I have no idea what's going on or how to qualify for the next round.

The problem is that no olympiads have been posted on the forum, which must be done before adding them to the resources. Also, there haven't been any other than math olympiads yet in the resources...

You can't add the problems to the resources, only moderators can.

Hmm.. I think we could do with having resources based on the replies :) .
What do you say 'mod's? :roll:

Well, yes, but as I said, there is a lack of physics competitions in the forum. Just look at the resources, you'll see lots of international and even more national competitions. How many such physics contests have been posted?

We have 1! The Physics Olympiad, and there's around 140-150 problems from just preliminaries alone, many of which have already been posted.


At the national curling championships, there are three teams of four players each. After the championships are over, the very courteous participants each shake hands three times with every member of the opposing teams, and once with each member of their own team.

How many handshakes are there in total?

There are 12 total people. They shake hands 3 times with 8 of the people, and 1 time with 3 people. That is $ 24\plus{}3\equal{}27$ handshakes. Multiply by 12 to get all the people, and then divide by two because each handshake consists of 2 people. This is just multiplying by 6:

$ 27*6\equal{}162$

For each participant, there are 8 opponents to shake hands with, and 3 team members to shake hands with, giving $3\times8+3=27$ handshakes for each individual participant.

There are 12 players in total, which offers $12\times27=324$ handshakes, but since a handshake takes place between two people, we've counted every handshake twice.

The final answer is $\dfrac{324}{2}=\boxed{162}$ handshakes.

just saying... you don't have to tell everyone to hide their answers, they probably won't see your message anyway

He's farming posts. He's been doing it for the past few minutes on several threads now ;)

@noah: and next time just report the post(s), which people have been doing.

EDIT: @below: yeah all the posts were reported.

Well, i don't know what happened, but I guess a mod deleted the posts, as the post count jumped from 3 to 7

Another way to solve this:
If everyone shook hands only once, there would be ${12 \choose2} = 66$ handshakes. Since for each player there are $3$ teammates and $8$ opponents, $\frac{3}{11}$ of these handshakes should be counted once, while $\frac{8}{11}$ of them should be counted three times.

The answer is $\frac{3}{11}*66+3*\frac{8}{11}*66=162$

[hide=Alternative Solution]
Use casework:
[hide=Case 1 (handshakes with opposing teams)]
Every member in the first team shakes hands with every member in the other $2$ teams, so there are $8+8+8+8=32$ handshakes. Every member in the second team shakes hands only with the third team, since the first team already shook their hands. This gives us $4+4+4+4=16$ handshakes, getting a total of $32+16=48$ handshakes. But each member shakes $3$ times with their opponents, making a grand total of $48\times3=144$ handshakes.
[/hide]
[hide=Case 2 (handshakes with own team)]
Each member shakes $3$ hands with their team, so there are $\binom{4}{2}=6$ handshakes per team. So $6\times3=18$.
[/hide]

Adding both cases together gives us $144+18=\boxed{162}$ total handshakes.
[/hide]


A triangle is constructed like that below, but with 1, 2, 3, ... , 2000 as the first row. Each number is the sum of the two numbers immediately above. Find the number at the bottom of the triangle.

1   2   3   4   5
  3   5   7   9
    8   12  16
      20  28
        48
Have fun!

Err...are you sure this belongs in HSB?  :maybe:

Look for a pattern, then prove it by induction.

[hide="dude hint"]i think that looking at this for smaller values helps us generalise and come upon the answer[/hide]
oh and this one was the easiest of the 3 i just posted. please answer one i haven't gotten please please please and any random moderator can move my posts if they believe they have to

here is my solution...

[hide]let the numbers at the bottom be $a_{1,1}, a_{1,2}, a_{1,3}, ...$.  Then let $a_{2,i}$ be the numbers in the second row. we have $a_{2,i}= a_{1,i}+a_{1,i+1}$.  Similarly, $a_{3,i}= a_{1,i}+2a_{1,i+1}+a_{1,i+2}$.  A general formula for $a_{m,i}$ is as follows:

$a_{m,i}= \sum_{k=0}^{m-1}{m-1 \choose k}a_{i+k}$

Hence we must find $\sum_{k=1}^{2000}{1999 \choose k-1}k$.

i leave that to you.
 [/hide]

dude
WINNER WINNER WINNER
that is the correct answer

[quote="rnwang2"]It looks like the bottom number will be $(n+1)\cdot 2^{n-2}$, when there are $n$ numbers in the first row.[/quote]

Dude ok now prove that based on Teki-Teki's work (or some other way).

[hide="Proof by Induction"]
Assume that for a triangle with $n$ rows, the number at the bottom is $(n+1)2^{n-2}$.  Thus, we must prove that the last number on the triangle with $n+1$ rows is $(n+2)2^{n-1}$.  We know that it works for 1 row, since that number will be $1$ and $(1+1)(2^{1-2})=1$.  Now, it is easy to see that on the nth row, the numbers will differ by $2^{n-1}$.  Thus, if you put the number $2^{n-2}(n+1)+2^{n-1}=2(2^{n-2})+2^{n-2}(n+1)=(n+3)2^{n-2}$ next to the $2^{n-2}(n+1)$ and add them, you will get the last number on a triangle with $n+1$ rows.  When you add them, we find that it is $2^{n-2}(n+1)+2^{n-2}(n+3)=2^{n-2}(2n+4)=2(2^{n-2})(n+2)=2^{n-1}(n+2)$, which is what we want. [/hide]

[quote="rnwang2"]Ok, probability, this is for you.
[hide]
Based on Teki-Teki's result

$a_{m,1}= \sum_{k=0}^{m-1}{m-1 \choose k}a_{1,k+1}$,

$=\binom{m-1}{0}\cdot 1+\binom{m-1}{1}\cdot 2+\ldots+\binom{m-1}{m-1}\cdot m$.

Let this sum equal $S$.

We have

$2S=\left(\binom{m-1}{0}+\binom{m-1}{1}+\ldots+\binom{m-1}{m-1}\right)\cdot (m+1)$

$=2^{m-1}(m+1)$.

Dividing both sides by 2, we obtain the result

$2^{m-2}(m+1)$.[/hide][/quote]
dude probability
that's so cule but i am so scared of summation
maybe we could all learn a bit from looking at this post
it makes (me at least) understand dealing with the symbols a bit more

[quote="rnwang2"][quote="funcia"]dude probability
that's so cule but i am so scared of summation
maybe we could all learn a bit from looking at this post
it makes (me at least) understand dealing with the symbols a bit more[/quote]
I'm not probability... :dry:[/quote]
uh yeah i know that
i was pointing to your post and summoning probability to look at it
maybe i should be more precise
hey i better go to sleep
school
thanks for your help rnwang


inequalities
search
geometry unsolved
geometry

I was given tthe following problem:

   If $a , b , c$  are the sides of a triangle , prove that :

             $\sum \frac{a-b}{a+b} < \frac{1}{16}$

 I tried to search , but I found nothing-only two problems sent by manglio.

[u] Babis[/u]

What keywords did you use? I searched for *\frac{a-b}{a+b}* and found

http://www.mathlinks.ro/Forum/viewtopic.php?t=57240
http://www.mathlinks.ro/Forum/viewtopic.php?t=55632
http://www.mathlinks.ro/Forum/viewtopic.php?t=51448

Stronger version at http://www.mathlinks.ro/Forum/viewtopic.php?t=22786 .

  darij

Thank you Darij !

  I'm sure I do not know how to search. I wrote << iberoamerican olympiad AND inequality AND sides a , b, c >>.

  [u]Babis[/u]


If $ \displaystyle \sum_{n \equal{} 0}^{\infty} \cos^{2n} \theta \equal{} 5$, what is the value of $ \cos{2\theta}$?

$ \textbf{(A)}\  \frac15 \qquad \textbf{(B)}\  \frac25 \qquad \textbf{(C)}\  \frac {\sqrt5}{5}\qquad \textbf{(D)}\  \frac35 \qquad \textbf{(E)}\ \frac45$

Solving the infinite series gives $\cos^{2}{\theta}=4/5.$
$\cos{2\theta}=2cos^{2}{\theta}-1=8/5-1=3/5$ [b]D[/b]

[quote="Farenhajt"]This was posted before.

[hide]The answer is D[/hide][/quote]Really? I didn't come across it using the search function...?  :huh:

[b]Solution:[/b] $$1 + \cos^2\theta + \cos^4\theta + \ldots = 5 \Rightarrow \frac{1}{1-\cos^2\theta} = 5 \Rightarrow \cos^2\theta = \frac{4}{5}$$ $$\Rightarrow \cos 2\theta = 2\cos^2\theta - 1 = \frac{8}{5} - 1 = \boxed{D}$$

Since the summation is an infinite geometric series, $\frac{1}{1-\cos^2 \theta} = 5 \implies \cos^2 \theta = \frac{4}{5}.$ Thus, $\cos 2\theta = 2\cos^2 \theta - 1 = \boxed{\frac{3}{5}}.$ So the answer is $\textbf{(D)}.$


calculus
integration
function
calculus computations

Find the continuous function $f(x)$ and constant number such that $\int_{0}^{x}f(t)\ dt=e^{x}-ae^{2x}\int_{0}^{1}f(t)e^{-t}\ dt.$

First set $x=0:$ $0=1-a\int_{0}^{1}f(t)e^{-t}\,dt$

So $\frac1a=\int_{0}^{1}f(t)e^{-t}\,dt$ and the original equation becomes

$\int_{0}^{x}f(t)\,dt=e^{x}-e^{2x}$

Differentiate that to get $f(x)=e^{x}-2e^{2x}$

And by integration we find that $a=\frac1{3-2e}.$

(Note that $a$ is a negative number.)

Needless to say, your answer is correct.


Wasn't sure where to post... not really in high school...

How much of the other books do the classic books (vol. 1 & 2) cover? Will there be anything I miss out on, if, say, I get the classics but not intro to algebra? Is there a ton of information in those number theory and counting books that can't be found in the classic books?

I have Volume 2 so I can't vouch for or against the other editions but I can tell you that Volume 2 gives a very nice treatment of each of the topics.

The Intro/Intermediates offers a larger depth of material while the edition books only give introductions. Don't get me wrong, the edition books are really helpful in getting you started in "challenging" math but the Intro/Intermediate books are where you want to go if you're interested in something.


calculus
integration
real analysis
real analysis unsolved

If $ f(x)\geq0$

does $ \int_{0}^{\plus{}\infty}f(x)dx<\plus{}\infty$

imply

$ \int_{0}^{\plus{}\infty}f^2(x)dx<\plus{}\infty$
?

Under what conditions can we say that $ \int_{0}^{\plus{}\infty}f^2(x)\,dx \leq \left(\int_{0}^{\plus{}\infty}f(x) \,dx\right)^2$?

No. Take $ f(x) \equal{} \begin{cases}1/\sqrt x,\quad 0 < x\le 1, \\
0\qquad x > 1\end{cases}$

$ L^1(\mathbb{Z})\subset L^2(\mathbb{Z})$.
$ L^2([0,1])\subset L^1([0,1])$.
On $ [0,\infty)$, which has both infinite measure and infinitely divisible measure, neither inclusion holds.


Για ποια τιμή του θετικού ακεραίου $x$, είναι ο $y$ ακέραιος;

$y=\sqrt{1597x^{2}+1}$

Gia να είναι ακέραιος ο y θα πρέπει $1597x^{2}+1=q^{2}$
ή ισοδύναμα $q^{2}-1597x^{2}=1$ που είναι εξίσωση Pell  :wink: .
Άρα αν $(q_{0},x_{0})$ is the least nontrivial solution τότε
όλες οι λύσεις (q,x) δίνονται από τη σχέση
$q+y\sqrt{1597}= \pm (q_{0}+x_{0}\sqrt{1597})^{n}$  :wink:

[quote="silouan"]
Άρα αν $(q_{0},x_{0})$ is the least nontrivial solution τότε
όλες οι λύσεις (q,x) δίνονται από τη σχέση
$q+y\sqrt{1597}= \pm (q_{0}+x_{0}\sqrt{1597})^{n}$  :wink:[/quote]

H legomenh "vasikh" lisi tis eksiswsis Pell gia to parapanw provlima (simfwna me tin antistoixi thewria gia tin euresi tis) einai (me "merikous" ipologismous):

$(q_{0},x_{0})=(519711527755463096224266385375638449943026746249, \ \ 13004986088790772250309504643908671520836229100)$

Einai ontws mi tetrimmeni lisi :)

Alexandros

[quote="silouan"]
όλες οι λύσεις (q,x) δίνονται από τη σχέση
$q+y\sqrt{1597}= \pm (q_{0}+x_{0}\sqrt{1597})^{n}$  :wink:[/quote]

γιατι - ? τα $x,y$ ειναι θετικα


Prove that the polynomial x200*y200+1 cant be expressed in the form P(x)*Q(y) where P(x) and Q(y) are polynomials of x and y respectively.

Solution: Suppose  x200*y200+1≡ P(x) *Q(y).Then by the equalities:
P(0)*Q(0)=1
P(0)*Q(y)=1
P(x)*Q(0)=1
We get P(x)*Q(y)=1 which is not possible.

How does one get those three equalities???

uff :( I don't know how to superscript the 200-s,they are the powers of x and y.

By letting x=y=0, then x=0, then y=0 in the original polynomial.

By the way, I noticed that you usually attach the powers of a variable right next to it, like x5. If you could just use the ^  symbol, to make it x^5, it would be much easier to read! Also, you could use TeX and type the same thing as [tex]x^5[/tex]. You can hover over that last x^5 to see the code.


geometry
parallelogram
analytic geometry
geometry unsolved

In the quadrilateral $ PQRS$, $ A$, $ B$, $ C$ and $ D$ are the midpoints of the sides $ PQ$, $ QR$, $ RS$ and $ SP$ respectively, and $ M$ is the midpoint of $ CD$. Suppose $ H$ is the point on the line $ AM$ such that $ HC\equal{}BC$. Prove that $ \angle BHM\equal{}90^\circ$.

easy to know ABCD is a parallelogram
CN=CH=BC=AD  triangle BHN is a right triangle
done！

WOW!!! Great solution!

During the contest, I used coordinate geometry. :oops:  But I didn't finish my solution. (I started in the last 3 minutes.)


inequalities
induction
inequalities solved
n-variable inequality
cauchy schwarz
Tournament of Towns

For positive reals $a_1, a_2, \cdots , a_n$ show that
\[\left(1+ \frac{a_1^2}{a_2}\right)\left(1+ \frac{a_2^2}{a_3}\right)\cdots \left(1+ \frac{a_n^2}{a_1}\right) \geq (1+a_1)\cdots (1+a_n)\]

Nice problem.

I don't see an immediate solution but maybe this "theorem" might help :

If a1, a2, ..., an > 0 then

(1 + a1)(1 + a2)...(1 + an) \geq (1 + g)n

where g is the geometric mean of a1, a2, ..., an.

I'm not sure whether it will really work ...

The best ( :? ) way I could come up with so far is this:

we actually multiply the brackets on both sides and we get some sums (sums of products of 1,2,3,..,n factors). We use rearrangement ineq to show that each sum on the left is >= its corresponding sum (composed of products with the same number of factors) on the right. I'm positive it works, but I'm also pretty sure there must be a nice solution out there...

Do you remember the following similar problem :

(1 + a13)(1 + a23)...(1 + an3) \geq (1 + a12a2)(1 + a22a3)...(1 + an2a1) ?

You can prove it with the above lemma.

It comes down to prove that (1 + x)(1 + y) \geq (1 + xy).

This follows immediately from the lemma I mentioned.

Probably something similar can be done here.

Use the fact that for $x,y > 0$ we have
\[1+\frac{x^2}y \geq \frac{(1+x)^2}{1+y}\]

Aha ... Thanks.

It's strange that I didn't see that, neither did Grobber.


let $ G\equal{}\{ A\equal{}\begin{bmatrix}a&b\\c&d\end{bmatrix}| a,b,c,d \in \mathbb{R}, \mbox{and} \,\,\,\det(A)>0\}$ Prove that $ G$ is not contractible

I know that one way to prove a space is not contractible is to show that $ \pi_{1}(G)\neq 1$ but I am getting stuck to find a familiar space which is homeomorphism to $ G$.   Any idea will be highly appreciated .

One can use the QR decomposition of matrices:
Every $ A\in GL_n$ can be continuously decomposed in $ A\equal{}QR$ with $ Q$ orthogonal and $ R$ an upper triangular matrix with positive diagonal entries. $ (Q,R)$ is continuous in $ A$ because we can construct them as follows: $ R$ is the (transposed) cholesky-factor of $ A^T A\equal{}R^T(Q^TQ)R\equal{}R^T R$. So $ R$ is continuous in $ A$ (just have look at the cholesky-algorithm). Now $ Q\equal{}AR^{\minus{}1}$ is continuous in $ A$ too. By construction we get homeomorphisms $ GL_n \leftrightarrow O_n\times X_n$ where $ X_n: \equal{}\{A\in\mathbb{R}^{n\times n} | A_{ii}>0, A_{ij}\equal{}0 \text{ for } i>j\}$.
The restriction of this to $ G$ leads to an homeomorphism $ G\to SO_2\times X_2$.

$ X_n$ deformation retracts to a point since it is obviously homeomorphic to $ \mathbb{R}_\plus{}^n\times\mathbb{R}^{\tfrac{n(n\minus{}1)}{2}}$. So $ G \simeq SO_2 \cong S^1$ hence $ \pi_1(G)\equal{}\mathbb{Z}$ and in particular $ G$ is not contractible.


Let x,y,z,t be real numbers and:
x^2+y^2=9
z^2+t^2=4
xt-yz=6
Find the greatest value of P=xz

hello, from the equation system we got $ x\equal{}\pm\frac{3}{2}\sqrt{4\minus{}z^2}$ and so we have
$ P(z)\equal{}\pm\frac{3}{2}z\sqrt{4\minus{}z^2}$. It is easy to show that $ \max_{\minus{}2\le z\le 2}{P(z)}\equal{}3$.
Sonnhard.

[hide="Vectors"]Consider the vectors $ u \equal{} \langle x,y \rangle, v \equal{} \langle z,t \rangle$. We're given that $ |u| \equal{} 3, |v| \equal{} 2, |u \times v| \equal{} 6$. Since $ |u \times v| \equal{} |u| |v| \sin \theta$ where $ \theta$ is the angle between the vectors, we can find that $ \sin \theta \equal{} 1$ and thus the vectors are orthogonal. Now write them as $ u \equal{} \langle 3 \cos \alpha , 3 \sin \alpha \rangle, v \equal{} \langle 2 \sin \alpha, \minus{} 2 \cos \alpha \rangle$ for some $ \alpha$. Then $ P$ is $ 6 \cos \alpha \sin \alpha \equal{} 3 \sin 2\alpha$, which clearly has maximum $ 3$.[/hide]

[hide="Algebra"]Multiply the first two equations and set this equal to the square of the second equation. After plenty of simplifying we end up with $ (xz \minus{} yt)^2 \equal{} 0$, so $ xz \equal{} yt$. Now by AM-GM, $ 9 \equal{} x^2 \plus{} y^2 \geq 2xy$, $ 4 \equal{} z^2 \plus{} t^2 \geq 2zt$, so multiplying gives $ 9 \cdot 4 \geq 4xyzt$. From this we deduce $ 9 \geq (xz)^2$ and $ xz \leq 3$. Equality holds when $ x \equal{} \pm y, z \equal{} \pm t$ and $ xz$ is positive.[/hide]

[quote="Dr Sonnhard Graubner"]hello, from the equation system we got $ x \equal{} \pm\frac {3}{2}\sqrt {4 \minus{} z^2}$ and so we have
$ P(z) \equal{} \pm\frac {3}{2}z\sqrt {4 \minus{} z^2}$. It is easy to show that $ \max_{ \minus{} 2\le z\le 2}{P(z)} \equal{} 3$.
Sonnhard.[/quote]
Can you show me why $ x \equal{} \pm\frac {3}{2}\sqrt {4 \minus{} z^2}$, please? :?:

$ (x^2\plus{}y^2)(z^2\plus{}t^2) \equal{} (xt\minus{}yz)^2 \plus{} (xz\plus{}ty)^2 \Rightarrow xz\plus{}ty \equal{} 0$

$ (x^2\plus{}y^2)(z^2\plus{}t^2) \ge 4|xytz| \equal{} 4 (xz)^2 \Rightarrow xz \le 3$


A real number $a_1$ is given. The sequence $(a_n)$ is defined by $a_{n+1}=a_n^2-a_n+1$
Prove that for all positive integers $n$ the following inequality holds:

$\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}<\frac{1}{a_1-1}$

$\displaystyle a_{n+1}-1=a_n(a_n-1)\Rightarrow \frac 1{a_{n+1}-1}=\frac 1{a_n-1}-\frac 1{a_n}\Rightarrow \frac 1{a_n}=\frac 1{a_n-1}-\frac 1{a_{n+1}-1}$, which gives us the possibility to use a telescopic sum. We get $\displaystyle\sum_{i=1}^n \frac 1{a_i}=\frac 1{a_1-1}-\frac 1{a_{n+1}-1}$.

I do think, however, that we need more constraints on $a_1$. For example, it should be $\ne 1$. In fact, it should be $>1$.


Let $ m$ and $ n$ be two positive integers. If there are infinitely many integers $ k$ such that $ k^2\plus{}2kn\plus{}m^2$ is a perfect square, prove that $ m\equal{}n$.

Consider two cases:
(a) $ m > n$:
Let $ k^2 \plus{} 2kn \plus{} m^2 \equal{} S^2$,then $ S^2 \minus{} (k \plus{} n)^2 \equal{} m^2 \minus{} n^2$.Since $ m > n$,we conclude that $ S > (k \plus{} n)$,hence $ S\geq k \plus{} n \plus{} 1$,so
$ m^2 \minus{} n^2\geq 2(k \plus{} n) \plus{} 1$,now to obtain contradiction,it is enough to take $ k > \frac {m^2 \minus{} n^2 \minus{} 1 \minus{} 2n}{2}$.
(b) $ m < n$,as we do it in $ (a)$,if we take $ k > \frac {n^2 \minus{} m^2 \plus{} 1 \minus{} 2n}{2}$ and obtain a contradiction.
So we prove that $ m \equal{} n$.

if $ x,a\in\mathbb{Z},a\neq0$ and $ x^2+a$ is a perfect square,then $ x^2+a$ is not in the segment ${ }((|x|-1)^2,(|x|+1)^2)$ (because the only perfect square in this segment is $ |x|^2$) therefore $ |a|\geq2|x|-1$
let $ A=k^2+2kn+m^2=(k+n)^2+m^2-n^2$
if $ m^2-n^2\neq0$ and A is a perfect square for some $ k$ then we have:
$ |m^2-n^2|\geq2|k+n|-1\geq2|k|-2|n|-1\Rightarrow{|k|\leq\frac{|m^2-n^2|+2|n|+1}{2}}$
which is impossible,because $ A$ should be a perfect square for infinitely many $ k$ therefore $ m^2=n^2\Rightarrow{m=n}$
@Erken:I think your solution need small correction...maybe S be a perfect square for k<0...

If $ m > n$,  for every $ k >\frac{m^2 \minus{} (n \plus{} 1)^2}{2}$ , $ (k \plus{} n)^2 < k^2 \plus{} 2kn \plus{} m^2 < (k \plus{} n \plus{} 1)^2$ if $ m>n$ , for every $ k>\frac{m^2\minus{}(n\minus{}1)^2}{2}$, $ (k\plus{}n\minus{}1)^2<k^2\plus{}2kn\plus{}m^2<(k\plus{}n)^2$. For sufficiently large $ k$ ,  $ k^2\plus{}2kn\plus{}m^2$  isn't a perfect square.For only finite $ k$, $ k^2\plus{}2kn\plus{}m^2$ is a perfect square.


There are two questions here:

1st is...

The velocity of sound in air is 343 m/s. A train moving away from a detector at 29 m/s blows a 350 Hz horn. What frequency is detected by a stationary train?

I got the correct answer of 322.7 using dopplers equation.

Now part 2 is...

What frequency is detected by a train moving away from the first train at a speed of 15.6 m/s?

Plugging in ---->

f' = [350*343] / [343+15.6] = 334.77

It seems correct to me, but when I submitted my answer online I was incorrect.

Please help! Thanks!

Remember that the first train is moving, so the velocity of the first train, relative to the second [i]moving[/i] train is $ 29 \minus{} ( \minus{} 15.6) \equal{} 44.6\;m/s$ (assuming the trains are moving in opposite directions). We input that value to get

$ f' \equal{} \left(\frac { v}{v \plus{} v_{s,r}}\right)f \equal{} \left(\frac {343 \frac {m}{s}}{343 \frac {m}{s} \plus{} 44.6 \frac {m}{s}}\right)350\;Hz \approx \boxed {309.7\;Hz}$

If the velocities of the trains are two dimensional, simply add the vectors: $ v_{s,r} \equal{} \vec{v_{A}} \plus{} \vec{v_{B}}$


algebra
polynomial
trigonometry
system of equations
algebra proposed
Hi

Find all triples $ (x,y,z)$ of real numbers that satisfy the system of equations
\[ \begin{cases}x^3 \equal{} 3x\minus{}12y\plus{}50, \\ y^3 \equal{} 12y\plus{}3z\minus{}2, \\ z^3 \equal{} 27z \plus{} 27x. \end{cases}\]

[i]Razvan Gelca.[/i]

[quote="tuandokim"]This is my solution:
we have 2 cases:
case1:$ x\ge 2$,from 1 we have $ y\le 4$,from 3 we have $ z\ge 6$ so from 2 we have $ y\ge 4$ so x=2,y=4,z=6
case2:$ x\le 2$,from 1 we have $ y\ge 4$,from 3 we have $ z\le 6$ so from 2 we have $ y\le 4$ so x=2,y=4,z=6
Q.e.d[/quote]
nice solution

Another solution could be just setting

x=a+2, y=b+4, z=c+6.

If one crunches the numbers, and observes the trivial inequality, he/she will find a=b=c=0, so x=2, y=4, z=6.

set $ z\equal{}3u$ and take $ x$ from the last equation into the first one, so that it turns into

$ (u^3\minus{}3u)^3\equal{}3(u^3\minus{}3u)\minus{}12y\plus{}50$ and $ y^3\equal{}12y\plus{}9u\minus{}2$...

take $ y\equal{}w\plus{}4$, and this system turns to

$ (u^3\minus{}3u)^3\minus{}3(u^3\minus{}3u)\minus{}2\equal{}\minus{}12w$ and $ w^3\plus{}12w^2\plus{}36w\equal{}9u\minus{}18$...

this can be rewritten as

$ (u\minus{}2)(u^4\plus{}u^3\minus{}3u^2\minus{}2u\plus{}1)^2\equal{}\minus{}12w$ and $ w(w\plus{}6)^2\equal{}9(u\minus{}2)$...

from this it's easy to conclude that $ u\equal{}2, w\equal{}0$, or equivalently, $ z\equal{}6, y\equal{}4, x\equal{}2$  :D

Great solutions, but how did you guys get that 2, 4, 6 worked? was it just guess?

Another solution involves using the cos(3*theta) formula to kill the cubic, which makes it immediately evident why 2 4 and 6 are solutions.

Rewrite the equations as the following:    $ 12(4 - y) = (x - 2)(x + 1)^{2}$   $ (1)$,      $ 3(z - 6) = (y - 4)(y + 2)^{2}$ $ (2)$,      $ 27(x - 2) = (z - 6)(z + 3)^{2}$ $ (3)$.        $ (2)$, $ (3)$ gives  $ (y - 4)(x - 2)\geq{0}$ (Or $ y = - 2$, which leads to a contradiction), but from $ (1)$ we get:
$ (4 - y)(x - 2)\geq{0}$. (Or $ x = - 1$, which also leads to a contradiction). So $ 0\leq{4 - y)(x - 2)\leq{0}}$. It means $ y = 4$ or $ x = 2$, but both of them lead to an unique solution: $ (x,y,z) = (2,4,6)$.

[quote="ZerAn"]Great solutions, but how did you guys get that 2, 4, 6 worked? was it just guess?
[/quote]

I have the same question.

Here's a slightly more deterministic solution.  Substitute $y=2r, z=3s$ to make more of the coefficients match.  Then we get $x^3-3x=50-24r$
$r^3-3r=(9s-2)/8$,
$s^3-3s=x$.

The solution $x=r=s=2$ pretty much jumps out.  We note that
$x^2-3x-2=(x-2)(x+1)^2$, so

$(x-2)(x+1)^2=24(2-r)$,
$(r-2)(r+1)^2=9(s-2)/8$,
$(s-2)(s+1)^2=(x-2)$.

If any of $x,r,s$ equals 2, then they all do.  If not, multiplication yields that
$(x+1)^2(r+1)^2(s+1)^2=-27$, a contradiction.

Let $x=2a$, $y=4b$ and $z=6c$ (this is motivated by observing $a=b=c=1$).  Then, the givens rearrange to \begin{align*}
		a(4a^2-3) &= 25-24b \\
		b(4b^2-3) &= \frac{9c-1}{8} \\
		c(4c^2-3) &= a
	\end{align*} Note $c < 1 \implies a \le 1 \implies 25-24b \le 1 \implies b \le 1 \implies \frac{9c-1}{8} \le 1 \implies c \ge 1$ and $c > 1 \implies a > 1 \implies 25-24b > 1 \implies b < 1 \implies \frac{9c-1}{8} < 1 \implies c < 1$.
So $a=b=c=1$ is the only solution; i.e. $(x,y,z) = (2,4,6)$.

==========================================================

This bounding approach is pretty motivated by the polynomial $f(x) = 4x^3-3x$, which behaves very nicely with respect to bounds; in fact, I was trying to get a solution using the identity $\cos 3x = 4\cos^3 x - 3 \cos x$ which is why I attempted the bounding in the first place.

[quote="ZerAn"]Great solutions, but how did you guys get that 2, 4, 6 worked? was it just guess? [/quote] Many substitutions later.  Not realizing that this was essentially bounding, I tried a lot of different substitutiotns and one of which eventually led to solving $u-v = \tfrac{u^3-50}{3}$, $v-w = \tfrac{v^3+128}{192}$, $w-u=\tfrac{w^3}{27}$.  At this point I decided to start looking for integer solutions and finally noticed $(u,v,w) = (2,16,-6)$ which gave $(x,y,z) = (2,4,6)$.  In retrospect I think I should have tried the guessing first.

So I guess the main difficulty of the problem was finding the triple in the first place, and afterwards it was just a matter of careful manipulations.

[hide=Solution]
Suppose that $x>2$. Then on $(2, \infty)$ the function $x^3-3x$ is increasing so we have $y<4$ by the first equation. Also the third equation yields that, from $x>2$, $z^3-27z>54 \implies z>6$. But then the second equation tells us that $y^3-12y>16 \implies y>4$. Contradiction.

Now suppose that $x<2$. From equation 3, $z^3-27z<54 \implies z<6$. Then equation 2 yields $y<4$. Using this in the first equation we have $x^3-3x=50-12y>2 \implies x>2$. Contradiction again.

It is easy to check that $x=2, y=4, z=6$ works so $(x, y, z)=\boxed{2, 4, 6}$
[/hide]

[hide=comments]
Now I want to discuss motivation, which was pretty tricky for this problem honestly. Honestly when I first attempted this problem I guessed that $x, y, z$ had to be all integers. I tried substituting floors and what not to prove that this was the case, but it turned out too messy. I proceeded to try to just guess integer solutions, starting from $x=1, 2$, etc. and soon stumbled on $(2, 4, 6)$. I thought about using $x=2a, y=4b, z=6c$ but tried dumb things first like bounding on just $x, y, z$ and it worked ridiculously nicely.

I would also be strongly interested in seeing a solution using $\cos 3x = 4\cos^3 x - 3 \cos x$, it seems like the most motivated way to solve this problem, and it would be still kinda doable if $x, y, z$ weren't all integers
[/hide]

[b]Answer:[/b] The only triple which satisfies our condition is $(x, y, z)=(2, 4, 6)$.

(Proof) Suppose $(x-2)(y-4)(z-6) \ne 0,$ as otherwise we get $(x, y, z)=(2, 4, 6)$ as claimed. Note that the given equations may be re-written as:

\[ \begin{cases} -12(y-4)=(x-2)(x-1)^2, \\ 3(z-6)=(y-4)(y+2)^2, \\ 27(x-2)=(z-6)(z+3)^2. \end{cases}\] Taking the product and cancelling the common (non-zero) factor $(x-2)(y-4)(z-6),$ we see that both sides have an opposite sign unless $(x-1)(y+2)(z+3)=0$. Evidently, the latter cannot occur as $$x=1 \Longrightarrow y=4 \Longrightarrow z=6 \Longrightarrow x=2,$$ a contradiction! Similar arguments apply for the other two factors.

We claim the only solution is $(x,y,z)=\boxed{(2,4,6)}$, which can easily be checked to work.

To show that this is the only solution, we have the following key lemma.

[u]Lemma:[/u] We have
\begin{align*}
x\ge 2 &\iff x^3-3x\ge 2 \\
y\ge 4 &\iff y^3-12y\ge 16 \\
z\ge 6 &\iff z^3-27z\ge 54,
\end{align*}
and similar statements with all the inequality directions reversed.

[u]Proof:[/u] The proof is ``you just check it''. $\blacksquare$

The solution to the problem is now easy. We see that
\begin{align*}
x\ge 2 &\iff z^3-27z\ge 54 \\
&\iff z\ge 6 \\
&\iff y^3-12y\ge 16 \\
&\iff y\ge 4 \\
&\iff x^3-3x\le 2 \\
&\iff x\le 2,
\end{align*}
so $x=2$. This easily gives that $(x,y,z)=(2,4,6)$, as desired.

We have
\begin{align*}
x\ge 2 &\implies \frac{z^3-27z}{27} \ge 2 \implies z^3-27z-54\ge 0 \implies z\ge 6 \text{ or } z=-3. \\
 z\ge 6 &\implies y^3-12y = 3z-2 \ge 16 \implies y^3-12y-16 \ge 0 \implies y\ge 4 \text{ or } y=-2. \\
y\ge 4 &\implies x^3-3x-50=-12y \le -48 \implies x^3-3x-2\le 0 \implies x\le 2 \text{ or } x=-1.
\end{align*}
It is easy to check that none of $z=-3$ or $y=-2$ or $x=-1$ produce valid solutions. Therefore, $x\ge 2 \implies x\le 2$. We can make a similar inequality chain to show that $x\le 2\implies x\ge 2$. Therefore, $x=2$, which produces the solution $(x,y,z)=(2,4,6)$.

Substituting $x=a+2$, $y=b+4$, $z=c+6$, we get the equations $a(a+3)^2=-12b$, $b(b+6)^2=3c$, $c(c+9)^2=27a$. So, $a,b$ are of opposite sign, $b,c$ are of the same sign, and $c,a$ are of the same sign. The only way this can happen is when $a=b=c=0$, so our only solution is $(2,4,6)$.

The only solution is $(x, y, z) = (2, 4, 6)$. Note that if $x > 2$, then we have $y < 4$ and $z < 6$ by monotonicity, which imply $x < 2$, contradiction. The case is similar for $x>2$.

The solution is $(2,4,6)$. Rearrange the equations to $(x+1)^2(x-2)=-12(y-4), (y+2)^2(y-4)=3(z-6), (z+3)^2(z-6) = 27(x-2)$. Multiplying, we see that if $(x-2)(y-4)(z-6)$ is nonzero, the product of 3 perfect squares is negative, which is absurd. Then one of them is $0,$ and it is easy to check this implies the solution claimed.

We claim that $(x, y, z) = (2, 4, 6)$ is the only solution, which clearly works.

Let $x = 2 + a$, $y = 4 + b$, and $z = 6 + c$. Substituting these expressions into the system, rearranging, and factoring gives $$-12b = a(a+3)^2,$$ $$3c = b(b+6)^2,$$ $$27a = c(c+9)^2.$$ Now, if $a, b, c$ are all non-zero, then multiplying the three equations above yields $$((a+3)(b+6)(c+9))^2 = -12 \cdot 3 \cdot 27 < 0$$ which contradicts the Trivial Inequality. Thus, at least one of $a, b, c$ is 0, implying that all three variables are equal to $0$. This clearly works and corresponds to the triple $(2, 4, 6)$, which suffices. $\blacksquare$

We claim that the only solution is $(2,4,6)$. Clearly this works.

[color=#00f][b]1. Let $x=a+2$, $y=b+4$, and $z=c+6$.[/b] [/color]

Plugging these in and expanding the equation gives that
\[a^3+6a^2+12a+8=(3a+6)-(12b+48)+50\]
\[\iff{}a^3+6a^2+9a=-12b \iff{}a(a+3)^2=-12b,\]
and
\[b^3+12b^2+48b+64=(12b+48)+(3c+18)-2\]
\[\iff{}b^3+12b^2+36b=3c \iff{}b(b+6)^2=3c,\]
and
\[c^3+18c^2+108c+216=(27c+162)+(27a+54)\]
\[\iff{}c^3+18c^2+81c=27a \iff{}c(c+9)^2=27a.\]

[color=#00f][b]2. $(0,0,0)$ is the only possible solution for $(a,b,c)$[/b] [/color]
Multiplying the three equations we found above, we find that either $abc=0$ or $K^2<0$, where $K=(a+3)(b+6)(c+9)$. Since the latter is obviously untrue, we know that at least one of $a$, $b$ or $c$, is $0$. However, setting one of them to $0$ results in the other also being forced to equal $0$, proving [color=#00f][b](2)[/b][/color].

Therefore, plugging $(a,b,c)$ back into $(x,y,z)$ gives that $(2,4,6)$ is the only solution, finishing the problem.

Great problem!

    Notice the equality case (2,4,6), hence we are motivated to let $x=a+2,y=b+4,z=c+6$ and try to show they're all 0. We get $$a^3+6a^2+12a+8=(3a+6)-(12b+48)+50\iff a^3+6a^2+9a=-12b\iff a(a+3)^2=-12b,$$$$b^3+12b^2+48b+64=(12b+48)+(3c+18)-2\iff b^3+12b^2+36b=3c\iff b(b+6)^2=3c,$$
    $$c^3+18c^2+108c+216=(27c+162)+(27a+54)\iff c^3+18c^2+81c=27a\iff c(c+9)^2=27a.$$ Multiplying them together, the LHS has sign abc while the RHS has opposite sign -abc; hence abc=0. Now, caseworking on any one of them being 0, we get that each of a,b,c=0, as desired. $\blacksquare$

The only triplet is $(x, y, z) = (2, 4, 6)$, which works. To prove this is the only triplet, let $x = a + 2, y = b + 2, z = c + 2$. Then our equations rewrite as
$$\begin{aligned} a(a + 3)^2 &= -12b, \\ b(b + 6)^2 &= 3c, \\ c(c + 9)^2 &= 27a. \end{aligned}$$
If any of $a, b, c$ equals $0$, it's easy to see they must all equal $0$ (which works). Else,
[list]
[*]If $a > 0$, then $b < 0$ from the first equation, and $c < 0$ from the second equation, which violates the third equation
[*]If $a < 0$, then $b > 0$ from the first equation, and $c > 0$ from the second equation, which violates the third equation
[/list]
Thus, $(a, b, c) = (0, 0, 0) \implies (x, y, z) = (2, 4, 6)$, as desired.

[hide=solution]
First, suppose that $x\geq2.$ So, from equation 1, we have $y\leq4$ and from equation 2, we have $z\leq 6.$ Therefore, from equation 3, we have $x\leq 2.$ So, $x = 2,y = 4,z = 6.$ Now, suppose that $x\leq 2.$ So, from equation 1, we have $y\geq4$ and from equation 2, we have $z\geq6.$ Therefore, from equation 3, we have $x\geq2.$ So, $x = 2,y = 4,z = 6.$ In conclusion, the only solution is $(x,y,z) = \boxed{(2,4,6)}.$
[/hide]

We claim that the only triple is $(x,y,z) = \boxed{(2,4,6)}$, which clearly works.

Substitute in $(x,y,z)=(a+2, b+4, c+6)$ to get

\begin{align*}
a(a+3)^2 &= -12b,\\
b(b+6)^2 &= 3c,\\
c(c+9)^2 &= 27a.
\end{align*}

If $a=0$, then we have $b=c=0$, which produces our claimed solution.

If $a>0$, we have $b<0 \implies c<0 \implies a<0$, contradiction.

If $a<0$, we have $b>0 \implies c>0 \implies a>0$, contradiction.

Thus, there are no other solutions than the one claimed.

[quote=USA TST 2009/7 wrote:]Find all triples $ (x,y,z)$ of real numbers that satisfy the system of equations
\[ \begin{cases}x^3 = 3x-12y+50, \\ y^3 = 12y+3z-2, \\ z^3 = 27z + 27x. \end{cases}\][/quote]
Fun(ny?) Problem
realised never typed this up
re-write the equations as follows
$(x-2)(x+1)^2=-12(y-4)$
$(y-4)(y+2)^2=3(z-6)$
$(z-6)(z+3)^2=27(x-2)$
if $x \neq 2, y \neq 4, z \neq 6$ then multiply all of the equations,
we get one side positive while other negative, contradiction!.
Hence forcing $x=2,y=4,z=6$

We claim that the only solution is $\boxed{(x,y,z)=(2,4,6)}$.
Let $x=a,y=2b,z=3c$. Substituting them, we get the following equations:
\[ \begin{cases}a^3-3a=50-4b, \\ b^3-3b=\frac{9c-2}8, \\ c^3-3c=a. \end{cases}\]
Note that the function $f(x)=x^3-3x$ is a cubic polynomial function. $f'(x)=0$ has two solutions, namely $x=\pm1$.
$f(1)=-2,f(-1)=f(2)=2$. If $x<-1$ then $f(x)<f(-1)=f(2)=2$ and if $-1<x<1$, then $2=f(-1)>f(x)>f(1)=-2$. If $1<x<2$ then $f(x)<f(2)$.
This means that if $x<2$ then $f(x)\leq f(2)$ and if $x>2$ then $f(x)>f(2)$.
So, if $a>2$, then $50-24b>2$, giving $b<2$. If $b<2$, then $\frac{9c-2}8\leq2$, which means $c\leq2$. If $c\leq2$ then $a\leq2$, contradiction.
If $a<2$, then $50-24b<2$, giving $b>2$. If $b>2$, then $\frac{9c-2}8>2$, which means $c>2$. If $c>2$ then $a>2$, contradiction.
This means $a=2$ is forced, giving $b=2$ and $c=2$, which clearly work.
So, $(x,y,z)=(2,4,6)$ is the only solution.

Answer is $(2, 4, 6)$.
Substitute in
\[(x, y, z) = (\alpha + 2, \beta + 4, \gamma + 6)\]
\[(\alpha + 2)^3 = 3\alpha - 12\beta + 8\]
\[(\beta + 4)^3 = 12\beta + 3\gamma + 64\]
\[(\gamma + 6)^3 = 27\alpha + 27\gamma + 216\]
\[\implies\]
\[\alpha(\alpha +3)^2 = -12\beta\]
\[\beta(\beta + 6)^2 = 3\gamma\]
\[\gamma(\gamma + 9)^2 = 27\alpha\]
\[\implies\]
\[\alpha\beta\gamma(\alpha +3)^2(\beta + 6)^2(\gamma + 9)^2 = -972\alpha\beta\gamma\]
FTSOC, assume that $\alpha\beta\gamma \neq 0$.
Then dividing by $\alpha\beta\gamma$ results in
\[(\alpha +3)^2(\beta + 6)^2(\gamma + 9)^2\ = -972\]
which is a contradiction due to the trivial inequality.
Hence, $\alpha\beta\gamma = 0$. Plugging $0$ into $\alpha$, $\beta$, or $\gamma$, we can clearly see that it makes the rest of the variables $0$, so our answer for $(x, y, z)$ is $(2, 4, 6)$.

We claim our only solution is $\boxed{(x,y,z)=(2,4,6)}$. Motivated by this answer, we notice our equations can be rearranged to be
\begin{align*}
(x-2) \cdot ((x-2)+5)^2 = -12(y-4) \\
(y-4) \cdot ((y-4)+10)^2 = 3(z-6) \\
(z-6) \cdot ((z-6)+15)^2 = 27(x-2).
\end{align*}

Multiplying these, we see that we must have one of $x=2$, $y=4$, or $z=6$ to avoid a trivial inequality contradiction. However, any one of the these imply the other two, which finishes. $\blacksquare$

We claim that $(x, y, z) = (2, 4, 6)$ which clearly works.

Note the system of equations rearranges to
\begin{align*}
(x + 1)^2 (x - 2) &= -12(y - 4) \\
(y + 2)^2 (y - 4) &= 3(z - 6) \\
(z + 3)^2 (z - 6) &= 27(x - 2).
\end{align*}
Multiplying yields \[(x + 1)^2 (y + 2)^2 (z + 3)^2 (x - 2)(y - 4)(z - 6) = -972(x - 2)(y - 4)(z - 6).\]
Now if $x = 2$, $y = 4$ or $z = 6$ then it is easily seen that $(x, y, z) = (2, 4, 6)$ by direct substitution.
Otherwise \[(x + 1)^2 (y + 2)^2 (z + 3)^2 = -972,\] a contradiction.

The only solution is $\boxed{(2, 4, 6)}$.

Consider rearranging as,
\begin{align*}
(x-2)(x+1)^2 &= 12(4 - y)\\
(y-4)(y+2)^2 &= 3(z - 6)\\
(z-6)(z+3)^2 &= 27(x - 2)
\end{align*}
Multiplying we have,
\begin{align*}
(x-2)(y-4)(z-6)(x+1)^2(y+2)^2(z+3)^2 &= -927(x-2)(y-4)(z-6)
\end{align*}
Clearly if all of the factors $(x-2)$, $(y-4)$ and $(z-6)$ are nonzero we have no solutions. However if we have $x = 2$, $y = 4$ or $z = 6$ we find the claimed solution so we are done.

The hardest part of this problem is finding the solution itself.

We claim that the only solution to this system of equations is $(x,y,z) = (2,4,6).$ This can clearly be shown to work just by plugging in.

Now we show that this is the only solution. Inspired by the equality case, we set $x = p + 2, y = q + 4,$ and $z = r+6.$ After plugging in and reducing the equation, we get the following systems of equations:
\begin{align*}
p(p+3)^2 &= -12q \\
q(q+6)^2 &= 3r \\
r(r+9)^2 &= 27p.
\end{align*}
If any of $p,q,r$ are equal to $0,$ then all of them must be $0$ simply by plugging in, giving us the solution claimed at the beginning of our solution. Henceforth assume that $p,q,r$ are nonnegative.

The first equation rearranges to
\[
-\frac{p}{12q} = \left(\frac{1}{p+3}\right)^2,
\]
so
\[
-\frac{81p}{q} = 972 \cdot \left(\frac{1}{p+3}\right)^2.
\]
Meanwhile, the second and third equations multiply to
\[
qr((q+6)(r+9))^2 = 81pr,
\]
which, since $r \ne 0,$ rearranges to
\[
\frac{81p}{q} = ((q+6)(r+9))^2.
\]
Adding our two equations together, we get
\[
972 \cdot \left(\frac{1}{p+3}\right)^2 + ((q+6)(r+9))^2 = 0.
\]
Since the square of a real number is nonnegative, all of the squares in the above expression are $0.$ In particular, $\frac{1}{p+3} = 0,$ which is impossible.

Therefore, $p = q = r = 0,$ and so the only solution to our system of equations is $\boxed{(x,y,z) = (2,4,6)},$ as claimed.


Το βιβλίο [b]ΙΜΟ COMPENDIUM [/b]με όλα τα θέματα ολυμπιάδων ,τα shortlist (Λυμένα)και πλήθος άλλων ασκήσεων κυκλοφόρηρε και έφτασε και στην Ελλάδα.
 Ένας γνωστός το διαθέτει (σε .... ελληνικό στιλ) στην τιμή των 25 ευρώ, καλοφτιαγμένο .Η τιμή στο ΑΜΑΖΟΝ είναι 80 - 100 ευρώ. Αν κάποιος δεν διαθέτει χρήματα για το πρωτότυπο , ας γράψει να του δώσω τη διεύθυνση για να το βρει φθινότερο.
 Το βιβλίο είναι 750 σελίδες.Υπολογίστε 750 Χ 0,04 = 30 ευρώ. Με 25 ευρώ είναι κάτω του κόστους. Βάλτε και 3 ευρώ το δέσιμο,... , συμφέρει! Αυτό το βιβλίο δεν πρέπει να λείπει από κανέναν φίλο των ολυμπιάδων και των διαγωνισμών.
 Ο γνωστός που τα φτιάχνει για το ελληνικό κοινό , θέλει τα καλά βιβλία να είναι προσβάσιμα σε όλους τους μαθητές  και όχι μόνο σε λίγους που έχουν την οικονομική δυνατότητα  να τα αποκτήσουν.Μην τον ρωτήσετε τι κέρδος έχει , γιατί αρπάζεται - με το δίκιο του - και δε θα ξαναδώσει σε κανένα !!!

    Κάθε καλοπροαίρετος ας το σκεφτεί.Εγώ σας δίνω μόνο την πληροφορία και αξιοποιείστε την.

 Μπάμπης

 ### Σιλουανέ - Γιώργο κλπ, το Σάββατο θα είμαι Αθήνα. Αν σας ενδιαφέρει , να πάρω τηλέφωνο να σας φέρουν και σας  ένα , μαζί με το δικό μου.###

ΚΥΡΙΕ ΣΤΕΡΓΙΟΥ ΘΑ ΜΠΟΡΟΥΣΑΤΕ ΝΑ ΜΟΥ ΣΤΕΙΛΕΤΕ ΣΤΟ ΣΠΙΤΙ ΜΟΥ ΤΟ ΒΙΒΛΙΟ?
ΠΕΣΤΕ ΜΟΥ ΜΟΝΟ ΠΩΣ ΘΑ ΣΑΣ ΠΛΗΡΩΣΩ Η ΠΕΙΤΕ ΜΟΥ ΤΟ ΤΗΛΕΦΩΝΟ
Η ΠΛΗΡΗΣ ΔΙΕΥΘΥΝΣΗ ΜΟΥ
ΑΜΒΡΟΣΙΟΣ ΑΝΤΩΝΙΟΥ ΣΑΒΒΙΔΗΣ
ΣΥΡΟΥ 8 ΗΡΑΚΛΕΙΟ ΑΤΤΙΚΗΣ 14122
ΣΑΣ ΕΥΧΑΡΙΣΤΩ ΠΟΛΥ (ΠΑΡΑΚΑΛΩ ΟΠΩΣ ΑΠΑΝΤΗΣΕΤΕ ΠΟΛΥ ΣΥΝΤΟΜΑ)

Σε ποιον πρέπει να απευθυνθούμε για το βιβλίο?

Λοιπόν , αγαπητοί μικροί και μεγάλοι φίλοι. Χωρίς τα καλά βιβλία  ολυμπιάδων δεν υπάρχει ελπίδα για διάκριση ως χώρα και ως άτομα.Όλα τα καλά βιβλία είναι διαθέσιμα.
     [color=blue]α) Engel: Problem solving strategies

     b) Mathermatical circles- Leningrand olympiads

     c) PrasoloV : Geometry

     d) Zeits: Kraft and ......

     e) IMO compendium

     f) Moskow Olympiads

    g) Tournament of towns: 1980  - 1998( Τα υπόλοιπα είναι στο δίκτυο)και άλλα.[/color]

 Ανοίξτε τα μάτια σας ! Χρησιμοποιώ τη φράση αυτή για να πάρετε πολύ σοβαρά αυτά τα βιβλία. Θα δώσω στον κύριο αυτόν και όλα τα BALTIC WAY, με τις ολυμπιάδες  USA  και  CANADA.

 [hide] Κρατείστε το τηλέφωνο , διότι σε κανα - δυο μέρες θα σβήσω το μήνυμα αυτό για ευνόητους λόγους.[/hide]


Do you guys know any freeware borlandc editor that has a simple "compile file" (kind of like "CTRL-F9" on the old Ms-Dos interface) button? I tried PSPad, but besides the fact that I always had to name the file (it didn't know to compile the file in the current window) I couldn't get it to work properly with the compiler (ie no output was shown).

Try DJGPP for Windows (or GCC as a compiler + RHIDE as an IDE). The look and feel of RHIDE is exactly like the old Borland C. It's the main thing being used at major contests for highschool. (IOI, BOI, etc.)

I don't like the look & feel of the old BorlandC, that's the thing. Otherwise I would use the build in editor :D

Thanks! Dev-cpp is exactly what I was looking for :)


What is the equivalent of $ \frac{\pi}{32}$ radians?
A. 5°6’25”
B. 5°15’12”
C. 5°37’30”
D. 5°45’45”
E. NOTA

[hide="Hint"]Recall that $ \pi^\mathrm{r} \equal{} 180^\circ$.

Therefore, $ \frac {\pi^\mathrm{r}}{180^\circ} \equal{} \frac {180^\circ}{\pi^\mathrm{r}} \equal{} 1$.[/hide]

[hide="Solution"]$ \frac {\pi}{32}\cdot\frac {180^\circ}{\pi} \equal{} \frac {180}{32}^\circ$. I'll leave you to put it in terms of minutes and seconds.[/hide]

 :lol: sorry wasn't thinking. That's what you get for listening to music while doing math, I suppose.

Um, $ \pi$ radian is 180 degrees...

So it's $ \frac{180}{32}\equal{}5.625$

Answer: A

The answer is not A. $ 5.625$ is a decimal, and we're asked for a hexagesimal, not a decimal.

$ \frac{\pi}4\equal{}45^{\circ}$

$ \frac{\pi}8\equal{}22^{\circ}30'$

$ \frac{\pi}{16}\equal{}11^{\circ}15'$

Try cutting that in half once more.

A minute is $ \frac 1{60}$ of a degree and a second is $ \frac 1{60}$ of a minute.

$ 5.625\equal{}5\plus{}\frac a{60} \plus{} \frac b{3600}$

$ a\equal{}37$ and $ b\equal{}30$.

So the answer is $ \textbf{(C) }5^\circ 37'30''$.


calculus
integration
trigonometry
Euler
limit
calculus computations

what is :D

$ \int_0^1\frac {dx}{x} \minus{} \int_0^\infty \frac {\cos\,x \; dx}{x}$

I didn't actually calculate it but I would take a wild guess and say it is Euler gamma! :)

Maybe $ \sqrt {\frac {2}{\pi}}$ or $ \sqrt {\frac{\pi}{2}}$.

I did some numbers in Mathematica and it showed up to be exactly what I said. The integrals can be set to have $ L$ as the lower bound (instead of $ 0$) and then they are the same as

$ Ci(L)\minus{}\ln(L)$

where $ Ci(x)$ is the cosine integral. And the result comes from the fact that $ Ci(x)$ has a Taylor representation of the form:

$ Ci(x) \equal{} \gamma \plus{} \ln(x) \plus{} \sum ^\infty_{k\equal{}1}\frac{\left(\minus{}x^2\right)^k}{2k(2k)!}$

So we have:

$ \lim_{L \rightarrow 0} (Ci(L)\minus{}\ln(L)) \equal{} \lim_{L \rightarrow 0}\left(\gamma \plus{} \sum ^\infty_{k\equal{}1}\frac{\left(\minus{}x^2\right)^k}{2k(2k)!}\right) \equal{} \gamma$

The expression as written is undefined.  You can't simply replace the lower limits on the integrals by $ L$ and then take $ L$ to $ 0$.  The sum of limits is equal to the limit of the sums only if the limits are convergent to begin with; in this case they are not, so the operation is invalid.  If you chose the lower limit of one to be $ L$ and the other to be $ 2L$, for example, you would come up with a different value.

But this is basically a problem of how the question is defined.. I just did something that should be written in the first place since we all know what misan meant.