trigonometry
inequalities
real analysis
real analysis solved

Here is a nice problem.Find all values of $\alpha$ such that the series $\sum_{n=1}^{\infty}{{a_n}^{\alpha}}$ converges, where:
[i]a)[/i]$a_{1}=\sin{x},a_{n+1}=\sin{a_{n}}$ for $n{\geq}1$ and $x\in{(0;\frac{\pi}{2})}$,
[i]b)[/i]$a_{1}=\arctan{x},a_{n+1}=\arctan{a_{n}}$ for $n{\geq}1$ and $x>0$.

Conjecture: the answer is the same in both places, with the series converging for $\alpha>2.$ I don't have the proof yet.

Yes you are right [b]Kent[/b].How did you guess it?

It can be shown that $\sin\frac{1}{n+\frac{k}{n}}\geq\frac{1}{n+\frac{k+1}{n}}$, $k=0,...,n-1$
So first series diverges for $\alpha=2$.

Something similar should work for second series too.

Moreover, we can establish $\sin\frac{1}{n+\frac{k}{n}}\leq\frac{1}{n+\frac{k+0.1}{n}}$. It gives the series converges for $\alpha>2$.

Mr Merryfield! You have a great intuition!

Sorry for the stupid question Myth but how do you deduce from your inequalities the result? :?  :blush:

First inequality shows that there are at least $n$ terms are greater than $\frac{1}{n+1}$, for example.

[quote="fedja"]This is a classical problem. Usually one asks to find an equivalent for $a_n$, which in both cases is some constant divided by $\sqrt n$, from where the answer follows immediately.  :)[/quote]
Why did I not noticed that :? ? It is in a direct connection with my inequalities.

My solution is the following:
Notice that $(a_{n}^{\delta}-a_{n+1}^{\delta}){\sim}{\delta}\frac{a_{n}^{\delta+2}}{6}$,so
the series converge for $\alpha>2$ and disverge for $\alpha{\leq}2$.

For a/ it was posted an equivalent is $a_n \sim \sqrt{\frac{3}{n}}$ with Cesaro theorem

$a_n^{\alpha}\sim \frac{C}{n^{\frac{\alpha}{2}}}$ the serie converge if

$\alpha >2$

[quote="Moubinool"]For b/ $a_n\sim \sqrt{\frac{3}{2n}}$[/quote]
We have already got the idea.  :D

[quote="Levon Nurbekian"]Yes you are right Kent.How did you guess it?[/quote]
Spreadsheet-assisted intuition. I knew the sequence tended to zero slowly, and suspected some power of $n.$ I used the spreadsheet to find $a_n$ out to $n=256,$ sampled it at $n=2^k,$ and tried several manipulations of those numbers before deciding that $a_n=O(n^{-1/2}),$ or at least that $\sqrt{n}a_n$ appeared to grow slower than any power of $n.$ None of it was proof, of course.


trigonometry
probability
geometry
AMC
AIME
geometric transformation
reflection

1. Let
$f_k(x)=\frac{1}{k}\left(\sin^k x + \cos^k x\right)$
for all positive integers $k$. Prove that
$f_4(x)-f_6(x)=\frac{1}{12}$
for all real x.

2. A circle of radius 1 is randomly placed in a 15-by-36 triangle ABCD such that the circle lies completely inside the rectangle. Find the probability that the circle will not touch diagonal AC.

3. Find the area of the region containing all points $(x, y)$ such that $x^2 + y^2 \le 100$ and $\sin (x + y) \ge 0$.

[hide]First one:
$1/4(\sin^4x+\cos^4x)-1/6(\sin^6x+\cos^6x)$.  After factoring the second term and then using the identity $\sin^2x+\cos^2x=1$, one gets $1/12(\sin^4x+\cos^4x)+1/6\sin^2x\cos^2x.$  If one then factors away the 1/12, they get $1/12(sin^2x+cos^2x)^2=1/12$[/hide]

[quote="mna851"]I don't believe the second one is a trigonometry problem.  At least it doesn't have to be, I just used similar triangles.[/quote]
Yes, I did the same. There is a trig solution though, as it's in a trig book.

[quote="chesspro"]2. A circle of radius 1 is randomly placed in a 15-by-36 triangle ABCD such that the circle lies completely inside the rectangle. Find the probability that the circle will not touch diagonal AC.[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=292973#p292973]AIME 2004A, #10[/url]

[hide]Notice that if you reflect (x,y) across the line $x+y=0$ to ($x_i$,$y_i$), then $x+y=-(x_i + y_i)$. Seeing as sin(x) is an odd function, reflecting across that line will change the sign of sine. So if a point works, its reflection across that line does not work.

Then, $x^2+y^2 \le 100$ describes a circle of radius 10 around the origin. This circle is symmetric with respect to the line $x+y=0$, and thus the other inequality takes out half of its area. The total area is $\frac{10^2 \pi}{2} = 50 \pi$.[/hide]


inequalities
geometric inequality
algebra solved
algebra

x,y,z,a,b,c>0
x+a=b+c  y+b=c+a   z+c=a+b
Prove that:
 \sum (x+1) \sqrt (zy)  \leq a(2c+1)+ b( \sqrt (2a 2 +2b 2 ) +1) -c(a-1)

Hi,

the sum on the left-hand side should be cyclic?!

Ok, this looks like a geometric inequality if you set x=2x' etc.
(despite of its inhomogenity).
So use the inverse transform a=2*(y'+z') etc. to get rid of the
three conditions. The term  \sqrt(2a^2+2b^2)-1 reminds me on
the length of a bisector of a triangle...

Eckard

The initial enequality is the sum of the two inequalities:
1) \sqrt(xy)+\sqrt(xz)+\sqrt(zy)\leq a+b+c
and
2) x\sqrt(yz)+y\sqrt(xz)+z\sqrt(xy)\leq ac+b\sqrt(2a2+2b2)

1) It is a well known inequalities because
a+b+c=x+y+z\geq \sqrt(xy)+\sqrt(xz)+\sqrt(zy)

2) As \sqrt(2a2+2b2)\geq a+b we will prove that
x\sqrt(yz)+y\sqrt(xz)+z\sqrt(xy)\leq ac+ab+b2.
Actually ac+ab+b2=(x2+y2+2z2+2xy+4xz+2yz)/4=
(x2+y2+z2+z2)/4 + (xy+xy+xz+xz)/4 + (xz+xz+yz+yz)/4 \geq x\sqrt(yz)+y\sqrt(xz)+z\sqrt(xy) as a result of Cauchy inequlity.


algebra
polynomial
calculus
integration
induction
function
least common multiple

(a) Let $ \varepsilon>0$. Define $ \varphi(x)\equal{}x\plus{}x(1\minus{}2x)(1\minus{}x)$. Then $ \varphi$ is monotone increasing with fixed points $ 0,\frac{1}{2},1$. Show that some iterate $ \varphi_{n}$ of $ \varphi$ is monotone increasing on $ [0,1]$ and $ |\varphi_{n}\minus{}\frac{1}{2}|<\varepsilon$ for $ x \in [\varepsilon,1\minus{}\varepsilon]$.

(b) Given $ \alpha$ with $ 0<\alpha<1$ and any $ \varepsilon>0$, show that there is a polynomial $ \psi$ with integral coefficients and no constant term such that $ 0 \leq \psi(x) \leq 1$ in $ [0,1]$ and $ |\psi(x)\minus{}\alpha|<\varepsilon$ for all $ x \in [\varepsilon,1\minus{}\varepsilon]$.

[quote="chungyc"](a) Let $ \varepsilon > 0$. Define $ \varphi(x) \equal{} x \plus{} x(1 \minus{} 2x)(1 \minus{} x)$. Then $ \varphi$ is monotone increasing with fixed points $ 0,\frac {1}{2},1$. Show that some iterate $ \varphi_{n}$ of $ \varphi$ is monotone increasing on $ [0,1]$ and $ |\varphi_{n} \minus{} \frac {1}{2}| < \varepsilon$ for $ x \in [\varepsilon,1 \minus{} \varepsilon]$.[/quote]
You should have no difficulty showing that $ \varphi$ is increasing. By an easy induction argument, so is $ \varphi_n$.

By elementary algebra, $ \varphi(x) \minus{} \textstyle\frac12 \equal{} (x\minus{}\frac12)(2x^2\minus{}2x\plus{}1)$. On the interval $ [\varepsilon,1 \minus{} \varepsilon]$, the maximum value of $ 2x^2\minus{}2x\plus{}1$ occurs at the endpoints, and is $ \delta: \equal{}2\varepsilon^2\minus{}2\varepsilon\plus{}1$, which is less than 1. So $ |\varphi(x) \minus{} \textstyle\frac12| \leqslant \delta|x\minus{}\frac12|$, and by iteration $ |\varphi_n(x) \minus{} \textstyle\frac12| \leqslant \delta^n|x\minus{}\frac12|$. Now you just have to choose $ n$ so that $ \delta^n<\varepsilon$.

[quote="chungyc"](b) Given $ \alpha$ with $ 0 < \alpha < 1$ and any $ \varepsilon > 0$, show that there is a polynomial $ \psi$ with integral coefficients and no constant term such that $ 0 \leq \psi(x) \leq 1$ in $ [0,1]$ and $ |\psi(x) \minus{} \alpha| < \varepsilon$ for all $ x \in [\varepsilon,1 \minus{} \varepsilon]$.[/quote]
Part (a) does this for $ \alpha\equal{}\textstyle\frac12$. For a general value of $ \alpha$, start with a polynomial such as $ p(x) \equal{} (4\alpha\minus{}1)x\plus{}(2\minus{}4\alpha)x^2$, that takes 0 to 0, 1 to 1, and ½ to α. Use (a) to choose a polynomial that approximates ½ really closely on a suitable interval, and compose it with p(x).

I would like to go on to the other parts of this problem.
The original problem is as follows:
Let $ I\equal{}[\minus{}1,1]$ and $ f$ a continuous real-valued function on $ I$ such that $ f(\minus{}1),f(0),f(1)$ are integers and $ f(1) \equiv f(\minus{}1) \mod 2$. Then given $ \varepsilon>0$, there is a polynomial $ P$ with integral coefficients such that $ |f(x)\minus{}P(x)|<\varepsilon$ for all $ x \in I$.

Parts (a) and (b) were the first 2 hints. The next 2 hints, which I have sort of solved, are as follows:
(c) Given a number $ 0<\alpha<1$ and any $ \varepsilon>0$, there is a polynomial $ \psi$ with integral coefficients (and no constant term) such that $ 0 \leq \psi \leq 1$ on $ [0,1]$ and $ |\psi(x)\minus{}\alpha|<\varepsilon$ for all $ x \in [\varepsilon,1\minus{}\varepsilon]$.

(d) Let $ P$ be a polynomial with integral coefficients, and suppose that $ P(\minus{}1)\equal{}P(0)\equal{}P(1)\equal{}0$. Let $ \beta$ be any real number. Then for each $ \varepsilon>0$, $ \beta P$ can be uniformly approximated to within $ \varepsilon$ on $ [\minus{}1,1]$ by a polynomial having integral coefficients and no constant term.

The last part (e) which I would like to ask is:
Reduce the statement of the original problem to (d) and the Stone-Weierstrass Theorem.

[quote="chungyc"]The last part (e) which I would like to ask is:
Reduce the statement of the original problem to (d) and the Stone-Weierstrass Theorem.[/quote]
So you are given a continuous function f on [-1,1] with f(-1)=a, f(0)=b and f(1)=a+2c, where a, b and c are integers. Let Q(x) be the polynomial $ Q(x) \equal{} (a\minus{}b\plus{}c)x^2 \plus{} cx \plus{} b$. Replacing f by f - Q, we may assume that a=b=c=0.

Next, given f with f(-1) = f(0) = f(1) = 0, use the S–W theorem to approximate f by a polynomial R(x) with rational coefficients, such that R(x) also vanishes at -1, 0 and 1. (That's a bit fiddly, but it can be done.)

Let N be the least common multiple of the denominators of the coefficients of R(x), so that N*R(x) has integer coefficients and vanishes at -1, 0 and 1. Let $ \beta \equal{} 1/N$, and apply (d) to the polynomial N*R(x) for that value of β.


Right now we are still in the sign-up phase.  Everyone is encouraged to sign up, not just South Carolinians.  You may sign up for any of three levels, listed below.  You may only sign up for one level.  Please sign up, and post any questions, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107018]here[/url].
[u][b]Levels:[/b][/u]
Level 1: Middle Schoolers and Below
Level 2: AIME Scores Less than 8
Level 3: AIME Scores Greater than or Equal to 8
If you are eligible for both levels 1 and 2, it is at your discretion to choose one.

[b][u]Format:[/u][/b]
25 Questions per level.

[b][u]Date:[/u][/b]
Late October to early November.  We will have more details as sign-up progresses.  Contest will last one month.

[b][u]Promotion:[/u][/b]
Please add a link to this contest in your signature, we'd really appreciate it.

[b][u]Moderators:[/u][/b]
The moderators of this contest are myself and mad_skillz_aops.  We will both be posting under the username [b]AoPS Kings Director[/b], and all PMs should be directed to that account.

This thread will be locked after mad_skillz_aops has a chance to post in it; please keep all questions to the other thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107018]here[/url].  Thanks!

Sign-Up List:

[b][u]Level 1:[/u][/b]
SplashD
kheg_k
bpms
13375P34K43V312
anirudh
ragnarok23
b-flat
daermon
daine
akalra1
slamchess
mthiyaga
meenamathgirl
goldendomer
D.P.L
klutzycherry14
now a ranger

[b][u]Level 2:[/u][/b]
lotrgreengrapes7926
cosinator
Xantos C. Guin
rzsolt
fireemblem13
Ignite168
ShoeFactory
Klebian
junggi
ch1n353ch3s54a1l
not_trig
PTynan89
robinhe
mustafa
maokid7
Captain Sugar
vihits13
azndemon43
Leao
NeverOddOrEven
AdorableAddict
pianoforte
chi^2
perfect628
DanK
groundhawk2006

[b][u]Level 3:[/u][/b]
mathgeniuse^ln(x)
solafidefarms
jb05
rem
tim1234133
E^(pi*i)=-1
Xevarion
Kurt Godel
K81o7
matt276eagles
me@home
Mathgod
Treething
thealchemist
lightrhee
JeriC
AnnieS1527
Inquisitor
Mathlete9366
Ilthigore

(Joe, can Karan edit this post if I lock the topic?  If not, I won't lock it, but no one post in this topic please!)

Er, I dunno :maybe: You can test it and see.

There will be 25 problems for each level.  You may only participate in one level officially but I will be happy to check your answers for other levels as long as you state in the title thats its unofficial.  Calculators will be allowed.  You will be given 1 month to complete this test.  The problems will be posted in late October or November.  When you finish, PM your answers to AoPS Kings Director.  If you have any questions, PM mad_skillz_aops.  Sign-ups are still allowed.

Yeah, seriously...I mean...I thought someone said it will run between October 15-October 20.

[quote="anirudh"]Maybe he didn't want to do it :rotfl:

No seriously, when is this going to happen. Do not delete this post please. Thanks.[/quote]
[quote][u][b]Date:[/b][/u]
Late October to early November. We will have more details as sign-up progresses. Contest will last one month. [/quote]
next time, read the whole thing before you post

Dude, it takes lots of work to run a competition, especially 3 different levels simultaneously.

Chill

The AoPS Kings topic was locked (not by me) due to continued spamming by iversonfan2007 and MysticTerminator despite repeated deletion of said spam.  I will open it up again, contingent upon NO SPAM!

The October date was an estimate, NOT AN EXACT DATE!  We just need a couple more level 3 problems and we will be done.

We will be closing the sign-ups after today ends.  If you haven't signed up, please do it today!

Locked due to continuous spam.  PM serious questions to me or mad_skillz_aops.


Let $P(x)$ be a polynomial of degree $n$ whose roots are $i-1, i-2,\cdot\cdot\cdot, i-n$ (where $i^2=-1$), and  let $R(x)$ and $S(x)$ be the polynomials with real coefficients such that $P(x)=R(x)+iS(x)$. Show that the polynomial $R$ has $n$ real roots. (R. Stanojevic)

if $i-1$ is a root of $P(x)$ isnt it necessary, that $-i-1$ is a root as well? and since $P(x)$ has only $n$ roots, that isnt possible or did i get something wrong?

[quote="peeta"]if $i-1$ is a root of $P(x)$ isnt it necessary, that $-i-1$ is a root as well?[/quote]

Only if the coefficients are real - which they are not in this case, obviously.

I have edited it.
Also see http://www.imo.org.yu/othercomp/Yug/YugTST04.pdf

Well, you saw from where did i get it  :(  Does anybody know the right statement

Oh the mistake is MINE - bad algebra.

Deleted the so called "counterexample".

:)  I`m happy that the problem is correct. Hope somebody have a solution :?

This is really hard for Me. No matter how am i trying i cannot reach anything. Will any smart person help

Here's an idea to get you started.

More generally, let
\[ P(x + yi) = R(x,y) + iS(x,y), \]
where $R(x,y)$ and $S(x,y)$ are polynomials with real coefficients.

In particular, for $n = 4$,
\[ P(x + yi) = (x + yi + 1 - i)(x + yi + 2 - i)(x + yi + 3 - i)(x + yi + 4 - i). \]
Expanding, if my algebra is correct, we get
\[ R(x,y) = x^4 - 6x^2 y^2 + y^4 + 10x^3 + 12x^2 y - 30xy^2 - 4y^3 + 29x^2 + 60xy - 29y^2 + 20x + 66y - 10. \]
So the problem, in this case, is to prove that the polynomial $R(x,0) = 0$ has four real roots.

Here's a plot of $R(x,y) = 0$ in the plane:
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=576[/img]

Here's a zoomed out view.  (There's some noise at the centre.)
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=577[/img]

This second view shows that the graph of $R(x,y) = 0$, when far from the centre, becomes eight spokes.  What are the angles of these spokes?  Why?  What happens in the general case?  And why does this prove the result?


geometry
incenter
circumcircle
conics
analytic geometry
vector
parallelogram

I have a question: What is the sufficient and nessecery conditions for two triangles having:
(1) the same incentre.
(2) the same othocentre.
(3) the same circumcentre.
(4) the same gravititianal centre.

There has just been a problem related to (4) in the National College Entrance Exam in China.

Hope I posted it in the correct forum.

for the incenter and circumcenter:
Projective geometry has theorems allowing one to express the condition that the 6 vertices are on a conic (Pascal's theorem) or that the 6 sides are tangent to a conic (Brianchon's theorem).   For your question, one needs an additional condition saying that the conic is a circle, which is equivalent to saying that the conic goes through the 2 "circular points at infinity" (i.e. projective solutions of $X^2 + Y^2 = 0$).  If you allow complex coordinates and projective changes of coordinate, these points are no different from any other 2 points, so the problem is to give conditions when a conic (given by 6 points or 6 tangent lines) goes through 2 given points.  Pascal's theorem answers the question for a conic given by 6 points.  In principle you can work out what this means in the standard projective plane with real coordinates (and so eventually in the metric Euclidean plane for which concepts like "circle" make sense), but I don't see immediately what the answer would be.

for the center of gravity:
the sum of the vertices of the triangles are equal.  So, for example, the sum of vectors
$A_1 B_1 + A_2 B_2 + A_3 B_3 = 0$.  The sum of any 2 of these vectors can be computed using a parallelogram, and its diagonal is equal to and parallel to the other vector.  Were you looking for some more intelligent condition?


A computer program requires a password of atleat 6 but no more than 8 characters.All letters (of the english alphabet,not case sensitive)and digits (0 to 9) may be used.The password must contain atleat one digit and one letter.How many different password are possible.

For $ 6$ characters (for $ 7$ and $ 8$ it's similar):
Without restrictions, there are $ (26\plus{}10)^6\equal{}36^6$ passwords, where $ 26$ is the number of letters in the English alphabet (I believe). Among them, there are $ 26^6$, which containt only letters and $ 10^6$, which contain only digits. So for there are $ 36^6\minus{}26^6\minus{}10^6$ good passwords.


superior algebra
superior algebra solved

Why is the p-adic expansion of a rational always periodic?

Assume the rational number is $p^{n}\frac{a}{b}$ with $p\not|ab$.
$p^{n}$ doesn't matter. It just induces a shift in the $p$-adic expansion. We can also assume that $0\leq a<b$ because adding integers doesn't change the periodicity of the expansion either.
$\frac{a}{b}=\frac c{p^{u}-1}$, where $b|p^{u}-1$ and $\frac{c}{a}=\frac{p^{u}-1}b$. For example let $u=\varphi(b)$.
Now $a<b\Rightarrow c<p^{u}-1$.
We can write $\frac{a}{b}=\frac c{p^{u}}\cdot\sum_{i\geq 0}\frac 1{p^{ui}}=0,(c)$ where we can add some $0$s in front of $c$ to make it have length $u$ in the $p$-adic expansion.


ARML
Vieta
geometry
AMC
USA(J)MO
USAMO
projective geometry

Well, ARML 2006 is over.  So, what did we think?  You can speak about anything: what you thought about the questions, how your experience was, what you think you should have done, etc.

Looking at ARML 2005 on the [url=http://www.arml.com/]website[/url], I'd say that was easier.  Of course, I wasn't under so much pressure.

I think the Individual Test was definitely easier. Everything except #6 and #8 was basically just a standard problem (even #6 was pretty much standard). I was surprised that only 9 people at the Las Vegas site made the tiebreaker (4 from WA, yay). Unfortunately, the heat seemed to get to us in the 2nd relay, and we lost our lead in the site.

The Las Vegas site was pretty disorganized this year (probably due to a series of misfortunes). The testing room yesterday was not only amazingly hot, but there was no microphone so the director had to like shout at people to be quiet. Also, the song contest and super relay never happened (did it at other sites?). It was pretty fun anyway, though.

The Las Vegas site coordinator [i]passed out.[/i]

1, 2, 3, 4, 5, and 7 were all quite basic, and 6 only required a little bit of thought and a firm grasp of Vieta's formulas.  8 was the only really challenging problem.  ... Considering that, I did surprisingly poorly.  I blame the heat.  :(

Given the heat, I'm glad that the Super Relay didn't happen. The heat also probably caused the disorganization, since Mr. Rusczyk passed out from it (I assume, I couldn't quite hear the provisional site coordinator who took over for him).

And we were trying to test in that same heat...

since only WA people have responded...
i'll do too! :D

yeah i basically agree with all of the above :D
LV was pretty disorganized...the song contest didn't happen;otherwise we would have gotten something...
and that stupid raffle thing never hit anyone from our team( except once) ;)

but again, the heat was a major problem...i could not think on the last 2 rounds especially on 3rd round of individual...my energy was spend on trying fan myself with a piece of paper instead of doing math!

tests were quite easy...Relay#2 killed us though...otherwise we would have been in the top 10 easily!

[quote="Kent Merryfield"]You wouldn't have gone unopposed had there been a song contest ....

I asked jmerry about the facilities in use at ARML Las Vegas in 2000 when he was there. He didn't recognize the "Paradise Campus"; he said they used two side-by-side auditoriums for the individual and relay rounds, then piled everyone into one for the awards ceremony.

Frankly, my reaction to the room was that the Paradise Campus looked like a junior high school, in which case the room we used would have been the "cafetorium." Even if the airconditioning could have kept up, it would still have been a bad room.

Question for you Washington guys: what does your whole score line look like? My understanding was that you beat us by one point. Our line:

Individual: 8+15+14+10+13+6+13+0
Team: 28
Relay: 14+6
Power: 29
Total: 156[/quote]

This is from recollection:

Individual: 82 (don't know the breakdown)
Team: 28 (32, +4 later from #9 dispute)
Relay: 18+4
Power: 27
Total: 159 (163)

I'm pretty sure about most of those.

[quote="paladin8"]The Las Vegas site was pretty disorganized this year (probably due to a series of misfortunes). The testing room yesterday was not only amazingly hot, but there was no microphone so the director had to like shout at people to be quiet. Also, the song contest and super relay never happened (did it at other sites?). It was pretty fun anyway, though.[/quote]

The song contest did happen, albeit in a rather ad hoc manner outside the cafeteria after awards.  I think we had parodies of Mambo No. 5 and All Star, and a rendition of "Aleph-Nought Bottles of Juice on the Wall", though I only caught the last two.  Next year, if the organizers aren't going to put something on, you could probably just organize it in advance via AoPS.

The super relay was going to happen, but they canceled it due to the heat.  I don't see any reason why it wouldn't have happened at other sites.

Finally, here are some score lines I wrote down in between rounds in the grading room.

Washington Gold
Team: 28 (changed to 32, apparently)
Power: 27
Individual: 10, 14, 14, 13, 11, 5, 13, 0.  80 total, unless something was changed later.
Relay: 18/4; 22 total.

Bay Area A
Team: 20 (changed to 28)
Power: 39 (woohoo!)
Individual: 10, 11, 10, 13, 13, 4, 13, 0.  74 total.
Relay: 16/12; 28 total.

Alabama
Team: 28 (changed?  I don't know.)
Power: 30
Individual: 10, 10, 12, 9, 15, 5, 11, 0.  72 total.
Relay: 16/8; 24 total.

Colorado
Team: 24 (changed?)
Power: 38 (leading the site after Friday night)
Individual: 9, 10, 10, 7, 11, 1, 12, 0.  60 total.
Relay: 18/8; 26 total.

Although the contest was fun as usual, I hope the Western site gets moved back to San Jose next year, because I think the heat in Las Vegas interfered greatly with the contest.  My performance on Sunday certainly increased once the bottled water arrived, and many other people were also negatively affected by the heat.
I hope that Mr.Rusczyk is OK.

I've just been lurking here, since I had to miss ARML this year (I've gone every year since 1995 -- it felt very odd not being there!), but I do remember the building used for the individual/relay rounds in 1995-2000: Classroom Building Complex A
[the little round building near CBC-C, which is where you had the team/power rounds):
http://maps.unlv.edu/cbc.html
 it's four little-bitty auditoriums wrapped around a common central area -- they could get around 6 teams into each one for the individuals, and then they tried to run everything simultaneously. And we all crammed into a single one for the awards.

At least one year, the air conditioning failed in one of the auditoriums, and they had to cram too many teams into the others to make it work.

I don't really understand why a huge school like UNLV doesn't seem to have an auditorium or lecture hall big enough to accomodate everyone.

Certainly, San Jose is a MUCH more convenient site for our teams, but the reality seems to be that many more other teams will come to Las Vegas (either because it's more centrally located, or because airfares are cheaper, or because there's more to do in Las Vegas than San Jose). I'm for whatever allows the most teams to participate. But, given all that,  we should give helpful feedback to ARML so that the difficulties with the site can be addressed --  an air conditioned location for the individual rounds when the temperature outside is 105 is certainly a start!

[quote="Sly Si"]Colorado
Team: 24 (changed?)
Power: 38 (leading the site after Friday night)
Individual: 9, 10, 10, 7, 11, 1, 12, 0.  60 total.
Relay: 18/8; 26 total.[/quote]

This is the right score breakdown for Colorado.  We didn't get team problem 9, so our team score remained at 24.

[quote="TMA"]I don't really understand why a huge school like UNLV doesn't seem to have an auditorium or lecture hall big enough to accomodate everyone. [/quote]
I've played "what if" to ask if our campus could host. (Not a serious suggestion, by the way - there may be other scheduling issues that would arise.) We'd have no problem with Friday night team and power rounds (I'm talking about the West when I say things like "Friday night") but for Saturday morning, the best I can come up with:

* Two side-by-side lecture halls each seating about 250. That sounds a lot like what TMA described of the previous Las Vegas setup - and if there are going to be 20+ teams, it wouldn't be big enough.

* A theater, used for plays. I don't know what the seating capacity is - I suspect not big enough.

* Our performing arts center. Besides the likely cost, there's a fatal flaw: steeply banked seats with no central aisles; a relatively small number of very long rows of seats. The proctors wouldn't be able to get in and out.

* The biggest ballroom of the student union. We'd have to set out rows of chairs, just like we had yesterday. I think it would be slightly bigger, and we could set up audio-visual equipment, and there's the whole climate advantage. That may be the best option. It's the room we're going to use for our local contest next year, which involves 150-200 students and table seating.

* The floor of our sports arena.

But we don't have anything like that big old auditorium that San Jose has.

I'm curious to hear what people have to say about the Power Round.

[quote="lingomaniac88"]Well, ARML 2006 is over.  So, what did we think?  You can speak about anything: what you thought about the questions, how your experience was, what you think you should have done, etc.[/quote]
I don't know what other arml-ites thought about it, but what did you who went think about the wording in the power round.  Because I can't quite match the language those origametry questions were in, I'd like to quote it.  Now it seems a little obvious to me that:

[b]B5: Given line $l$, point $P$ not on $l$, and point $Q$ not necessarily on line $l$, you can make a fold that places $P$ onto line $l$ while the crease passes through $Q$, [i]unless no fold satisfying these conditions exist.[/i][/b](italics mine)

B6 also had the "...you can make a fold...unless no fold satisfying these conditions exists" phraseology.  Wow.
Did anyone else think that the power round was essentially asking us to derive a whole "new" geometry given certain axioms (B1-B6)?  It all just seemed to remind me of Euclid.
In terms of favorite problems, I really like #6 on the individual round in terms of the neat/interesting solutions (although i didn't get it at the time) and again # 6 on the team round (which i didn't work in that round anyway, but afterwards.)

all i know is that the team round killed us :(

I have to agree with Midnight.  Penn State was a little too large for my taste, especially with all of the shadowy stuff up top and the giant black curtains which made it really kinda spooky.  I know that sounds weird but you had to be there.  I think that given what I've heard about the other sites, Penn State seemed to be the best since the temperature was nice and there was more than enough space.  Do any teams at the Iowa site have something to say about their location?

In the end, though, I can't really complain seeing as my team took 1st in the nation.  It helped having two USAMO winners on our squad for the Power and Team rounds.

The only way Penn State could have been better is if they made breakfast digestable, and they made the stadium one big freezer. That would have been cool.

BTW: Th3 numb3r 3 usually only r3plac3s E and not any oth3r l3tt3r.

[quote="Teki-Teki"]Penn State was a little too large for my taste, especially with all of the shadowy stuff up top and the giant black curtains which made it really kinda spooky.  I know that sounds weird but you had to be there.
[\quote]

Umm... what?? It looked like a normal sports stadium thing to me.

I thought it was better organized and whatnot than last year. The stadium was kind of over-doing it, and it was hard to see all the way to the "stage". It was comfortable and helpful for the relay and super-relay, though.

[quote="Teki-Teki"]Penn State was a little too large for my taste, especially with all of the shadowy stuff up top and the giant black curtains which made it really kinda spooky.  I know that sounds weird but you had to be there.
[/quote]

Umm... what?? It looked like a normal sports stadium thing to me.

I thought it was better organized and whatnot than last year. The stadium was kind of over-doing it, and it was hard to see all the way to the "stage". It was comfortable and helpful for the relay and super-relay, though.

[quote="jb05"] It was comfortable and helpful for the relay and super-relay, though.[/quote]

NC definitely had the most style on the super relay with your V formation thing (or whatever it's called). ;)

This was my 5th ARML as proctor.  Although the size of the Jordan Center may have made things a bit impersonal, I can not say enough great things about it. I loved the wide (and flat!) walkways and the good sightlines.  No tripping on legs. No contortions.  No sore back or sore knees (like the first four times).  :)

I mentioned this to ARML management right on the spot (as did another proctor - same thoughts.) . Problem: Jordan Center rents for 10000 dollars a day vs 3000 dollars at Eisenhower. Eisenhower was not available this year due to renovations.    :(

jb05, (team/roommate)

what I didn't like was that from where we were, there were no lights above us, so it made everything in the center of the arena too bright, and there were no lights on at the far end, so it was harder to see the stage.  plus there's the fact that the announcer was really really far from us.  it made me feel isolated.

joml88, our V formation may have looked cool (and if we had been geese it would have been aerodynamic) - but we didn't even get the problem right :blush:

if we had there wouldn't have been a tie. ;)

Hip - this was my first ARML so I guess I didn't really know how bad it has been, although with all of the complaining (which probably was justified, I'm not passing judgment) on the part of the teams at UNLV, Penn State seems to have worked just fine - the JumboTron clock was cool.

[quote="Three"]Ye3h...I didn't really wann3 meet m3ny aopsers since Three was a little busy on Fr3day...I d3n't think anyone really kn3ws who Three is... But I d3d go to P3nn State and m3t WarpedKlown, Iversonfan2005, joml88, MysticTerminator, and JBL. Maybe I passed by a f3w aopsers like Boy Soprano, solafidefarms...etc...3dk.[/quote]

What about me?

The curtains were there because the upper levels weren't being used.

On the subject of the power question, our team didn't use the sixth rule at all, which might account for something, and also 10A was pretty trivial once you started thinking about constructing equilateral triangles.

[quote="Boy Soprano II"]On the subject of the power question, our team didn't use the sixth rule at all, which might account for something, and also 10A was pretty trivial once you started thinking about constructing equilateral triangles.[/quote]

You weren't "supposed to" use rule 6 anywhere but in the question that specifically asked about it.

This was my first ARML. :)   I'm so glad that I went this year. The power topic was nice, though a bit vague...

I enjoyed the power round.  Of course, I didn't do too well {proofs are NOT my strong point}.  I liked how they had origami (finally a 'real' use for paperfolding). Most of my friends think I'm really good at it.  I'm just not so good at making polygons with 11 sides... There is a book that shows how to do that, and it supposedly works for any amount of sides.  So much for the idea that 17-gons are impossible to construct.  Good job Kunihiko Kasahara!

When people say that for some value of $n$, an $n$-gon is "unconstructable", they mean "unconstructable with a straight-edge and compass", i.e., with Euclid's axioms, not the origami axioms.

Also, 17-gons are famous because it [i]is[/i] possible to construct a regular 17-gon with a standard construction.


[quote="someperson01"]Hi. After a long period of inactivity I though it would be nice to host a mock round, as Mathcounts is steadily approaching. If enough people sign up (about 10 or so) I'll release the problems tomorrow.
So please sign up!
 :lol:[/quote]
I'm in...
How hard is it?

I'm in. Are you making up these questions, or are you "ripping them off" real competitions? I would like to see new problems. :P

Ok I'll do it, but only if these are original problems of an appropriate level.

[color=orange][If you're going to complain about others' leetspeaking, don't abbreviate OK either.][/color]

I'll do it. I also probably won't waste time doing old comps for the 100th time.

[quote="13375P34K43V312"]k I'll do it, but only if these are original problems of an appropriate level.[/quote]
I know right.
I'm sure we all enjoyed : http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112370
Greatest mock ever [/sarcasm]
I have a feeling this is going to be a good mock :)
The only great mathcounts mock I've seen is one that was written by a national mathcounts champion...

[quote="Fanatic"]A mi, me gusta.[/quote]
No comma. :wink:

@ someperson01 - That's 12 people. :)

[quote="fireninja0"]#30 on Sprint:
it says find length LC. L is a line and C is a point, so the length should be a the line perpendicular to L and going through C. Is this problem easier than it looks, because I think the problem is basically telling you the answer...
Maybe I drew it wrong...

P.S., weekend evenings are good for me (Countdown)[/quote]

Oh I assumed that L was the point where the perpendicular touched $\line{AB}$.

Thanks, sonny :wink:
Arg!! They're due today? I've been busy this week, with my school Mathcounts elimination and all. I hope someperson01 will still grade mine... :maybe:

i promise i will send them in by 8 ET completed or not :maybe:

[quote="anirudh"]For #10, are you sure that it works? :maybe:[/quote]

I think it's supposed to be OP. He said earlier that r is the radius.

And we were supposed to stop at 40 min. for sprint and 24 min. for target, right? Just making sure.

Most people got #6 on target wrong. The square makes +-1 on the right side, so -1,-2,-3,-4 work. Thanks for participating.
Top scores wil be posted, unless you object.

yeah if i am in the top i object but seeing how many problems were *incomplete* there is no hope.... :D

I dont really mind just post the top/all scores. I just wanted to know how many people participated and how many go to countdown. I missed like 3 problems because I  read them wrong. :blush:

is some person still alive? oh well finals

its not too late to koin now is it?
if it isn't, im in as well

i think due to the fact i can't contact the dead you might be able to submit answers

I sent in my answers all in latex, but 10 of them were unreadable. :mad:


If x is NOT the square of an integer, then the value of \[7+\sqrt{x}+\frac{1}{5-\sqrt{x}}\] is a rational number for only one positive integer x. What is x? and what is the value that you get?

Moo...

[hide]$7+\sqrt{x}+\frac{1}{5-\sqrt{x}}$
We rationalize the denom. and combine to get a numerator of:
$(25-x)(7+\sqrt{x})+5+\sqrt{x}= a+b\sqrt{x}$, where we want to solve for $x$ when $b=0$ (to assure the numerator is rational).

So $-x\sqrt{x}+26\sqrt{x}= b\sqrt{x}\Rightarrow x = 26$.

Plugging into the original equation before we combined to get a numerator, the expression is $2$.[/hide]

[hide]
Ok, so we can throw away the $7$ for now, since that doesn't affect the rationalness of the number. Then, the problem is equivalent to finding the rationalness of:
\[\sqrt{x}+\frac{5+\sqrt{x}}{25-x}=\frac{ (26-x)\sqrt{x}+5}{25-x}\]
and it follows that $x=26$, so the original expression is equal to $\boxed{2}$.[/hide]

(I was typing this up while the others posted, so don't complain.  :P )


Replace each of the boxes with a distinct number from $ \{1,2,3,4,5,6,7,8,9\}$ so that the sum is as close as possible to 1 without being greater than or equal to 1. What is the sum expressed as a common fraction?
\[ \dfrac{\Box}{\Box}\plus{}\dfrac{\Box}{\Box}\]

This sum is $ \frac ab\plus{}\frac cd\equal{}\frac{ad\plus{}bc}{bd}$. The closest we can get to one without being greater or equal to $ 1$ occurs in $ \frac{bd\minus{}1}{bd}$, where $ bd$ is as high as possible. It is obvious that $ 8\cdot9\equal{}72$ is the highest possible $ bd$ in this problem, and $ ad\plus{}bc\equal{}71$ is achievable when $ (a,b,c,d)\equal{}(7,8,1,9)$. Thus, our final answer is $ \boxed{\frac{71}{72}}$.


nataiejo kesi nemidoone???
man faghat midoonam IRAN 15 om shode
inaram motmaenam ke ,1 tala darim ,3 ta noghre ,1 bronze , 1 nafaram nemidonam chi shode :P

see here: [url]http://www.imo-official.org/team_r.aspx?code=IRN&year=2009[/url]
& [url]http://www.imo-official.org/year_country_r.aspx?year=2009[/url]


algebra
function
domain
abstract algebra
vector
polynomial
superior algebra

Can you find an example  of a simple domain not being skew field?

What is a simple domain?  Do you mean a simple module or a ring?

My guess is that he's looking for a noncommutative ring (with identity?) with no nontrivial two-sided ideals, no zero divisors, and some noninvertible elements.

This is pretty tricky because the simplicity condition cannot apply to left and right ideals, as that would imply that each element is invertible.

EDIT: My last comment was wrong, changed it to something more reasonable.

actually I can not find the definition of a simple domain.

[quote="LydianRain"]My guess is that he's looking for a noncommutative ring (with identity?) with no nontrivial two-sided ideals, no zero divisors, and some noninvertible elements.[/quote]

if this is the definition we are working with. then any noninvertible element $ a\in R$ will generate an ideal $ RaR\equal{} \left\{ \sum_{i\equal{}0}^n r_i a s_i \;\bigg|\; r_i,s_i \in R\right\}$... right?

Yes, so we must have $ RaR\equal{}R$.  Does this tell us anything?

[quote]My guess is that he's looking for a noncommutative ring (with identity?) with no nontrivial two-sided ideals, no zero divisors, and some noninvertible elements.[/quote]
It must be an infinite dimensional central simple algebra (the center of our ring has to be a field due to the simplicity). But I don't know such an example right now...
Moreover for any $ D$ division ring cointained in our ring (say) $ R$ the dimension(as a left/right vector space) of $ R$ over $ D$ must be infinite - otherwise we get that $ R$ is a division ring.

I think I saw this example in a book by Lam:

Let $ k$ be any field, and $ \sigma$ any automorphism of $ k$ of infinite order. Now consider the so-called [i]skew Laurent polynomials[/i] over $ k$: this ring consists of Laurent polynomials of the form $ \sum_{i\equal{}m}^n a_ix^i$ with $ a_i\in k$ and $ m\le n\in\mathbb Z$, with $ k$ acting on $ x$ by multiplication to the right via the rule $ xa\equal{}\sigma(a)x,\ \forall a\in k$.

It should be quite clear why this is a domain (but not a division ring), but as I recall, the proof for simplicity was a bit involved.

Nice, I was thinking at something similar but I didn't know how to prove the simplicity. Thanks.
For the proof: T. Lam - A first course in Noncommutative rings, page 46, theorem 3.18 (and Corollary 3.19).

BTW: Skew Laurent rings (and some generalisations) arise in noncommutative Iwasawa theory as noncommutative Iwasawa algebras


function
number theory
IMO
algebra
functional equation
IMO 1990

It suffices to construct such a function satisfying $f(ab)=f(a)f(b),\ \forall a,b\in\mathbb Q^+\ (*)$ (this implies $f(1)=1$) and $f(f(x))=\frac 1x,\ \forall x\in\mathbb Q^+\ (**)$.

All we need to do is define $f(p_i)$ s.t. $(*)$ whenever $x=p_i$ for some $i\ge 1$, where $(p_n)_{n\ge 1}$ is the sequence of primes, and then extend it to the rest of $\mathbb Q^+$ so that $(**)$ holds. Then it's clear that $(*)$ will automatically hold.

as Grobber said(he didn't prove it) we have $ f(ab) \equal{} f(a)f(b)$ , this  implies $ f(\prod_{i \equal{} 1}^np_i^{\alpha_i}) \equal{} \prod_{i \equal{} 1}^nf(p_i)^{\alpha_i}$ , hence we must deifne the function on all primes, let $ p_i$ denote the $ i \minus{} th$ prime number we define $ f$ as:
$ f(p_{2k \minus{} 1}) \equal{} p_{2k}\ ,\ f(p_{2k}) \equal{} \frac {1}{p_{2k \minus{} 1}}$
this function satisfies the problem , clearly.

$ f(f(y)) \equal{} f(1)/y$. This implies that $ f$ is injective. $ f(f(1)) \equal{} f(1) \longrightarrow f(1) \equal{} 1$

Therefore $ f(f(y)) \equal{} 1/y$. Let $ y\equal{}f(y)$, so $ f(x/y) \equal{} f(x)/f(y)$. Then $ f(1/f(y)) \equal{} f(1)/f(f(y)) \equal{} y$

From the original equation, letting $ y\equal{}1/f(y)$ implies $ f(xy) \equal{} f(x)f(y)$. A function on primes like Amir's works.

[quote="triplebig"]I don't understand how you can conclude that $ f$ is injective, can anyone please share some light?[/quote]
$ f(y_1) \equal{} f(y_2)$ $ \implies$ $ f(xf(y_1)) \equal{} f(xf(y_2))$ $ \implies$ $ \frac {f(x)}{y_1} \equal{} \frac {f(x)}{y_2}$ $ \implies$ $ y_1 \equal{} y_2$ (since $ f(x)\neq 0$)

[quote="Dijkschneier"][quote="Rofler"]So how do you extend to Q?[/quote]
Can sameone answer to this, please ?[/quote]

There is no need for extension : the problem is just for Q+ and we know that any positive rational may be written in a unique manner as the product of prime numbers raised to integer powers.

What do you want more ?

Sorry to revive this topic, but can someone please explain why this doesn't work? :

Note that from the above we've already established that $f(xy)=f(x)f(y)$ and $f$ is injective.
Also, from $ f(f(y)) = \frac{1}{y} $, we know that $ f $ is surjective, therefore $ f^{-1}(x) $ exists for all positive rationals $ x $.

So set $ f^{-1}(y)$ as $ y $ in $ f(f(y)) = \frac{1}{y} \Rightarrow f(y)f^{-1}(y) = 1 $

Thus set $ f^{-1}(x) $ as $ x $ and $ y $ as $ y $ into the original equation and we obtain:
$ f(f^{-1}(x)f(y)) = \frac{x}{y} $
$ \Rightarrow f^{-1}(x)f(y) = \frac{x}{y} $
$ \Rightarrow \frac{f(y)}{f(x)} = \frac{x}{y} $
$ \Rightarrow f(x) = \frac{1}{x} $ $\forall$  $ x \in \mathbb{Q}^{+} $
which is obviously not a solution to the equation.

Can someone please explain what went wrong? Thanks.

[quote="CPT_J_H_Miller"]Thus set $ f^{-1}(x) $ as $ x $ and $ y $ as $ y $ into the original equation and we obtain:
$ f(f^{-1}(x)f(y)) = \frac{x}{y} $
$ \Rightarrow f^{-1}(x)f(y) = \frac{x}{y} $.[/quote]
The implication is abusive; from $f(A) = B$ you infer $A=B$.

Argh... silly mistake... yes it should be $ f(f^{-1}(x)f(y)) = xf(f(y)) = \frac{x}{y} $.
Thanks!

I was able to determine the conditions for the function, but not able to construct it.
Out of curiosity, how many points would I get for this (on the actual thing I would probably spend time finding it since the conditions take a very small time to find, but...)?

:mad: Plugging in x = 1 we get f(f(y)) = f(1)/y and hence f(y1) = f(y2)
implies y1 = y2 i.e. that the function is bijective. Plugging in y = 1 gives
us f(xf(1)) = f(x) ⇒ xf(1) = x ⇒ f(1) = 1. Hence f(f(y)) = 1/y.
Plugging in y = f(z) implies 1/f(z) = f(1/z). Finally setting y = f(1/t)
into the original equation gives us f(xt) = f(x)/f(1/t) = f(x)f(t).
Conversely, any functional equation on Q+ satisfying (i) f(xt) = f(x)f(t)
and (ii) f(f(x)) = 1
x for all x, t ∈ Q+ also satisfies the original functional
equation: f(xf(y)) = f(x)f(f(y)) = f(x)
y . Hence it suffices to find
a function satisfying (i) and (ii).
We note that all elements q ∈ Q+ are of the form q = $n
i=1 pai
i where
pi are prime and ai ∈ Z. The criterion (a) implies f(q) = f($n
i=1 pai
i ) = $n
i=1 f(pi)ai . Thus it is sufficient to define the function on all primes. For
the function to satisfy (b) it is necessary and sufficient for it to satisfy
f(f(p)) = 1
p for all primes p. Let qi denote the i-th smallest prime. We
define our function f as follows:
f(q2k−1) = q2k, f(q2k) =
1
q2k−1
, k ∈ N .
Such a function clearly satisfies (b) and along with the additional condition
f(xt) = f(x)f(t) it is well defined for all elements of Q+ and it satisfies
the original functional equation. :P  :mad:  :mad:

[quote=MathPanda1][quote=Amir.S]as Grobber said(he didn't prove it) we have $ f(ab) = f(a)f(b)$ , this implies $ f(\prod_{i = 1}^np_i^{\alpha_i}) = \prod_{i = 1}^nf(p_i)^{\alpha_i}$ , hence we must deifne the function on all primes, let $ p_i$ denote the $ i - th$ prime number we define $ f$ as:
$ f(p_{2k - 1}) = p_{2k}\ ,\ f(p_{2k}) = \frac {1}{p_{2k - 1}}$
 this function satisfies the problem , clearly.[/quote]

What is the motivation for constructing such a function? Thank you very much![/quote]

First you try out some algebraic methods: none of them are fruitful. Then you note that the problem said construction, which implies a numbertheoretic approach. This immediately applies looking for multiplicity and a way to define f(1), which just happen to be related to each other. (Shows that you are on the right track). Then you obtain the relation:

$f(f(p)) = \frac{1}{p}$ for all primes.

So it is clear that you cannot manipulate the power of prime p to get from an exponent of 1 to -1 in two steps over $\mathbb{Q^{+}}$. So you have to manipulate the primes in some other method, with a group of elements acting as a medium. (In other words, f(f(p)) maps A $\rightarrow$ B $\rightarrow$ C, where you know that {p} = A, B is unknown, and {$\frac{1}{p}$} = C.

This suggests bipartitioning the set of primes, which suggests considering the sets {$p_{2k}$}, {$p_{2k-1}$}, {$\frac{1}{p_{2k}}$} and {$\frac{1}{p_{2k-1}}$}. Playing around with directed arrows that map elements between the sets gives you the function.

Cute problem, nice fit for a $P1$ :)

First we get the obvious substitutions over with- $P(1,x): f(f(x))=\frac{f(1)}{x}$. It's pretty trivial to see that this function is necessarily bijective, as $f(y)=f(t)=> f(xf(y))=f(xf(t))=>y=t$ and surjectivity is direct from $f(f(x))=\frac{f(1)}{x}$. Again, $P(1,x): f(xf(1))=f(x)=>f(1)=1=>f(f(x))=\frac{1}x$ by injectivity.

Here you could already try to hunt for a solution, but unless your intuition is [i]really[/i] strong, it'd be quite hard to already hit upon a direct solution- also do try and remember it's an IMO problem, you can't expect to be already done with it in 2 minutes (let's just ignore this year's $P1$ [hide=Note] Sorry I just had to get that meme in  :D [/hide]).

Playing around with the problem, let's put an extra $f$ around the problem to get $f(f(xf(y)))=f(\frac{f(x)}{y}) =>\frac{1}{xf(y)}=f(\frac{f(x)}{y})$. Replace $x$ with $f(x)$ to get $\frac{1}{f(x)f(y)}=f(\frac{f(f(x))}{y})=f(\frac{1}{xy})$.

Naturally, we'd now want to put in $y$ as $\frac{1}{x}$ to get $\frac{1}{f(x) f(\frac{1}x)}=1=> f(\frac{1}{x})=\frac{1}{f(x)}$.
Re-substitute to get $f(\frac{1}{x})f(\frac{1}{y})=f(\frac{1}{xy}) =>f(x)f(y)=f(xy)$.

Ok this seems like a pretty concrete result- perhaps now we'll be equipped enough to start our construction. Straight off the bat, we see that this must have something to so with the prime factorizations- firstly the function is multiplicative, secondly there doesn't seem to be any purely algebraic
solution, and, perhaps most importantly, $Q^+$ is our domain aka integer prime factorizations- this problem is calling for a number-theoretic way to attack the problem.

Hmm... so let's just put in $x=p_1^{\alpha_1}p_2^{\alpha_2}...p_l^{\alpha_l}, y=q_1^{\beta_1}q_2^{\beta_2}...q_k^{\beta_k}$ in our original problem- simplifying a little and spamming the multiplicative property, we get $f(f(q_1)^{\beta_1} \cdot f(q_2)^{\beta_2} \cdot ... \cdot f(q_k)^{\beta_k})=\frac{1}{q_1^{\beta_1}q_2^{\beta_2}...q_k^{\beta_k}}$. Hmm... so how do we get the primes to randomly appear in the denominator- oh wait this is trivial, just make a$\pmod{2}$ cycle of sorts- let the $j_{th}$ prime, $p_j$, map to $p_{j+1}$ if, say, the index is odd, and let it map to $\frac{1}{p_{j-1}}$ if even. Now all we have to do is put it back into the equation and check, and this construction indeed seems to work.

Hence we're done.

[hide=Note] I know my solutions are very free-flowing, but I feel that it's usually easier to follow along, kinda like a narrative, when a solution is written in this style, even though it compromises a little on the rigor[/hide]

Our function will satisfy $f(f(x))=\frac{1}{x}$ and $f(ab)=f(a)f(b)\forall a,b\in \mathbb{Q^{+}}$.

Consider the function that satisfies $f(ab)=f(a)f(b)$ and $f(p_{2k-1})=p_{2k}$, $f(p_{2k})=\frac{1}{p_{2k-1}}$, and $f\left(\frac{1}{p_{2k-1}}\right)=\frac{1}{p_{2k}}$,  where $p_n$ is the $n$th prime for all positive integers $k$. It's easy to see we can extend this to all positive rationals.

Now we will show that this satisfies $f(f(x))=\frac{1}{x}$. Call a number 1-over-prime if it can be written as $\frac{1}{p}$ for some prime $p$. Clearly $f(f(m))=\frac{1}{m}$ for all primes and 1-over-primes $m$. Now we have $f(f(x))$ can be written as $f(f(m_1))\cdot f(f(m_2))\cdots f(f(m_n))=\frac{1}{x}$ where the $m_i$ are some primes or 1-over-primes which multiply up to $x$. $\blacksquare$

Now we have $f(xf(y))=f(x)f(f(y))=\frac{f(x)}{y}$, as desired.

[quote=orl]Let $ {\mathbb Q}^ \plus{}$ be the set of positive rational numbers. Construct a function $ f : {\mathbb Q}^ \plus{} \rightarrow {\mathbb Q}^ \plus{}$ such that
\[ f(xf(y)) \equal{} \frac {f(x)}{y}
\]
for all $ x$, $ y$ in $ {\mathbb Q}^ \plus{}$.[/quote]

Let $P(x,y)$ be the assertion.
If $f(y_1)=f(y_2) \implies y_1=y_2$.
$P(x,1) \implies f(1)=1$.
$P(1,y) \implies f(f(y))= \frac{1}{y}.$ Inputting this to get,
$f(\frac{1}{y}=\frac{1}{f(y)}$.
$P(x,f(\frac{1}{t})) \implies f(xt)=f(x)f(t) \forall t \in \mathbb{Q^+}.$
Now the functions:
(a) $f(xt)=f(x)f(t)$; (b) $f(f(x))=\frac{1}{x}$ solve the equation.
 A function $f$ mapping through primes, satisfying (a) as,
$f(p^{n_1}_1p^{n_2}_1 \dots p^{n_k}_k) = [f(p_1)]^{n_1} [f(p_2]^{n_2} \dots [f(p_k)]^{n_k},$ where $p_j$ is the jth prime and $n_j \in \mathbb{Z}$. Such a function wil satisfy (b) for each prime. A possible solution is: $$\begin{cases} P_{j+1} & \text{ if j is odd, } \\ \frac{1}{P_{j-1}}  & \text{ if j is even.} \end{cases}$$
Clearly, $f(f(p))=\frac{1}{p}$. Thus $f$ satisfies.

wow quite hard for an imo/1 especially one that was 33 years ago

Note that any muliplicative function that satisfies $f(xy)=f(x)f(y)$ and $f(f(y))=1/y$ works, as $$f(xf(y))=f(x)f(f(y))=f(x)/y.$$

Let $p_i$ be the $i$th prime.\\

We claim that the following function works:\\

(i) If $i$ is odd, $f(p_i)=p_{i+1}.$\\

(ii) If $i$ is even, $f(p_i)=\frac{1}{p_{i-1}}.$\\

(iii) Otherwise, if $m$ and $n$ are relatively prime positive integers, $$f(m/n)=\frac{\prod_{p}f(p)^{v_p(m)}}{\prod_{p}f(p)^{v_p(n)}}$$

Clearly, this is multiplicative from (iii). Additionally, $$f(f(m/n))=f(\frac{\prod_{p}f(p)^{v_p(m)}}{\prod_{p}f(p)^{v_p(n)}})$$ $$=\frac{f(\prod_{p}f(p)^{v_p(m)})}{f(\prod_{p}f(p)^{v_p(n)})}=\frac{\prod_p f(f(p))^{v_p(m)}}{\prod_p f(f(p))^{v_p(n)}}=\frac{1/m}{1/n}=n/m.$$


AMC
AIME
induction
algebra
system of equations

Find a, b so that x^2-x-1 is a factor of ax^17+bx^16+1

[hide]

If that divides through, then  for some . If  are the roots of , then .

Now, notice that , and multiplying through by , we have  Repeating this, we get  Fibonacci, what a surprise. Do induction or whatever you will and you find , where  is the nth Fibonacci number.

Then we simply substitute this into  and get . Both the constant and the  term must equal zero, so now we have an easy system of equations in  and , and solving it yields  and .

[/hide]

By the way, the solution as Elemennop saw it suggests the possibility of an interesting connection between the Fibonacci numbers and Pascal's triangle.  Can you find it?


algebra
polynomial
inequalities
algebra proposed

Let $P(x)$ be a real polynomial such that the following inequalities are satisfied:
\[P(0) > 0, P(1) > P(0), P(2) > 2 \cdot P(1) - P(0), P(3) > 3 \cdot P(2) - 3 \cdot P(1) + P(0),\]
and
\[P(n+4) > 4 \cdot P(n+3) - 6 \cdot P(n+2) + 4 \cdot P(n+1) - P(n) \quad \forall n \in  \mathbb{N}.\]
Prove that for every natural number $n$ $P(n)$ is positive.

So strange...

We have $\Delta^4P(n)>0$, i.e. $\Delta^3P(n+1)-\Delta^3P(n)>0$ for all $n$, but $\Delta^3P(0)>0$. Therefore $\Delta^3P(n)>0$ for all $n$. Do the same steps two more times.

[quote="Myth"]So strange...
[/quote]

Maybe the same author as for IMO LL 89, problem 11, Cuba...


A soccer jersey must be numbered with a one- or two-digit integer.  Ellen has 8 each of the digits 1-8.  What is the maximum number of jerseys she can number if no jersey number is repeated?

[hide]She could do 8 for a single digit, and for every 2 digit number starting with some given digit, she can do 8 combinations. So for 8 numbers * 8 combinations=64 plus 8 1 digit numbers yields 72[/hide]

[hide]She can do 8 one digit numbers.
She has 7 sets of 1-8.
She can do:
12
13
14
15
16
17
18, etc for each set of numbers. The second one would be:
21
23
24
25
26
27
28.
She has 7*7+8=49+8=57[/hide]

[hide]
I get 64:
She can have 8 jerseys numbered 1 - 8
and then
8 * 7 = 56 jerseys which are two digit numbers
So 64.
[/hide]

she can number 8 single-digit jerseys.
then for a double digit, she has 8 choices for the first digit, and 7 choices for the second digit (since no repeating allowed). Therefore u have 8+8*7=8+56=64 number of possible numberings.

[hide="solution"]it can be _ _

there are 9 possibilities for the tens digit

there are 8 possibilities for the unit digit

9*8=[b]72[/b][/hide]

She only has 64 digits total to work with.  Can she possibly make more than 64 jerseys?  Can she possibly make 64 jerseys (or anywhere close to 64 jerseys) if she can't repeat the uniforms 1, 2, 3, 4, 5, 6, 7, or 8?

Always sanity-check your answer against the question when you are finished!

[quote="rrusczyk"]Always sanity-check your answer against the question when you are finished![/quote]
We have some pretty good problem-solvers, with answers that don't meet Richard's sanity-check.  I am quite curious if each of these problem-solvers made the same original mistake.  Or did the later solvers read previous solutions, and get led astray by those who came before?

[hide="One way to check the sanity of answer"]For those who are making , say 40,  (or more )  2 digit Jersey's ..(which may rquire 80 or more numbers..) There arent' that many numbers to sew..) .. I know you have already figured that out  but will mention it just in case. ..[/hide]

Oh!
[hide]
She has 8 sets of the numbers to sew on, so she has
1, 2, 3, 4, 5, 6, 7, 8
1, 2, 3, 4, 5, 6, 7, 8
1, 2, 3, 4, 5, 6, 7, 8
1, 2, 3, 4, 5, 6, 7, 8
1, 2, 3, 4, 5, 6, 7, 8
1, 2, 3, 4, 5, 6, 7, 8
1, 2, 3, 4, 5, 6, 7, 8
1, 2, 3, 4, 5, 6, 7, 8
I didn't understand the wording the first time.  I figured she was only allowed to use the digits 1-8 and could use them as much as she wanted.

For one set:
1, 2, 3, 4, 5, 6, 7, 8 = 8 jerseys
She still has 7 sets left and can make a maximum of 4 two-digit numbers from each set
ie 12, 34, 56, 78
So 8 + 7(4) = 36.[/hide]

[hide]8 one-digit numbers.
64 - 8 = 56 numbers left.. these must be used for two-digit numbers.
These two-digit numbers don't have to repeat because each number can be paired with 7 different numbers...
8 + 56/2 = 36?[/hide]

I also didnt pay attention to the fact that she had 8 of each digit.


function
limit
calculus
calculus computations

is it required that if the functions $ u(x)$ $ v(x)$ and $ w(x)$  are such that
$ u \leq v \leq w$ then $ \lim u \leq \lim v \leq \lim w$ holds only when $ \lim v$ exists even though it is known $ \lim u$ and $ \lim w$ exists

That wasn't quite intelligible.  If $ u \leq v \leq w$ and $ \lim u = \lim w$ then both are equal to $ \lim v$, and in particular $ \lim v$ exists.  If $ \lim u < \lim w$, then if $ \lim v$ exists we have $ \lim u \leq \lim v \leq \lim w$, but no guarantee that the limit exists.

yes that was exactly what i wanted.thanks :)


integration
logarithms
calculus
calculus computations

Define $ t^{2}\equal{}x\plus{}1$

so $ \int\frac{x\minus{}2}{x\sqrt{x\plus{}1}}\ dx$

      =$ 2\int \frac{t^{2}\minus{}3}{t^{2}\minus{}1}dt$
      =$ 2\plus{}2ln\frac{t\plus{}1}{t\minus{}1}$
      =$ 2\plus{}2ln\frac{\sqrt{x\plus{}1}\plus{}1}{\sqrt{x\plus{}1}\minus{}1}$

I have

$ I\equal{}\int \frac{x\minus{}2}{x \sqrt{x\plus{}1}}\ dx$.

$ u\equal{} \sqrt {x\plus{}1}$

$ du \equal{} \frac{1}{2} \frac{dx}{\sqrt{x\plus{}1}}$

$ I\equal{} 2\int\frac{u^{2}\minus{}3}{u^{2}\minus{}1}\ du$

$ \equal{}2\int(1\minus{}\frac{2}{u^{2}\minus{}1})\ du$

$ \equal{} 2(u \minus{} \ln \frac{u\minus{}1}{u\plus{}1})$

$ \equal{}2(\sqrt{x\plus{}1} \minus{} \ln \frac{\sqrt{x\plus{}1} \minus{}1}{\sqrt{x\plus{}1} \plus{}1}) \plus{}c$.


calculus
integration
function
calculus computations

Recursive formula:

$ \frac {2ax \plus{} b}{(n \minus{} 1)(4ac \minus{} b^2)(ax^2 \plus{} bx \plus{} c)^{n \minus{} 1}} \plus{} \frac {(2n \minus{} 3)2a}{(n \minus{} 1)(4ac \minus{} b^2)}\int\frac {1}{(ax^2 \plus{} bx \plus{} c)^{n \minus{} 1}} dx$

[hide]This should be OK... (principle is clear, but I could have made a mistake... It's too early for me to function properly ;)[/hide]


calculus
integration
calculus computations

Evaluate $ \int_0^1 \frac{2e^{2x}\plus{}xe^x\plus{}3e^x\plus{}1}{(e^x\plus{}1)^2(e^x\plus{}x\plus{}1)^2}\ dx$.

Let $ f(x) \equal{} \left(e^{x} \plus{} 1\right)\left(e^{x} \plus{} x \plus{} 1\right) \equal{} e^{2x} \plus{} (x \plus{} 2)e^{x} \plus{} x \plus{} 1$

Hence $ f'(x) \equal{} 2e^{2x} \plus{} xe^{x} \plus{} 3e^{x} \plus{} 1$

$ \therefore I \equal{} \int_{0}^{1}\frac {f'(x)}{\left[f(x)\right]^{2}}dx \equal{} \minus{} \left[\frac {1}{f(x)}\right]_{0}^{1} \equal{} \frac {1}{4} \minus{} \frac {1}{(e \plus{} 1)(e \plus{} 2)}$ :D

Incorrect, but you have almost solved it.


vector
trigonometry
analytic geometry
graphing lines
slope

I'm sorry for asking such a basic question, but I'm stumped. Here's a summary.
There is a ramp, but it is from the hotwheels company. It's attached to the edge of the table, and it is elevated by books. Naturally, the ramp is curvy. The end of the ramp is 2.6 cm higher than the edge of the table. There are two release points on the ramp. One is 11.7 cm higher than the end of the ramp, the other 6.3 cm higher than the edge of the ramp. The height from the edge of the ramp to the floor is 89.7 cm. Now, a ball bearing is rolled down from the first drop point of the ramp (11.7 cm), and it goes 50.51 cm away from the edge of the ramp, in a curve (since it's falling). Then, the same ball is released from the second point (6.3 cm) and moves a horizontal distance of 39.73 cm. How would I go about finding the horizontal vector of velocity of the ball the instant it leaves the ramp in both cases? If that didn't make sense, I'm asking how to find the horizontal velocity of a projectile, if no measurement of time is known.

One thing confusing me is the fact that the ball accelerates. But at what rate? Its downward acceleration is 9.8 m/s because of gravity, but what about horizontal? The vectors are all of different units, and I don't know how to cope with that. We have no data regarding time whatsoever. Please help D:

EDIT: With a little thinking, here's what I have. I think time is important, so I set about trying to find that. Now, this is where i'm uncertain, to say at the least. The ball traveled 89.27 cm from the edge of the ramp to the floor (vertical motion only). However, the drop heights must also be factored in. So, in reality, the ball was dropped from heights of 100.97 cm and 95.57 cm (This is where my assumption comes in, I am uncertain. The acceleration of the ball is still going to be $ 9.8\frac{m}{s^{2}}$ even if there's a ramp under it, so those distances must be added in for the overall drop of the ball. Then, using the equation of $ s \equal{} v_{0}t\plus{}\frac{1}{2}(9.8\frac{m}{s^{2}})(t^{2})$, and substituing in what I know, I get $ t \equal{} .454$

Now, I'm assuming that after the ramp has been traversed, there is no more horizontal acceleration. So, plugging values into $ s \equal{}\frac{r}{t}$, I get a value of $ 1.113\frac{m}{s}$for the horizontal component of the ball. Is this way of solving the problem correct, or did I leave something out?

One question before I start a solution: is the end of the ramp horizontal?

Yes. Sorry, I forgot to mention. That would change things...

The first thing I would like to say is that the time cannot be found. In fact you could make the ramp a kilometer, or just a foot long if you wish and have the same vertical range. The times taken to travel down these two ramps are obviously different.

The acceleration is not constant when the ball rolls down a curvy ramp. Lets consider the forces acting on the ball at any point, $ P$. We have it's weight, $ W$, acting vertically down, but we also have a normal reaction force from the ramp itself. This normal reaction, $ N$, acts perpendicularly from the ramp, hence why we use the term "normal". For now, let's assume the ramp at $ P$ makes an angle $ \theta$ from the horizontal. The ball is always moving along, or parallel to the ramp, and it does not have a component perpendicular to the ramp. Thus no component of the normal reaction acts along the direction of motion, and the only force acting in this direction is the component of weight parallel to the ramp. This component is $ F \equal{} mg\sin{\theta}$. So the acceleration is $ a \equal{} g\sin{\theta}$. Which of course is not constant, since the slope of the curve (and hence the angle $ \theta$) is not constant. If you are interested in finding the acceleration at any point then you could find an equation for the height of the curve at any horizontal distance along the ramp. Differentiating this equation gives the gradient $ m$ of the tangent to the ramp at any point, and the value of $ \theta$ can be found by the equation $ \theta \equal{}\tan^{\minus{}1}{m}$. this value can be plugged into the formula.

Now lets look at how we really tackle this problem, which is a lot less complicated than you are making it. Energy is conserved (we assume the ball is initially at rest). The potential energy lost in decending through a height $ h$ is $ E_{P}\equal{} mgh$. This energy is converted to kinetic energy in the ball, $ E_{K}\equal{}\frac{1}{2}mv^{2}$.

$ \Rightarrow\frac{1}{2}mv^{2}\equal{} mgh$

$ \Rightarrow v^{2}\equal{} 2gh$

$ \Rightarrow v \equal{}\sqrt{2gh}$

This equation gives the velocity at any height decended. It does not depend on the path taken (as long as friction is negligable). Because the end of the ramp is horizontal, we know that the horizontal component of velocity at the end of the ramp must be $ v \equal{}\sqrt{2gh}$, since there is no perpendicular component.

For the $ 11.7\,cm$ release point the final velocity is $ v \equal{}\sqrt{2\times 9.8\times 0.117}\equal{} 1.5\, ms^{\minus{}1}$.

For the $ 6.3\,cm$ release point the final velocity is $ v \equal{}\sqrt{2\times 9.8\times 0.063}\equal{} 1.1\, ms^{\minus{}1}$.

So the whole part about the horizontal range is reduntant. Alternately you could completely ignore the whole part about the ramp as follows:

Consider the vertical motion of the ball after it has just left the ramp. Let's say it is initally a height $ s_{y}$ above the ground. The inital velocity in the vertical direction is $ u_{y}\equal{} 0$ after leaving the ramp as it leaves the ramp horizontally. The acceleration is $ g$.

$ s \equal{} ut\plus{}\frac{1}{2}at^{2}$

$ \Rightarrow s_{y}\equal{}\frac{1}{2}g t^{2}$

$ \Rightarrow t^{2}\equal{}\frac{2s_{y}}{g}$

$ \Rightarrow t \equal{}\sqrt{\frac{2s_{y}}{g}}$

We have now found an expression for the time taken for the ball to hit the ground after leaving the ramp.

Lets say the ball travels a horizontal distance of $ s_{x}$ and it's (horizontal) velocity as it leaves the ramp is $ u_{x}$. This velocity does not change as no forces act horizontally after it leaves the ramp.

$ \Rightarrow u_{x}\equal{}\frac{s_{x}}{t}$ where $ t$ is the time taken for the flight. But this time was found above!

$ \Rightarrow u_{x}\equal{}\frac{s_{x}}{\sqrt{\frac{2s_{y}}{g}}}\equal{} s_{x}\sqrt{\frac{g}{2s_{y}}}$

So this is another equation for the velocity, without even having to bother considering the ramp. (you have actually given way too much information!).

In both cases, $ g \equal{} 9.8\,ms^{\minus{}2}$ and $ s_{y}\equal{} 0.897\,m$

For the first ball: $ u_{x}\equal{} 0.5051\sqrt{\frac{9.8}{2\times 0.897}}\equal{} 1.2 ms^{\minus{}1}$

For the second ball: $ u_{x}\equal{} 0.3973\sqrt{\frac{9.8}{2\times 0.897}}\equal{} 0.93 ms^{\minus{}1}$

These results should in theory be the same as those found above, but you may notice that they are less. This is probably because the ball bearing gains rotational kinetic energy when it rolls down the slope, and also energy is lost through overcoming frictional forces. Both these effects mean that not all of the potential energy lost will be converted to linear (or translational) kinetic energy. So the speed when it reaches the end of the ramp will be an overestimate (as above). By the way, the height of the ramp and the horizontal range are in direct correspondance. As soon as you set a height of release, the horizontal range is fixed and will always be the same (as long as you keep all other variables constant). This means that you can predict the horizontal distance the ball travels from how high the point of release is above the end of the ramp. It also works the opposite way too. If you know the horizontal distance covered, then you can automatically find the the point of release above the end of the ramp.

So, we haven't learned anything about force. However, I do understand the first half of your explanations, I just haven't learned the formulas yet. The second part is what I was trying to do, except that I messed it up, like usual. Anyway, thanks a lot for your help, I understand it perfectly now, including why I was wrong, which is as useful as knowing the right answer.


topology
advanced fields
advanced fields unsolved

Prove that in a complete metric space X, every nonempty open subset is of Category II as a subset of itself.
(where being Category I means that A is a countable union of nowhere dense sets, and Category II means everything not Category I)

i managed to prove that A is of category II in X, (if A is of Category I in X, X is complete, according to baire's theorem we get Int A = {}, but A is open in X, so IntA=A != {}.)

but i cant prove that A is of category II in A (as a subset of istelf)

any help is welcome!

thanks

Assume $A$ is a complete metric space and $O$ is open in $A$. Now assume $O$ is of category $I$. We can write $O=\cup_{i=1}^{\infty}{E_{n}}$ where the $E_{n}$ are nowhere dense. This also means $O= \cup_{i=1}^{\infty}{F_{n}}$ where $F_{n}$ are the closure in $O$ of the $E_{n}$. Each $F_{n}$ has the form $O \cap A_{n}$ where $A_{n}$ is closed in $A$, and we can assume $\forall n, A_{n} \subseteq Cl(O)$.
Now if $x \in O \cap A_{n}$, then $x \notin Int(A_{n})$ since the $E_{n}$ are nowhere dense. If $x \in Cl(O)-O$, then we can assume every open set containing $x$ intersects $A-Cl(O)$, whence $x \notin Int(A_{n})$ (otherwise $Cl(O)=O$, meaning $O$ is closed, which in particular says $O$ is complete and of category $II$ by Baire's theorem), and $Int(A_{n})=\emptyset$. Now $(\cup_{i=1}^{\infty}A_{n}) \cup O^{c}=A$. By Baire's theorem, $A=Int(O^{c})$, which is absurd since $O$ is nonempty.


Determine the smallest positive integer $ n$ such that $ 5n\plus{}1$ and $ 7n\plus{}1$ are perfect squares.

One (not particularly enlightening) attack:
[hide]Any perfect square is congruent to $ 0, 1$ or $ 4$ modulo 8.  If $ 5n \plus{} 1 \equiv 0, 1, 4 \pmod 8$ then $ n \equiv 3, 0, 7 \pmod 8$, while if $ 7n \plus{} 1 \equiv 0, 1, 4 \pmod 8$ then $ n \equiv 1, 0, 5 \pmod 8$.  It follows that both can be squares only if $ n \equiv 0 \pmod 8$.  Also, any perfect square is equivalent to $ 0$ or $ 1$ modulo $ 3$.  If $ 5n \plus{} 1 \equiv 0, 1\pmod 3$ then $ n \equiv 1, 0\pmod 3$, while if $ 7n \plus{} 1 \equiv 0, 1\pmod 3$ then $ n \equiv 2,0 \pmod 3$.  It follows that both can be squares only if $ n \equiv 0 \pmod 3$.  Thus, we're looking for multiples of 24, and 24 happens to work.[/hide]

Numbers of this type are (unsurprisingly) rather rare: the second is $ 3432$, while the third is $ 487344$, and there are no others smaller than twelve million.  So, my challenge: can anyone find all solutions?

This is a generalized pell equation...it has a procedure to solve...


algebra
function
domain
Ring Theory
superior algebra
superior algebra unsolved

A qualifier problem is giving me trouble:

Let $ R$ be a domain with quotient field $ K.$ Suppose $ P,Q$ are prime ideals of $ R$
with $ P \neq P\plus{}Q \neq Q.$ Prove

$ R_P \cap R_Q \equal{} R$ (taken in $ K$) $ \iff$ $ P,Q$ are the only maximal ideals of $ R.$

[$ R_P$ is the localization of $ R$ at the prime ideal $ P.$]

The converse is a standard problem. The forward direction, however, seems
like it should fall out easily, but I cannot make much headway. One idea
I have wanted to push is that $ PR_P,QR_Q$ are the unique maximal ideals of $ R_P$
and $ R_Q$ (resp.) and then apply some sort of correspondence argument.

Any help would be appreciated.

This seems more elementary.
If $ M\not\equal{}P,Q$ is a maximal ideal of $ R$,
then $ M\not\subset P$ and $ M\not\subset Q$, and assume $ M\subseteq P\cup Q$.
Take $ x\in M\setminus P\subseteq Q$ and $ y\in M\setminus Q\subseteq P$.
This gives the contradiction $ x\plus{}y\in M\setminus (P\cup Q)$.
So $ M\not\subset P\cup Q$ and we find some $ r\in M\setminus (P\cup Q)$.
Then $ r\not\in R^\times$ and we get $ r^{\minus{}1}\in (R_P\cap R_Q)\setminus R$.

Conversely take $ {a\over b}\equal{}{c\over d}\in (R_P\cap R_Q)\setminus R$ with $ a,b,c,d\in R,b\not\in P,d\not\in Q$.
Assume $ R\equal{}Rb\plus{}Rd$. Then $ 1\equal{}ub\plus{}vd$ for some $ u,v\in R$
and $ a\equal{}a(ub\plus{}vd)\equal{}uab\plus{}vad\equal{}uab\plus{}vbc\equal{}b(ua\plus{}vc)$,
which gives the contradiction $ {a\over b}\equal{}ua\plus{}vc\in R$.
So $ Rb\plus{}Rd\subsetneq R$ and we find some maximal ideal $ M\supseteq Rb\plus{}Rd$ of $ R$.
Then $ b\not\in P,d\not\in Q$ gives $ M\not\equal{}P,Q$.


[hide]this is a simple 3-4-5 triangle multplied by 4

4*4=16[/hide]

The Pythagorean Theorem states: $ a^2 \plus{} b^2 \equal{} c^2$. This means that $ a$ and $ b$, the legs(or the shorter sides) squared is the same as when you take $ c$, the hypotenuse(or the longest side, opposite from the right angle) and square it.

In the picture, we see that the hypotenuse has a length of $ 20$, and one of the legs has a length of $ 12$. We can substitute these numbers into our theorem.

$ a^2 \plus{} b^2 \equal{} c^2$

$ 12^2 \plus{} b^2 \equal{} 20^2$

$ 144 \plus{} b^2 \equal{} 400$

$ 144 \plus{} b^2  \minus{} 144 \equal{} 400 \minus{} 144$

$ b^2 \equal{} 256$

$ \sqrt{b} \equal{} \sqrt{256}$

$ b \equal{} 16$

So our answer is $ \boxed{16}$


What is the remainder when the sum of all the numbers created with 4,5, and 6 (i.e. 4, 5, 6, 45, 46, 54, 654, ...)  is divided by 3?

EDIT: Solve this without a calculator and without writing all possible numbers down or even adding all numbers mentally. :P

Masoud Zargar

is this "EMT" thing going to be a contest sort of thing or is it just a huge line of problems?

if you want to build it into a contest, i can sticky it

[hide="answer"]0

um....

if I'm getting it right, 4 + 5 + 6 =15 and when it's divided by 3, the remainder is  always 0

 [/hide]

Oh, sure.  I'll be very thankful if you sticky this.  Well, I think this can be a contest.  I'll take track of the points and place them up after every round, which is every 10 problems.  The points will be given to the first person who gives a complete solution. Thanks in advance. :)

Masoud Zargar

Since hopefully, this will turn out to be a contest, this will be the last problem without points.  As a result, I will give the answer.

Well, according to my calculations, the sum of all the numbers that can be created with 4,5, and 6 would be $(4+5+6)\cdot 245=3675\rightarrow (3+6+7+5)/3=7$. And therefore, the remainder is 0.

Btw, I don't know if this has already been discovered, but in order to figure out what the sum of all the numbers that can be created with 2 numbers, you can add the numbers, and then multipy them by 12.  For example, with 1 and 2: $(1+2+12+21)=36\rightarrow (1+2)\cdot 12=36$.  This also works with 3 numbers, however, this time, you add the 3 numbers and multiply them by 245.  I can figure out this for any number of numbers we have to deal with. :D

Your method seems to give the correct number of remainder too.  Nice job! However, if you want to work with their sum, you can try my method.

Masoud Zargar

[quote="chess64"]Yeah, maybe about 10 problems in one topic, then we could sticky it. :)[/quote]

Okay thanks.  So do I start a new topic called "Eager Minds of Tomorrow Round #1?"

thats  fine...btw do you know how to prove the 3 digit thing or is it just a  pattern?

It's just I pattern I have noticed.  I just take n numbers, find the sum of all possible numbers that can be made with those numbers, and then I divide it by the sum of those n numbers.  hence, I get a constant that works for the sum of all possible numbers that can be made with n numbers.

Masoud Zargar

Yes, that because they can always be expressed (through the sum of the digit-counting technique) as the product of the sum of the numbers.  4+5+6 is divisible by 3, hence the sum of all numbers that can be made by those digits is divisible by 3.  We can easily verify this by the symmetry of the numbers.


what is the apotham for a regular hexagon with an area of 50.3???

Divide the number by 6 and set it equal to the formula for the area of an equilateral trianglegiven the height.  A= (h2 3^1/2)/3  then solve for "H"


geometry
3D geometry
pyramid
sphere
function
analytic geometry
geometry unsolved

Find the locus of points laying inside a Cube, such, that in each of 6 pyramids, which verteces are considered points and bases are cube sides, is possible to inscribe sphere.

sorry for my bad English...

Russian variant: ''Найдите геометрическое место точек, лежащих внутри куба, таких, что в каждую из 6 пирамид, вершиной которых является рассматриваемая точка, а основанием- грань куба, можно вписать сферу.''

If ABCDE is a pyramid with square base ABCD and top vertex E, then it is possible to inscribe a sphere in it if and only if the sum of distances from E to the lines AB and CD is the same as the sum of distances from E to AD and BC. If ABCD is the unit square in the $ xy$ plane, we get the equation
\[ \sqrt{x^2\plus{}z^2}\plus{}\sqrt{(1\minus{}x)^2\plus{}z^2}\equal{}\sqrt{y^2\plus{}z^2}\plus{}\sqrt{(1\minus{}y)^2\plus{}z^2},\]
which has only 2 solutions: $ x\equal{}y$ and $ x\equal{}1\minus{}y$ (the function $ t\mapsto \sqrt{t^2\plus{}z^2}$ is strictly convex when $ z>0$). Looking at the pair $ xz$ of coordinates in the same way, we conclude that the answer is the union of 3 main diagonals of the cube.

Hope I didn't make any stupid mistake :)


Show that the unit disk $ x^{2}+y^{2}\leq 1$ cannot be partitioned into two congruent subsets.

Let $ \alpha$ and $ \beta$ be two sets forming such a partition and let $ f: \alpha \to \beta$ be an isometry. Let $ O$ be the centre of the disk. Consider the diameter $ MN$ of the disk which is perpendicular to the line $ Of(O)$. Then $ \left|Mf(O)\right| > 1$ and $ \left|Nf(O)\right|> 1$, so that $ M,N \in \alpha$. Now $ \left|f(M)f(N)\right| = \left|MN\right| = 2$ so $ f(M)f(N)$ is a diameter of the disk. Since $ \left|f(O)f(M)\right| = \left|f(O)f(N)\right| = 1$, $ f(O)$ must be the midpoint of $ f(M)f(N)$ and hence $ O = f(O)$. Contradiction.


function
inequalities proposed
inequalities

yes.very nice. but it is easy,isn't it?
let $f(x_1,x_2,+\cdots+x_n)=x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}{\leq}\frac{9}{8n}(x_{1}+x_{2}+\cdots+x_{n})^{2}$.it can be seen that when $f(x_1,x_2,+\cdots+x_n)$ is max.the $x_i$ must be $1$ or $2$
assume the number of $2$ is $k$
now we should only prove $4k+(n-k) \leq \frac{9n}{8}(k+n)^2$
but it's easy  :)

[quote="zhaobin"]yes.very nice. but it is easy,isn't it?
let $f(x_1,x_2,+\cdots+x_n)=x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}{\leq}\frac{9}{8n}(x_{1}+x_{2}+\cdots+x_{n})^{2}$.
[/quote]
I think you wanted to write $f(x_1,x_2,\cdots,x_n)=x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}{-}\frac{9}{8n}(x_{1}+x_{2}+\cdots+x_{n})^{2}$,and
how can you show that it admits its maximum when points are $1$ or $2$.(I know one way
using convexity but I am interested in yours).

[quote="Megus"][quote="zhaobin"]well.my solution for it is not so nice.
I consider the $x_1$ be variable first.and consider $x_2,???,x_n$ be constant.then $x_1$  must be 1 or 2. :)[/quote]

isn't it follows straight from the convexity of the function ?[/quote]
yes.it is ;)


Consider the problem:
There are two urns, both empty.There are two permissible operations: adding equal number of balls to both the urns, and halving the number of balls in either of the urns. Show that, if after an arbitrary number of operations, there are a , b, balls in the two urns respectively; a and b can take all possible values.

[hide="It seems a little easy for pre-Olympiad..."]
From an initial $(a_0, b_0) = (0, 0)$ we may apply the transformations

$(a_n, b_n) \to (a_n + k, b_n + k)$
$(a_n, b_n) \to ( \frac{a_n}{2}, b_n)$

Suppose we have a goal $(a_g, b_g)$ which we wish to reach.  WLOG $a_g \ge b_g \ge 0$.  Let us define an intermediate goal $(a_i, b_i) = \left( a_g, 2^k b_g \right)$ where $k$ is the maximum integer such that $a_g \ge 2^k b_g$.

Let $d = a_i - b_i$, which is positive.  Transform

$(0, 0) \to (2d, 2d)$ (first transform)
$(2d, 2d) \to (2d, d)$ (second)

Because $k$ is maximum, $d \le b_i$, hence $a_i - 2d = b_i - d \ge 0$.  Transform

$(2d, d) \to (a_i, b_i)$ (first transform)
$(a_i, b_i) \to (a_g, b_g)$ (second)

:)
[/hide]

You are of course right, and I agree this might be easy for pre-olympiad but then this problem was asked in the Regional Mathematical Olympiad, India in the year 1991.
Infact the original problem was the reverse of this,
you start with an arbitrary number of balls,the two operations are remove  an equal number of balls from both the urns, or double the number of balls in any urn. Show that you can make the urns empty after a finite number of moves.

So my idea originally for the solution for the problem I posed was:
Assume that you can reach any arbitrary number of balls in the urns, then using the inverse operations i.e douling instead of halving, removing instead of adding, transform this problem into an already known one, this method may be redundant in such simple problems, but  might be useful in some other problems.


Hi,

I'd like to suggest a BBcode that can embed pdf's into the forum. I'm kinda tired of having to post pdf's as a attatchment. Also, when you attach files, they always appear at the bottom, while sometimes, it is more meaningful to have them within the posts, when you want them.

Doing so should probably be pretty easy, I think.

Dojo

That's won't happen right now. In the future we will have the attachments be posted in the post at the users' discretion (including within hide tags -- which I know already is a most-requested feature.)


geometry
3D geometry
counting
distinguishability
rotation

(wow, didn't take long for the what's-your-predicted-score topic to appear.  i'm amused. :D )

i probably got 25, assuming i explained #4 well enough.

I think I got a 51455 because I was stupid on number 2, and I made a slight mistake on number 3. I might get a 2 on number 2, or a 3 on number 3...

I feel like I probably got a 25, though problem 1 may be a bit of a trouble spot.  I basically showed that there were four "classes" of cubes (one where all colors comprise 4 mutually tangent lines, one were the reds are all mutually tangent, one where the yellows are, and one where the blues are), and then I showed that a cube in each class is indistinguishable from its classmates.  I didn't explicitly say that the classes were indistinguishable from one another, but I feel like my explanation of the four classes was enough for anyone to get that they couldnot be rotated into one another.

What do you guys think?

JB

[quote="K81o7"]I think 25...however i am a little worried...

On #3, I left my answer as $6\sqrt{20}-12\sqrt{3}$ instead of simplifying to $12(\sqrt{5}-\sqrt{3})$.  Do you think they'll care?

#1, I refer to diagrams a bit, and they might find my solution a little unclear.  I hope it works out...

2, 4, 5, I think my arguments were airtight...[/quote]
They won't care about the factoring (I don't think), but they might care about the sqrt20 because that isn't simplest radical form. Maybe not though.

[quote="13375P34K43V312"]Hopefully I will get 5 5 5 5 1, but I might have made a mistake on #2 and forgotten to check $\frac{5\pi}{4}$, I honestly don't remember. I might have gotten 2 on #5, because I proved that $f(1)<1$, and I guessed $\frac{\sqrt{5}-1}{2}$, even though the answer was the conjugate of that.[/quote]

I thought it said the answer was that.  Unless I misread the sign.

the answer was $\frac{1-\sqrt{5}}{2}$ I think.

I'm pretty confident I got full credit for 1, 2, 3, and 5.

But, unfortunately, due to a rediculously stupid mistake on #4, I got floor(${\frac{n^{2}-k+1}{n-k+1}}$) as my answer.

[quote="nr1337"]This will probably be my third 24 in a row.  I'd like to say that I'm getting a 25, but with my sloppiness, that won't happen.[/quote] Me too!!!!

hopefully, I scored between a 15 and a 25.  :lol:

25 if they don't take off for $k=0$ on 4

All my answers are right, but I'm thinking that somewhere along the way I lost a point from not justifying everything quite enough.


induction
inequalities proposed
inequalities

It seems just a simple induction exercise..  :wink:

[hide]
It's equivalent to
\[\prod_{k=1}^{n}\left( 1+\frac{1}{kn}\right) \leq 2\]
But we have $kn\geq n-1+k$ which follows from $(n-1)(k-1)\geq 0$. Therefore,
\[\prod_{k=1}^{n}\left( 1+\frac{1}{kn}\right) \leq \prod_{k=1}^{n}\left( 1+\frac{1}{n-1+k}\right)\]
\[=\frac{n+1}{n}\cdot \frac{n+2}{n+1}\cdot \cdots \cdot \frac{2n}{2n-1}=2\]
[/hide]


Hi all !
    Can anybody have patient with me to explain me how I become a USAMO participant ?(here are my questions I don't understand or don't know and you can base on it to explain me :
        1) How many tests before the USAMO I have to take and what are they about ?
        2) In each state, does it have its own "basic points" (in the sense that the points which decide you pass or fail)
        3) When do often those tests begin ?
        4) Do students at different grade do different tests before the USAMO? ....)
   Thanks in advance.  :)

I'll try to explain it:

First you either take the AMC 10 or AMC 12.  Both are 75-minute multiple choice 25 question tests.  Calculators are allowed.  You need to be in 10th grade or below to take the AMC 10 or 12th grade or below to take the AMC 12.  They are both scored the same way: 6 points for a correct answer, 2.5 points for a question left blank, and 0 for an incorrect answer - don't guess!!!  You need at least 120 points on the AMC 10 or 100 points on the AMC 12 to qualify for the AIME.  Alternately, you can be in the top 1% of people who take the AMC 10 or the top 5% of the people who takt the AMC 12.  The AIME is a 15-question 3-hour test.  It is not multiple choice test.  All answer are integer answers between 0 and 999.  No calculators are allowed.  If you do well enough on that, you get to take the USAMO.


Find all primes $p,q$ such that $pq|2^{p}+2^{q}$.


geometry circumcircle AMC AIME USA(J)MO USAMO

In triangle ABC, the incircle, which has a radius of 5, is tangent to sides AB, AC, and BC at Z, Y , and X respectively. The lines AX, BY , and CZ concur at P. If 5AP/XP + 5BP/YP + 5CP/ZP = 74, and 2 cosB = cosA + cosC, compute the area of ABC.

How to find the radius of the circumcircle of triangle ABC?

Is this problem from an AMC, AIME, or USAMO? If not, moderators, please kindly move this topic to Pre-Olympiad Forum. Thank you.


Prove that if p is an odd  prime, then every primitive root move p is a quadratic non - residue mod p.

Let $ g$ be the primitive root. Let us assume to the contrary that $ \exists y$ such that $ g \equiv y^2 \bmod p$. As $ g$ is a primitive root, $ \exists \ell, \ell \leq p\minus{}1$  such that $ y \equiv g^{\ell} \bmod p$. Now, by a little algebra we get $ g^{2\ell \minus{} 1} \equiv 1 \bmod p$. Using the fact that $ \ell \leq p\minus{}1$ and $ \text{ord}_{p}(g) \equal{} p\minus{}1$ we get $ p \minus{} 1 \equal{} 2\ell \minus{} 1$ which suggests that p is even which is a contradiction. QED.


The Smith family has 4 sons and 3 daughters.  In how many ways can they be seated in a row of 7 chairs such that all 3 girls sit next to each other?

Let the seats be:

_ _ _ _ _ _ _

Clearly, there are $ 5$ spots to put the group of 3 girls, G G G _ _ _ _, _ G G G _ _ _, _ _ G G G _ _, _ _ _ G G G _, _ _ _ _ G G G. Within the group of girls, there are $ 3!\equal{}6$ ways to arrange them. Similarly, the boys have to be seated in the remaining 4 seats, which can be done in $ 4!\equal{}24$ ways. Multiplying gives $ 5\times6\times24\equal{}\boxed{720}$


In  looking at the classic $L^{p}$ analysis problem that Eugene posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100074]here[/url], I noticed that he was using "\parallel" for the norm symbol.

I distinctly recall that "\|" makes a somewhat better-looking norm symbol, and that I was accustomed to using that. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90666]Here[/url] is an example of that. But now, "\|" causes odd things to happen, involving "mid," as in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100112]this example[/url].

Grobber is having the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99994]same problem.[/url]

Ah, it slipped me. Basically I've made the parser change || to \parallel but I forgot some people use \| for norm :D and I changed | into \mid so that's what happens.

I will fix it.

Okay, a quick fix is removing the mid parse for now.


geometry
3D geometry
sphere
calculus
algebra
polynomial
derivative

Here I come with my questions again. This time I've moved from calculus to linear algebra! (Thank you wonderful people).

"The plane $x - 2y + z = 0$ intersects the sphere $x^2 + y^2 + (z - 6)^2 = 56$ along a circle C. For which values of the constant $d$ does the circle C intersect the plane $x - 3y + 2z = d$".

This is not an as hard problem as I thought. You determine a general line of solutions to the two planes in parametric form. Then you just plug the values into the equation of the sphere and see for what $d$ the polynomial has real solutions.

The sphere with centrum in $(-3,1,-2)$ with radius 5 intersects the plane $2x - 2y + z + 1 = 0$ along a circle. Determine the point on the circle which is closes to the point $(9,1-8)$.

Can someone please help me with this? I don't understand at all how to get an expression for the distance between a given point on the circle and the point.

Let your point be $(x,y,z)$. We know that \[ (x + 3)^2 + (y - 1)^2 + (z + 2)^2 = 25 \] (since the point is on the sphere) and also \[ 2x - 2y + z + 1 = 0 \] (since the point is on the plane). Now, we need to minimize \[ (x - 9)^2 + (y - 1)^2 + (z + 8)^2 \] under these conditions. Do you see how to do that? :)

No :( That's my problem.  I need more clues.

You can eliminate some of the variables :)

I'm reaaally not seeing what to do. Sorry, I just can't see how to get the plane equation in there.

You could, for example, replace $z$ by $2y - 2x - 1$ everywhere - then there are just two variables left :)

I know this is starting to get ridiculous, but I don't see how this leads anywhere! After that you have x^2 and y^2 and xy and the systems are just so not linear that I just can't see how to get rid of any more variables.

Depends on how much calculus you know. Do you know multivariable calculus?

I have no clue where this is going... unless Arne is hinting for partial derivatives (in that case: shame on you! :P)

My solution: consider the 2 spheres:
$(x+3)^2+(y-1)^2+(z+2)^2=25$
$(x-9)^2+(y-1)^2+(z-8)^2=d$

Their intersection (if existing) will certainly fulfil the relation obtained by eliminating $y$:
$(x+3)^2-(x-9)^2+(z+2)^2-(z-8)^2=25-d$ which is $24x+20z+d-157=0$.

Now eliminate $z$ by interzecting with $2x - 2y + z + 1 = 0$, so you get $16x-40y+177-d=0$. Now you have $y(x)$ and $z(x)$, filling in the equation of the first sphere will give you the relation $d(x)$, and deriving to $x$ gives you the minimal and maximal $x$, which solves the problem.

A little calculation: maximum distance is attained for $x=-1-\frac{8\sqrt{65}}{39}$, mimal distance for $x=-1+\frac{8\sqrt{65}}{39}$. (computer calculation: minimal distance is 10.72392577)

[quote="Peter VDD"]I have no clue where this is going... unless Arne is hinting for partial derivatives (in that case: shame on you!) [/quote]
Ah, but you used at least some derivative. Can we show that the problem was correctly placed in the linear algebra section by not using calculus at all?

Some general principles: quadratic functions can be maximized by completing the square, and/or by diagonalizing matrices. And quadratic objects can be usefully expressed in linear (i.e., matrix) notation.

[quote="Kent Merryfield"][quote="Peter VDD"]I have no clue where this is going... unless Arne is hinting for partial derivatives (in that case: shame on you! :P) [/quote]
Ah, but you used at least some derivative. Can we show that the problem was correctly placed in the linear algebra section by not using calculus at all?[/quote]...
A simple 1-variable derivative of a simple expression is not really worth classifying this under calculus imo. I mean, any problem can have this...

But it is well possible this type of problem in general belongs rather to calculus/analysis... actually, in our univ the analysts say it's algebra and the algebraicians say it's analysis. :P

I'm sorry to say that is not the right answer (although I can't see what's wrong with the solution). The answer is the point $(5/3, 1/3, -11/3)$. Also, I highly doubt this is the intended method. This is from a past exam (and there are several questions that are almost exactly the same, hence why I am eager to see a solution) and not meant to be a very difficult problem (more difficult than average though). I asked my lecturer in the course though, and he didn't see how to solve it either.

This is from an introductory course in linear algebra that could well be taken before any calculus course at all (but they reccomend doing it the other way around because linear algebra is harder than one variable calculus).

I would still like to thank especially Peter for taking the time to finish such an annoying solution for me (but I'm still not really satisfied, no :().

I guess there is a "geometric approach". Try to visualize the situation. Let $P$ be the given point and let $\pi$ be the given plane. Let $O$ be the centre of the sphere and let $O'$ and $P'$ be the projections of $O$ and $P$ respectively onto $\pi$.

Now, am I right here - is the point we are looking for one of the intersections of the circle and the line $O'P'$? (This might be rubbish :D)

If that is correct, you don't need calculus.

[quote="Kalle"]I'm sorry to say that is not the right answer (although I can't see what's wrong with the solution). The answer is the point $(5/3, 1/3, -11/3)$.[/quote]Sorry, I took (9,1,8)... still, the method should work (and even greatly simplified!)

New solution: consider the 2 spheres:
$(x+3)^2+(y-1)^2+(z+2)^2=25$
$(x-9)^2+(y-1)^2+(z+8)^2=d$

Their intersection (if existing) will certainly fulfil the relation obtained by eliminating $y$: $z = 2x-\frac{157}{12}+\frac{d}{12}$.
Now eliminate $z$ by interzecting with $2x - 2y + z + 1 = 0$, so you get $y = 2x-\frac{145}{24}+\frac{d}{24}$.
Now you have $y(x)$ and $z(x)$, filling in the equation of the first sphere will give you the relation $d(x)$, which obtains maximum in $x=\frac53$ and minimum in $x=\frac{-11}{3}$. Filling in this value shows that the minimal distance is $\sqrt{73}$.

But I think Arne's solution is much shorter indeed...

Thank you, Peter. I do think this method is quite hard and messy to do without a calculator though.

Maybe I need to study more of this course (we haven't come very far, I am solving exercises far ahead =p), but what does the projection of a POINT onto a plane mean (don't you also need to specifiy a line?). As I said, maybe I need to take more of the course first. If Arne's solution is short, can someone work through it so I can check back later in the course, pleeaaase?  :P

If $\pi$ is a plane and $P$ is a point not on $\pi$, then the projection of $P$ onto $\pi$ is the unique point $P'$ which satisfies $PP' \perp \pi$.

Ok, so we have that $P' = (7, 3, -9)$ and $O' = (-1,-1,-1)$. So the line passing through these points can be written as: $x = -1 + 2t, y = -1 + t, z = -1 - 2t$ for a parameter $t$. This line intersects the sphere for $t = \pm\frac43$. Calculating the two points and taking the shortest distance gives the exact right answer!

But, I cannot understand your argument, Arne :( I am very very bad at seeing things geometrically. Could you take a stab at explaining how it works? I realize this rather simple problem has required like 20 posts. I'm sorry for being so dumb.

well, think of that circle in space, the plane in space, the point in space. Due to pythagoras theorem we can split this into a component IN the plane and one PERPENDICULAR to the plane. Running on your circle will leave the perpendicular component invariant, so we basicly only look at the component in the plane, which is the distance from P' to the circle.

And the minimal distance from a point to the circle is well known: distance to its center minus radius. :)

So, if anyone feels like doing the computations... Don't have time right now... And I'm sure I would make stupid computational mistakes anyway... :)

Oh, sorry, you did :D I didn't even notice! Hm, I'm really too busy at this moment...  :( that's why I didn't read your post properly.


Help me solve this problem.
Let's consider a string of non-negative integers, containing N elements. Suppose these elements are S1 S2 .. SN, in the order in which they are placed inside the string. Such a string is called 'funny' if the string $S_1+1, S_2, S_3, .. S_{N-1}, S_N -1$ can be obtained by rotating the first string (to the left or to the right) several times. For instance, the strings 2 2 2 3 and 1 2 1 2 2 are funny, but the string 1 2 1 2 is not. Your task is to find a funny string having N elements, for which the sum of its elements $(S_1+S_2+..+S_N)$ is equal to K. 2<=N<=10000 1<=K<=30000 GCD(N,K)=1

Lets say you rotate the string by c. Then $S_1 = S_{1+c} = S_{1+2c} = \ldots$. If c and n share a common factor, that will end up $= S_1+1$, impossible, so they are coprime. So, if we think about the indices mod c, we have:

\[ S_1 = S_{1+c} = S_{2+c} = \ldots = S_{n-c} = S_n - 1 \]
\[ S_n = S_{n+c} = \ldots = S_1 + 1 \]

That means there are m $S_1$s, where $mc\equiv -1\pmod{n}$, and n-m $(S_1 + 1)$s.
Total sum is $nS_1 + n - m = K$, so $S_1 = \frac{K+m}{n}-1$.
So just choose $m\equiv -K\pmod{n}$, c such that $cm\equiv -1\pmod{n}$, and fill in the sequence starting with $S_1 = \frac{K+m}{n}-1$.


Yes, this inequality isn't true. Probably the author forgot "+" beteen fractions on the right side (inequality "with +" is true).

yes, I'm sorry, there was + but after "\"

it should be

let $n>1$; for all $x_{1},\dots,x_{n},\ y_{1},\dots,y_{n}>0$,
\[\frac{\prod_{i=1}^{n}(x_{i}+y_{i})}{\sum_{i=1}^{n}(x_{i}+y_{i})}\quad\ge\quad \frac{\prod_{i=1}^{n}x_{i}}{\sum_{i=1}^{n}x_{i}}+\frac{\prod_{i=1}^{n}y_{i}}{\sum_{i=1}^{n}y_{i}}\]
could you write a proof ?

you can do it by induction.
base is when $n=2$ and the inequality is equivalent to(after multiplication) $(x_{1}y_{2}-y_{1}x_{2})^{2}\geq 0$
now assume the inequality holds for $n$
instead of $x_{n}$ plug $x_{n}+x_{n+1}$ in the inequality for n variables and do the same for $y_{n}$.$(x_{n}+x_{n+1}+y_{n}+y_{n+1})\frac{\prod_{i=1}^{n-1}(x_{i}+y_{i})}{\sum_{i=1}^{n+1}(x_{i}+y_{i})}\geq (x_{n}+x_{n+1}) \frac{\prod_{i=1}^{n-1}x_{i}}{\sum_{i=1}^{n+1}x_{i}}+(y_{n}+y_{n+1}) \frac{\prod_{i=1}^{n-1}y_{i}}{\sum_{i=1}^{n+1}y_{i}}$
now $LHS$ of the inequality for $n+1$ variables is $\geq$
$\frac{(x_{n}+y_{n})(x_{n+1}+y_{n+1})}{(x_{n}+y_{n}+x_{n+1}+y_{n+1})}*[(x_{n}+x_{n+1})\frac{\prod_{i=1}^{n-1}x_{i}}{\sum_{i=1}^{n+1}x_{i}}+(y_{n}+y_{n+1}) \frac{\prod_{i=1}^{n-1}y_{i}}{\sum_{i=1}^{n+1}y_{i}}]$
now it's enough to proove that $\frac{(x_{n}+y_{n})(x_{n+1}+y_{n+1})(x_{n}+x_{n+1})}{(x_{n}+y_{n}+x_{n+1}+y_{n+1})}\geq x_{n}x_{n+1}$ and $\frac{(x_{n}+y_{n})(x_{n+1}+y_{n+1})(y_{n}+y_{n+1})}{(x_{n}+y_{n}+x_{n+1}+y_{n+1})}\geq y_{n}y_{n+1}$ which is trivial.
i didn't analize the equality cases but i think that if $n\geq 3$ that all $y$ must be zero(or arbitrarily close to zero)..

omg.i tried \pro and \product...
thanks :lol:

[quote="nayel"][quote="randomizer"]I used the inequality
\[\sum_{i=1}^{n}\frac{a_{i}b_{i}}{a_{i}+b_{i}}\le\frac{AB}{A+B}\]
where $A=a_{1}+\dots+a_{n}$, $B=b_{1}+\dots+b_{n}$.

Maybe someone else will give you a direct proof of your inequality.[/quote]

how do you prove it?[/quote]
you can prove it using induction,its a well-known problem:
first prove it for $n=2$ the rest is easy