How many solutions does the equation $ \sqrt{3x}\equal{}x\sqrt{3}$ have?

1) x=0
2) we may divide by sqrt3, so we get x=1
2 solutions

Square both sides.

$ 3x \equal{} 3x^2$

Quadratics have two solutions.

Of course, one must be careful to look for double roots.

Yes, the fundamental theorem of algebra really states: there exists, for a degree n polynomial, at most n roots. The multiplicities of all roots must add up to n.

For example $ x^2\minus{}2x\equal{}\minus{}1$ has only a solution $ x\equal{}1$, although it's quadratic. This happens when its discriminant is 0.

hello, from the range of definition we get $ x\geq 0$, squaring both sides of the
equation we have $ 3x\equal{}3x^2$ or $ 0\equal{}x(x\minus{}1)$, from here we get the solutions
$ x_1\equal{}0$ or $ x_2\equal{}1$, which both fulfill our equation.
Sonnhard.

But that's a critical mistake. Quadratics need not have 2 solutions. A better way is to eye the solutions 0 and 1, then conclude that the answer is 2.

[quote="AIME15"]It's a countdown, you know quadratics have $ 2$ solutions, so you guess $ \boxed{2}$...[/quote]

Look at post #6. You would make a mistake in this case.

But you know that it's not going to be the square of a binomial, just by looking at it...

Generally when they ask for how many solutions there are in a quadratic in a countdown, you'd guess $ 2$ and most of the times that'll work.  Of course, if you're wrong, the opponent basically has the answer :P

Or, hit the buzzer and solve it in the three seconds you have

Also, the question would have to say "how many distinct roots" to be clear, since one could interpret it two ways if there were a double root. This discussion is getting pointless, though...

Either it's 0 or 1.
There are 2 solutions.


νδο ο αριθμος  99899...9 με ν+3 ψηφία
[u]δεν [/u] διαιρείται με το 37.

 :)

Ο αριθμός μετά από πράξεις γράφεται    999 *  10^ν  -1.
 όμως το 37 | 999 και άρα το 37 | 999* 10^ν και δεν διαιρεί το
 999 *  10^ν  -1  :P


[youtube]cXz6UJL5oDA[/youtube]

very inspirational,
jorian hoover

wow
just wow
i'll definitely try to do it

i know.

i showed it to my mom today, and she's like "this makes you feel bad about arguing with me over a restaurant" and i felt really bad

so i put it on here

jorian

This has also been posted on IBScrewed.

Here's a link to that page. The address is listed on there.

[url]http://www.ibscrewed.net/forum/viewtopic.php?t=13019[/url]

Maybe we could get lots a people from aops to do this if we put this in as many forums as we could, without crossing the line of "spam"....


Consider a series starting with digits 1 and 2 and then arriving at the next number by multiplying the previous two digits.

So next one is 1*2 that is 2.

Next is 2*2 that is 4.

Next is 2*4 that is 8 so on ....

thus you have

1 2 2 4 8 32 24 6 4 8 24 24 32 16 8 8 8........

The question is: how many times the digit 9 will appear during the compilation of the first 200 digits??

 :?:  :?:

[quote="LordoftheMorons"]$0$
Also, I believe that your sequence is incorrect.[/quote]

I am not sure if 0 is the right answer and here is how I have derived the series...

1 2 2 4 8 32 24 6 4 8 24 24 32 16 8 8 8........

$1*2 = 2$
$2*2 = 4$
$2*4 = 8$
$4*8 = 32$

Till now the series is...
1 2 2 4 8 32

Now multiple $8*3$ and get $24$. Thus, the series updates to...
1 2 2 4 8 32 24

Now multiple $3*2$ and get $6$. Thus, the series updates to...
1 2 2 4 8 32 24 6

Now multiple $2*2$ and get $4$. Thus, the series updates to...
1 2 2 4 8 32 24 6 4

Now multiple $4*2$ and get $8$. Thus, the series updates to...
1 2 2 4 8 32 24 6 4 8

Now multiple $6*4$ and get $24$. Thus, the series updates to...
1 2 2 4 8 32 24 6 4 24

and so on....

[quote="LordoftheMorons"]Oh, you only multiply the first digit? I see. I still can't see how a $9$ would be introduced into the series, though.[/quote]

Its not that I only multiple the first digits... The multiplecation is digit by digit and go on increasing the series for 200 multiplications.

I also doubt, but still cannot point to it why 9 will not be there?

[quote="lotrgreengrapes7926"][hide="solution"] $0$
The largest number you can get is $9*9=81$ So the only 9's attainable are 9 and 49, which come from $3*3$ or $7*7$. 3 is prime so there must be a number ending in 3 and a number beginning in 3 next to each other. There are no numbers ending in 3 that are attainable. Going to $7*7$, since 7 is prime, there must be a number ending in 7 and a number beginning in 7 next to each other. Again, no numbers ending in 7 are attainable. *Note: I am trying to find the first 9, so 27 is not attainable because it requires $9*3$.  :D [/hide][/quote]

Thanks !


function
geometric sequence
algebra unsolved
algebra

find all function $f(x): N \to  N$satisfy :
$f(m-n)f(m+n)=f(m^2) \forall m>n$

This proof works if $n$ can be equal $0$....
By putting $n=0$ we obtain $f(m)^2=f(m^2)$, and by putting $n=1$ we get $f(m-1)f(m+1)=f(m^2)=f(m)^2$ so \[ \frac{f(m+1)}{f(m)}=\frac{f(m)}{f(m-1}  \] which means that $f(1), f(2)\cdots$ form a geometric sequence!!! Then, function $f$ is defined as followes: \[ f(n)=ab^{n-1}  \] where $a$ is natural (cos' $f(1)$ is natural) and $b$ is natural, as for each $n$ term $f(n)$ is natural...and as $f(n)^2=f(n^2)$, $a=1$...
Of course if $0$ isn't natural... problem gets complicated ;)  then by putting $n=m-2$ and $n=m-1$ we obtain \[ \frac{f(2)}{f(1)}=\frac{f(2m-1)}{f(2m-2)} \] and \[ \frac{f(3)}{f(2)}=\frac{f(2m-2)}{f(2m-3)} \] so if we define $f(1)=a$, $f(2)=f(1)q=aq$ and $f(3)=f(2)w=aqw$ then $f(4)=f(2^2)=f(1)f(3)=a^2qw$ and so on, we can construct next terms step by step... we can easily prove, that function defined like this has a form( for$n\geq4$) \[ f(n)=a^2q^{[\frac{n-1}{2}]}w^{[\frac{n-2}{2}]} \] now all wehave to do is to find out which functions of this form are "good"... I'll do it ...maybe tomorrow ;)


trigonometry AMC USA(J)MO USAMO inequalities calculus vector

A man with height [b]h[/b] fires a shell with a gun with initial speed [b]v[/b] at an angle [b]A[/b] above the horizontal . The shell hits a target which is a distance of [b]s[/b] from the man . If there is only one  possible value for [b]A[/b] for which this occurs , show that

$v^2 = g (-h +\sqrt{h^2 +s^2}) $

wow, that is some nice algebraic insight... and a very nice question!

This is the type of question where those USAMO (and up) skills pay off, of which I don't have many.

Nice solution Kunny.

[quote="kunny"]Let $T$ the time required for hitting target by the shell, then we have $s=v(\cos A)T\Longleftrightarrow T=\frac{s}{v\cos A}$.

The hight of the shell at the time $t=T$ can be obtained by $y=h+v(\sin A)T-\frac{1}{2}gT^2=h+s(\tan A)-\frac{1}{2}g\frac{s^2}{v^2\cos ^ 2 A}$. Since $y\geq 0$, we get

$v^2\geq \frac{1}{2}gs^2\cdot \frac{1+\tan ^ 2 A}{h+s\tan A}=\frac{1}{2}g\cdot \frac{s^2+(s\tan A)^2}{h+s\tan A}$

$=\frac{1}{2}g\left(s\tan A+h+\frac{h^2+s^2}{s\tan A+h}-2h\right)$

$\geq \frac{1}{2}g\left(2\sqrt{(s\tan A+h)\cdot \frac{h^2+s^2}{s\tan A+h}}-2h\right)=g(\sqrt{h^2+s^2}-h)$, we are done.[/quote]

What happened to the inequality, kunny? I followed your solution but you showed $v^2 \geq g(\sqrt{h^2+s^2}-h)$; this isn't precisely what was asked for.

[quote="shyong"] If there is only one  possible value for [b]A[/b] for which this occurs[/quote]
If only 1 possible value of [b]A[/b] is allowed, then this must be the A such that the maximum horizontal distance is achieved. Maybe this has something to do with it :\

To Dr.No:my solution is exactly same as that of Spoon's.

kunny

[quote="kunny"]Let $T$ the time required for hitting target by the shell, then we have $s=v(\cos A)T\Longleftrightarrow T=\frac{s}{v\cos A}$.

The hight of the shell at the time $t=T$ can be obtained by $y=h+v(\sin A)T-\frac{1}{2}gT^2=h+s(\tan A)-\frac{1}{2}g\frac{s^2}{v^2\cos ^ 2 A}$. Since $y\geq 0$, we get

$v^2\geq \frac{1}{2}gs^2\cdot \frac{1+\tan ^ 2 A}{h+s\tan A}=\frac{1}{2}g\cdot \frac{s^2+(s\tan A)^2}{h+s\tan A}$

$=\frac{1}{2}g\left(s\tan A+h+\frac{h^2+s^2}{s\tan A+h}-2h\right)$

$\geq \frac{1}{2}g\left(2\sqrt{(s\tan A+h)\cdot \frac{h^2+s^2}{s\tan A+h}}-2h\right)=g(\sqrt{h^2+s^2}-h)$, we are done.[/quote]

Thats really nice kunny !  :D  :D  :D  I face some problems when trying to group [b]v[/b] out since I use calculus to find the max of [b]A[/b] and then plugging that [b]A[/b]
inside really makes it so ugly .... Your method of solving have solve my problem ! Thanks  :D

[quote="Spoon"]hmm... I don't believe you can solve for this A analytically, someone correct me if I'm wrong!

Also, kunny I didn't post a solution  :blush:. Were you refering to my previous post about A being the ange yielding maximum horizontal distance?[/quote]

Actually it can be found that the distance is maximize when
$tanA = \frac {v}{\sqrt{v^2 + 2gh}}$

I'd very much like to see it, if you could please?

from law of conservation enery ,

1/2 *m*V^2 = 1/2*mv^2 + mgh

V^2 = v^2 + 2gh ------(1)    ( here , V is the velocity just before the shell strikes the target or ground  )

then suppose the time taken is [b]t[/b] then

V=gt + v  and  d=vcos[b]A[/b]t----(2)

drawing it in vector , and you will have a triangle PQR with QP = v , PR = gt , QR = V .
Drop a perpendicular from Q to PR and let the intersecting point be M . Then angle PQM = [b]A[/b] . Let the area of the triangle PQR be S then

S = 1/2 * vsin(90-A)gt = 1/2 * vgtcosA=1/2 *dg    (from (2) )

to maximize [b]d[/b] , you need to maximise S which tell us that triangle PQR must be right angle at Q . Hence

tan A = PQ/QR = v / V = v / {sqrt (v^2 + 2gh )}     ( from (1) )   ...Q.E.D

Very simple,shyong.as if Feynman's explanation. :)

kunny


Express $ 42\frac{6}{7}\%$ as a common fraction.

This can be re-written as $ \frac{300}{7}$ Since this is a percent, put it over 100, giving you $ \frac{300}{700}\equal{}\frac{3}{7}$


probability
trigonometry
arithmetic sequence
number theory

Well, I took another shot at the 12A recently to see how much I've improved (I became very interested in all types of problem solving shortly after I took the test) and found I am now much better. All of which is thanks to AoPS and every member on these boards ([i]YES[/i], even sharkman). Thanks guys.

[hide="Problem #7"]Let [i]a[/i], [i]b[/i], [i]c[/i], [i]d[/i], and [i]e[/i] be five consecutive terms in an arithmetic sequence, and suppose that $ a+b+c+d+e=30$. Which of the following can be found?
(A): $ a$
(B): $ b$
(C): $ c$
(D): $ d$
(E): $ e$
[hide="Correct answer:"][i]C[/i]

Ow, I thought it was [i]A[/i]. I was under the apparent misconception that $ a+(a+n)+(a+2n)+(a+3n)+(a+4n)=30$. Or is the answer that $ c$ is the mean, and $ c=\frac{30}{5}$?[/hide][/hide]
[hide="Problem #12"]Integers [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i], not necessarily distinct, are chosen independently and at random from $ 0$ to $ 2007$, inclusive. What is the probability that $ ad-bc$ is even?
(A) : $ \frac{3}{7}$
(B) : $ \frac{7}{16}$
(C) : $ \frac{1}{2}$
(D) : $ \frac{9}{6}$
(E) :$ \frac{5}{8}$
[hide="Correct answer:"][i]E[/i]

I got this. HOWEVER, I brute-forced this answer using a 13 or 14 line program that randomed [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i]. It did the calculation then stored to a constantly updating variable for even or odd respectively 300 times. It got a probability of .634 or something which was really close to [i]E[/i].  Can someone show how to not have to BS this problem? Thanks in advance.[/hide][/hide]
[hide="Problem #16"]How many three-digit numbers are composed of three distinct digits such that one digit is the sum of the other two?

(A) : $ 96$
(B) : $ 104$
(C) : $ 112$
(D) : $ 120$
(E) : $ 256$
[hide="Correct answer:"][i]C[/i]

My work: .... :read: $ \Rightarrow$  :stretcher: . The only number I could pit out was $ 96$, and I was pretty sure that was wrong. What am I missing?[/hide][/hide]
[hide="Thoughts:"]
Overall, of the 21 problems I completed, 3 were wrong, but I have already taken this test. Most of the problems I seem to be having trouble with include counting, and complex permutations/probability. I am partly finished with [i]Introduction to Number Theory[/i] (AoPS). Will [i]Introduction to Counting and Probability[/i] be highly beneficial?[/hide]
That's all for now, I still want to try to bust the trigonometry problem ([i]#17[/i]) on my own.

[quote="igiul"]
Ow, I thought it was [i]A[/i]. I was under the apparent misconception that $ a+(a+n)+(a+2n)+(a+3n)+(a+4n)=30$. .[/quote]
That's not a misconception.  But, [hide]there you have 2 variables in one equation.  So we can't pinpoint $ a$ or $ n$.
Howver, simplify that and you get $ 5a+10n=30$  Thus, $ a+2n=6$.  Since $ a+2n=C$, we have solved for C.

You can also intuitively solve this based on the fact that the sequence has to be "balanced" around 30/2 (EDIT: what!? I meant balanced around the middle).[/hide]

Oh wow, so I [i]was[/i] right...almost. Thank ya sir.

[quote="igiul"][hide="Problem #12"]Integers [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i], not necessarily distinct, are chosen independently and at random from $ 0$ to $ 2007$, inclusive. What is the probability that $ ad-bc$ is even?
(A) : $ \frac{3}{7}$
(B) : $ \frac{7}{16}$
(C) : $ \frac{1}{2}$
(D) : $ \frac{9}{6}$
(E) :$ \frac{5}{8}$
[hide="Correct answer:"][i]E[/i]

I got this. HOWEVER, I brute-forced this answer using a 13 or 14 line program that randomed [i]a[/i], [i]b[/i], [i]c[/i], and [i]d[/i]. It did the calculation then stored to a constantly updating variable for even or odd respectively 300 times. It got a probability of .634 or something which was really close to [i]E[/i].  Can someone show how to not have to BS this problem? Thanks in advance.[/hide][/hide][/quote]

[hide="Step 1"] In order for $ ad$ to be odd, both $ a, d$ must be odd, so the probability of $ ad$ being odd is $ \frac{1}{4}$.  Same with $ bc$.  [/hide]
[hide="Step 2"] In order for $ ad-bc$ to be even, either both $ ad, bc$ are even or both are odd. [/hide]
[hide="Step 3"] The probability that $ ad, bc$ are both even is $ \frac{3}{4}\cdot \frac{3}{4}$, and the probability that $ ad, bc$ are both odd is $ \frac{1}{4}\cdot \frac{1}{4}$.  [/hide]

[quote="igiul"]
[hide="Problem #16"]How many three-digit numbers are composed of three distinct digits such that one digit is the sum of the other two?

(A) : $ 96$
(B) : $ 104$
(C) : $ 112$
(D) : $ 120$
(E) : $ 256$
[hide="Correct answer:"][i]C[/i]

My work: .... :read: $ \Rightarrow$  :stretcher: . The only number I could pit out was $ 96$, and I was pretty sure that was wrong. What am I missing?[/hide][/hide][/quote]

Well, your problem statement is not correct.  One digit is supposed to be the [b]average[/b] of the other two.  $ A$ is the correct answer for the sum problem.


[hide]On a clock, 2 plus eleven is 1.[/hide]

[quote="Cartman"]When can you add two to eleven and get one as the correct
answer?[/quote]
[hide]Try mod 1, mod 2, mod 3, mod 4, mod 6, and mod 12.
Or, here's a proof that
$2+11=1$
$a=b$
$a^{2}=b^{2}$
$a^{2}-b^{2}=0$
$a+b=0$
$a=-b$
$1=-1$
$2=0$
$1=0$
$12=0$
$11=-1$
$2=2$
$2+11=1$[/hide]

[quote="davidyko"]lol @ proof.
But yeah, either on a clock, or in other bases.[/quote]

Not other bases, just other moduluses. (Moduli?)


I think our chapter competition is on the 3rd. How should I prepare? I have limited time (other important things, such as homework and stuff), and I have absolutely no idea as to how to prepare.

[quote="Zenox"]Get good sleep the night before, and practice until saturday, starting now - doesn't matter if it's online here, with a teacher, a friend, or by yourself with a book.[/quote]

Get a good sleep the night before the competition is really important, and do not panic before the competition! You still have several days to get ready. Good luck for the competition! :P

What you two said is important, but I think he wants to know how to practice for the competition. I would suggest doing a lot of previous chapters and states (found at [url]http://www.mathcounts.org[/url]), and regularly going here and doing Mathcounts level problems.

whatever you do, just don't cram the night before.
i have personal experince, and Ubemaya knows exactly what i am talking about.................

[quote="Ubemaya"]Seeing as it's in 5 days, you shouldn't be preparing anymore -- resting would be the best way to prepare.

But seriously, I doubt you need to worry about making States or doing well unless you fall into a coma on Friday or something...[/quote]

Yeah I agree, if you want to prepare don't start 5 days before...........

@vishalarul:  you have done mathcounts before, havent you?  don't you go to miller?  if you do, then you've done more than enough preparation.

[quote="i_like_pie"]No, he and I go to CMS.[/quote]

Looks like you guys have a strong team.....

uhh ohh
that's at least x-1(where x>10 and is an integer)teams better than ours.........we're going to lose
1. Miller
2. Harker
3. Hyde
4. Cupertino
5. Kennedy
6. Challenger Berryesa
7. William Hopkins
8. (reserved)
9. (reserved)
10.(reserved)
......
x. my team

[quote="anirudh"]Maybe what I'm doing -- go to the AMC 10 website and download those worksheets covering types of problems and doing them without the choices and brushing up on stuff you're not good at.[/quote]
where? i couldn't see them anywhere.

[quote="ProtestanT"][quote="i_like_pie"]No, he and I go to CMS.[/quote]
thats as clear as me saying i go to a school in central new jersey.
u no, im gonna count how many schools in this country have CMS as its abreviation. I know it stands for Cupertino MS in ilikepie's case, but there are many more. like Community MS in my district, etc

edit- on the first 10 results from googling CMS middle school got 9 middle schools' websites
i estimate there will be over 30 Middle schools called CMS
there are actually 2 middle schools with the string "Community Middle School" in it.

if any1's wonderin, i go to a school with abreviation GMS, which is actually short for TGMS[/quote]yep, i agree. all three middle schools in Carmel are abbreviated with CMS.

Clay MS, Creekside MS, Carmel MS.  :P

wow...too much santaclara chapter competition...they should send a lot more teams, considering how big and compeitive our chapter is

[quote="anirudh"]Our chapter competition is extremely competitive!

That's why I need to prepare quickly.

AMC 12's are the way for me.[/quote]

Where is your chapter???

The way to prepare:

Do a ton of mathcounts stuff:

Like problem of the day, go figure challenge, team round with your team, past sprints and targets, stuff like that...

P.S. Squeeze in an AMC 10. But if you get perfects on AMC 10, try doing National sprints and targets.

[quote="anirudh"]No. AMC 10's last 5 are waay harder than MC national sprints/targets. About 10-15 are targets and 1-7 are sprints

Try amc 12's.[/quote]
1. AMC 10s last five are on par with national targets.
2. If you did believe that, why would you be recommending amc 12s?

[size=75][color=darkred]Edited by nebula42.[/color][/size]

@bpms: I meant AMC12's. Sorry. I accidentally typed 10. :blush:

[quote="#H34N1"]Maybe what I'm doing -- go to the AMC 10 website and download those worksheets covering types of problems and doing them without the choices and brushing up on stuff you're not good at.[/quote]
what did you mean by this.....?

There's a webpage with a list of worksheets that are downloadable on the amc website.

[quote="funcia"][quote="davidlizeng"]don't cram too much on the last few days, else you go brain dead................
three food tips:
1. chew gum. Gum has been known to increase the attemtion span and concentration up to 50%.
2. eat bananas. I have no idea why, but eating bananas on average increases test score. It might be the potassium.
3. no fatty foods. Starting the day before, DO NOT eat fatty foods like pizza, chips, etc. Fat clogs arteries that decrease the amount of blood flow to the brain.
one exercise tip:
1. do a workout before the test. Only a small workout is needed, such as Ubemaya's Workout (one pushup and one situp) just to warm up the body.

I know these tips may sound crazy, but they really do help and have been proven statistically.

Personal experience: at Miller Math Marathon, i followed all four steps and ended up x-th place. Judging by the fact that i expected to bomb the test and by how dumb i am, i think they really worked.[/quote]
YAY!!! easy workouts

what do you do if you aren't allowed gum?

i've heard bannanas and chocolate are good but whatever[/quote]

chew it anyway...
once the test starts, there's noway you'll be forced to stop just to spit out gum. personally, i don't care if i get in trouble for chewing gum..........


modular arithmetic
number theory proposed
number theory

Find all pairs of possive integers $ (x;y)$ such that $ y^x\minus{}1\equal{}(y\minus{}1)!$

A less general problem was posted before:
http://www.mathlinks.ro/viewtopic.php?t=187830
but onfortunately no proof was produced. Could you please let us know what is the source of the problem?

See http://www.mathlinks.ro/viewtopic.php?t=150674 (containing a small but easy to fix error).

a similar solution to what ZetaX snet the link is:

let $ d\mid y$ and $ d < y$ then from the equation we have:

$ d\mid (y \minus{} 1)! \minus{} 1$

but on the other hand:

$ d < y\Rightarrow d\leq y \minus{} 1\Rightarrow d\mid (y \minus{} 1)!$

thus:

$ d \equal{} 1$

so $ y$ must be a prime number...

now according to WILSON theorem we have:

$ (y \minus{} 1)!\equiv \minus{} 1\pmod y$

so we have:

$ (y \minus{} 1)! \minus{} 1 \equal{} y^x\Rightarrow \minus{} 1 \minus{} 1\equiv 0\pmod y\Rightarrow y\mid 2\Rightarrow y \equal{} 2$
------------------------------
also another solution which belongs to my friend (ML id:only math) is the following one:

$ (y \minus{} 1)! \minus{} 1 \equal{} y^x$

$ \Rightarrow (y \minus{} 1)! \minus{} 2 \equal{} y^x \minus{} 1$

$ \Rightarrow (y \minus{} 1)! \minus{} \left(y^x \minus{} 1\right) \equal{} 2$

$ \Rightarrow (y \minus{} 1)! \minus{} (y \minus{} 1)\left(y^{x \minus{} 1} \plus{} y^{x \minus{} 2} \plus{} \ldots \plus{} 1\right) \equal{} 2$

but note that $ y \minus{} 1\mid LHS$ so we must have $ y \minus{} 1\mid RHS\Rightarrow y \minus{} 1\mid 2$
--------------------------------

the problem was also posted here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=183998[/url]

BaBaK Ghalebi i do not really understand your solution. you wrote that by wilson-theorem we have $ (y\minus{}1)! \equiv \minus{}1 \bmod{y}$, where $ y \in \mathbb{P}$ and using this in the equation we get $ (y\minus{}1)! \minus{} 1 \equiv \minus{}2 \not\equiv 0 \bmod{y}$. But the equation is $ (y\minus{}1)! \plus{} 1 \equal{} y^x$ and by wlson we get $ (y\minus{}1)! \plus{} 1 \equiv 0 \bmod{y}$.

oh,what a stupid mistake,both of my solutions are wrong,those solutions are for the following equation:

$ (y\minus{}1)!\minus{}1\equal{}y^x$

sorry :oops:

I'm sure the following solution is valid:
as I said in my previous post,according to wilson theorem we get that $ y$ must be prime,now let $ y > 5$ and $ x\not = 0$,note that:

$ (y - 1)! = (y - 1)\times (y - 2)\times\ldots\times (\frac {y - 1}2)\times\ldots\times 2\times 1$

$ \Rightarrow y + 1\mid (y - 1)!$

now according to our equation we have:

$ 1\equiv ( - 1)^x\pmod{y + 1}$

so $ x$ must be even,so let $ x = 2k$ where $ k\in\mathbb{Z}$ now we have:

$ (y - 1)! = (y^k - 1)(y^k + 1)$

$ \Rightarrow (y - 2)! = (y^{k - 1} + y^{k - 2} + \ldots + y + 1)(y^k + 1)$ (*)

but note that $ y - 1$ is a composite number so $ y - 1\mid (y - 2)!$ thus according to (*) we get that:

${ 0\equiv (\underbrace{1 + 1 + \ldots + 1}_{k\textrm{ times }})(1 + 1)}\pmod {y - 1}$

$ \Rightarrow 0\equiv 2k\pmod {y - 1}$

$ \Rightarrow 0\equiv x\pmod {y - 1}$

$ \Rightarrow x\geq y - 1$

but it's easy to see that:

$ y^{x}\geq y^{y - 1} > (y - 1)! + 1$

which is a contradiction,so we must check the cases $ y = 2,3,5$ and also the case $ x = 0$ which gives us the solution:

$ (y,x) = (2,1),(3,1),(5,2)$


LaTeX
inequalities unsolved
inequalities

Show that if x,y,z>0 and x+y+z=1 then
64<=(1+1/x )(1+1/y )(1+1/z )

Also, can anyone help me with the coding of mathematical symbols?  Is there a place for a tutorial on this page somewhere?  Thanks!

The first solution that comes to mind is using Holder and obtaining that $ RHS^{1/3}>\equal{}1\plus{}1/ \sqrt[3]{xyz}$ so it suffices to prove that $ 1\plus{}1/ \sqrt[3]{xyz}>\equal{}4\equal{}>1>\equal{}3\sqrt[3]{xyz}\equal{}>x\plus{}y\plus{}z>\equal{}3\sqrt[3]{xyz}$ which is true by AM GM inequality.

As far as latex is concerned (although I do't know it very well myself) you can start here: http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

And where I used $ \equal{}>$ i meant it in both directions, but I don't know how to type it!

Also possible solution to this simple question is:

$ \prod (1 \plus{} \frac {1}{x}) \equal{} \prod (1 \plus{} \frac {x \plus{} y \plus{} z}{x}) \equal{} \prod (1 \plus{} 1 \plus{} \frac {y}{x} \plus{} \frac {z}{x})\geq \prod 4 \sqrt [4]{\frac {yz}{x^{2}}} \equal{} 4^{3} \equal{} 64$

[quote="salonikios"]

And where I used $ \equal{} >$ i meant it in both directions, but I don't know how to type it![/quote]

Go \iff to get $ \iff$.


geometry
rectangle
combinatorics proposed
combinatorics

Here is a cute one :

Is it possible to colour all grid points in the plane white and red so that no rectangle with vertices on grid points of one colour and sides parallel to the grid lines has area from the set $\{1,2,4,8, \cdots , 2^n,\cdots, \}$?

Pierre.

[hide=Solution]
Yes. Using Cartesian coordinates, color the point $(x,y)$ white iff $x+y\equiv 0\pmod{3}$, otherwise color it red.

Suppose to the contrary that such rectangle exists.
Let $(p,q),(r,q),(p,s),(r,s)$ be the four vertices of the rectangle. We have $|(p-r)(q-s)|\in \{ 1,2,4,\cdots \}$.
Since $3\nmid p-r$, the points $(p,q),(r,q)$ can't be simultaneously white. So the four vertices must be red.
Also, we have $p+q\not\equiv r+q\pmod{3}$. This means $(p+q)+(r+q)\equiv 1+2\equiv 0 \pmod{3}$.
Similarly, $(p+s)+(r+s)\equiv 0\pmod{3}$. But this leads to $q\equiv s\pmod{3}\implies 3\mid q-s$ which is impossible.
[/hide]


Eto, nakon IMO-a dolazi i nešto sitno za nas (većim dijelom) umirovljene natjecatelje - IMC se ove godine održava u Budimpešti od 25. do 30.7.

Nas iz Zagreba dolazi 10, kao što sam napomenuo već u jednom topicu, a u jednom drugom sam nam spomenuo i imena:

Nikola Adžaga, druga godina
Goran Dražić, treća godina
Nikola Grubišić, četvrta godina
Željko Kereta, treća godina
Rudi Mrazović, četvrta godina
Marin Mišur, druga godina
Melkior Ornik, prva godina
Petar Sirković, druga godina
Saša Stanko, druga godina
Luka Žunić, druga godina

(leader nam je Matija Bašić)

Kao da sam negdje pročitao da dolazi i četvero ljudi iz Ljubljane, a hana je nešto spominjala da ide i njih četvero.

Eto, toliko. Više nemam što pametno da kažem, pa se idem povući da gledam Seinfeld.

Jedan oduševljeni čitatelj vam želi sreću na takmičenju i sretan put  :wink:

I ovaj čitalac se oduševljeno pridružuje svim lijepim željama

Cestitke svim osvajacima medalja (i onima koji su to zamalo postali)

Čestitke svima: čini mi se da je svako nešto donio, od Frst Prajza do Onorabl Menšna, tako da ste opravdali naša očekivanja  :lol:

Hvala na čestitkama, bilo je jako lijepo, a i rezultatom sam zadovoljan (zeznuli su mi score u završnim rezultatima  :roll:, ali oprostit ću im kad mi pošalju moju zasluženu treću nagradu :winner_third:).


Let a,b be integer integers such that ab-1 divides a^{2}+b^{2}.Show that:
\frac{a^{2}+b^{2}}{ab-1}=5

Very very very well-known method to solve it...

Does anyone try it?
PS:ZetaX,do you know Erich Ludendorff?

very very very very very easy:[url]http://en.wikipedia.org/wiki/Erich_Ludendorff[/url]:wink:

From history lessons, I'm not that old to be able to have lived with him ;)


Is there something wrong with the problem?

$ (x^2\plus{}y^6)(x^8\plus{}y^4)\equal{}x^{10}\plus{}x^2y^4\plus{}x^8y^6\plus{}y^{10}>x^{10}\plus{}y^{10}$

So apparently the result is true for any x,y...

[quote="v235711"][quote="Pain rinnegan"]Let $ x,y > 0$ such that $ x^5 \plus{} y^5 \equal{} 2$ . Prove that :

$ (x^2 \plus{} y^6)(x^8 \plus{} y^4) \plus{} 2\ge x^{10} \plus{} y^{10}$[/quote]

 this isn´t stronger than cauchy, since there is obviously no equality case...[/quote]

The correct version

$ (x^2 \plus{} y^6)(x^8 \plus{} y^4) \ge x^{10}\plus{}y^{10}\plus{}2$ with $ x^5\plus{}y^5\equal{}2$


greatest common divisor
number theory
least common multiple
prime factorization
relatively prime

Prove that the product of the greatest common factor and lowest common multiple of two numbers is equal to the product of the numbers.

[hide="Hint"]Prime factorization[/hide]

[hide]Let the numbers be $A$ and $B$. Let $g$ denote the GcF of $A$ and $B$. Then let $A$ and $B$  equal $gx$ and $gy$, respectively, where $x$ and $y$ are relatively prime integers. This means that the LcM of $A$ and $B$ is $gxy$.

$gx\cdot gy=g\cdot gxy$, so the statement is true.[/hide]

[hide="A formal construction of the proof by prime factorization"] Let $e_{p}(n)$ denote the exponent of the prime $p$ in the prime factorization of a positive integer $n$.  I give some unproven lemmas involving this function.

[b]Lemma:[/b]  $e_{p}(ab) = e_{p}(a)+e_{p}(b)$

[b]Lemma:[/b]  $e_{p}(a) = e_{p}(b) \forall p \in \mathbb{P}\Leftrightarrow a = b$

[b]Lemma:[/b]  $e_{p}( gcd(a, b) ) = min( e_{p}(a), e_{p}(b) )$

[b]Lemma:[/b]  $e_{p}( lcm(a, b) ) = max( e_{p}(a), e_{p}(b) )$

Lastly, recall that given two numbers $x, y$,

$min(x, y)+max(x, y) = x+y$

Then the proof is rather simple, and takes the following form:

$e_{p}( gcd(a, b) )+e_{p}( lcm(a, b) ) = e_{p}(a)+e_{p}(b) \forall p \in \mathbb{P}$ [/hide]

For this to work, you must allow for zero exponents.

$(a,b)\cdot[a,b]=a\cdot b$

[hide]Let $(a,b)=k$. Then we have $a=m\cdot k$, $b=n \cdot k$ and $[a,b]=p\cdot k$.
$k^{2}\cdot p = k^{2}\cdot m \cdot n$. We divide by $k$ so that $p=m \cdot n$. Replacing, $k^{2}\cdot m \cdot n = k^{2}\cdot m \cdot n$ which is true and i'm stuck.[/hide]

[quote="icx"]$[a,b]=p\cdot k$.
$k^{2}\cdot p = k^{2}\cdot m \cdot n$. [/quote]

Er, you're using the statement you're trying to prove.


[quote="Valentin Vornicu"][quote="jelyman"]Can one of you please post an announcement that spans all of the forums telling everyone to use the spoiler button? I understand that the spoiler button either wasnt at mathlinks or wasnt used much, but its appreciated tremendously here.
Thanx.[/quote]Basically people do not want in the advanced section the spolier. It is a general belief that at that level you do not need to hide information in that manner, as you can refrain yourself from seeing if you wish so.
So what I am asking is that people should post without spoilers in the Advanced Section and College Playground. It would be highly appreciated.

PS you can call me Valentin, I like that better than Vornicu :)[/quote]

I'm not sure who are you talking about when you say its a 'general belief', but this is definitely not true of the AoPS people, and as you don't have spoiler at Mathlinks I do not understand what you mean. At AoPS, in every problem forum including the advanced forum, the spoiler function is extremely appreciated and used.
We would all appreciate it if you continue using the system that has worked so well for us.

[quote="rrusczyk"][quote="rrusczyk"]Working on a lot of these little things.  In the MathLinks view, users cannot (yet) even put things into spoiler (so don't blame them for not doing it!) without explicitly typing the tags.  What we're looking into is making 'reveal all spoilers' an option in the Options panel, so that people who hate having to highlight answers won't have them hidden in the first place.  This should make everyone somewhere close enough to happy.[/quote]

How do people feel about this proposal?  This wouldn't change anything for those who don't like having to highlight text to read it - they could set their personal settings to 'reveal all spoilers' and thus they wouldn't have to highlight anything.[/quote]
I like that a lot, if the people who use this option will still put things in the spoilers (it would be easy to forget to use spoilers, if you never saw them).

[quote="rrusczyk"][quote="rrusczyk"]Working on a lot of these little things.  In the MathLinks view, users cannot (yet) even put things into spoiler (so don't blame them for not doing it!) without explicitly typing the tags.  What we're looking into is making 'reveal all spoilers' an option in the Options panel, so that people who hate having to highlight answers won't have them hidden in the first place.  This should make everyone somewhere close enough to happy.[/quote]

How do people feel about this proposal?  This wouldn't change anything for those who don't like having to highlight text to read it - they could set their personal settings to 'reveal all spoilers' and thus they wouldn't have to highlight anything.[/quote]

Yes, this is a very good idea.

[quote="rrusczyk"][quote="rrusczyk"]Working on a lot of these little things.  In the MathLinks view, users cannot (yet) even put things into spoiler (so don't blame them for not doing it!) without explicitly typing the tags.  What we're looking into is making 'reveal all spoilers' an option in the Options panel, so that people who hate having to highlight answers won't have them hidden in the first place.  This should make everyone somewhere close enough to happy.[/quote]

How do people feel about this proposal?  This wouldn't change anything for those who don't like having to highlight text to read it - they could set their personal settings to 'reveal all spoilers' and thus they wouldn't have to highlight anything.[/quote]
that'd be awesome

This is implemented now.  We will be making a global announcement about this and other items next week.

All old AoPS users have been set to have spoilers still hidden, while old ML users have been set to have them revealed by default.  Go to Options above to change your default.


This problem was just shown to me by a friend. I don't know if I can restate it as clearly as possible.

In a creek, three structures and paths between them. At one side of a creek is a resthouse and at the other side and at the other side a butterfly dome and a hut.  A pond is also to be made  beside the creek.  Suppose that you want to create a path in this order; resthouse, hut, pond and dome build a single bridge, connecting the three structures, and where will you put the bridge and the pond it so that the total distance of your path is minimum?

See skeleton drawing below: (Ignore the asterisk, its for spacing)

*******[b]Resthouse[/b]

______________________________________________
               Creek
______________________________________________

************************************[b]Dome[/b]

[b]Hut[/b]

[quote="DanK"]Well, you should connect the resthouse and the hut with a straight line, and then connect the hut and dome with a straight line.  That gives you the shortest distance for those three.  Then put the bridge over the creek on the line from the resthouse to the hut, and put the pond anywhere on the line from the hut to the dome, and you have your locations on the shortest possible path...[/quote]

but you should put the pond beside the river...

[quote="scorpius119"][hide]
Label the resthouse $R$, the hut $H$, the creek $c$, and the dome $D$. The shortest path can be found by first finding the shortest distance $RH$ and then finding the shortest path from $H$ to $D$ that touches $c$.

The first part is easy: the shortest distance is just a straight line, so the bridge should be put at the intersection between $\overline{RH}$ and $c$.

For the second part, one will have to touch $c$. Now suppose $c$ is horizontal and $R$ is above $c$. Then one will have to move back down in order to reach $D$. If one moves up instead, the end of the path will be at a point $D^\prime$ which is found by reflecting $D$ over $c$. So for any path from $H$ to $D$ that touches $c$, there is a path of equal length from $H$ to $D^\prime$ that touches $c$.

The minimum distance from $H$ to $D$ touching $c$ is equal to the minimum distance from $H$ to $D^\prime$ touching $c$. But [i]all[/i] paths from $H$ to $D^\prime$ touch $c$, so this is just a straight line, like the first part. The pond should be placed at the intersection between $\overline{HD^\prime}$ and $c$.
[/hide][/quote]

This is a good explanation, i think im getting the point. But the problem is, i made a mistake with my drawing. The hut should be slightly lower than the dome.. How are we going to compute it then? I have redrawn the sketch, please see the problem again.

Um, I think his method should still work...

I got it already, through translation and reflection.


geometry
3D geometry
sphere
calculus
calculus computations

What is the solid angle subtended by a cone of semivertical angle $ \alpha$.In general  ho to find the solid angle

Draw a sphere of radius $ 1$ centered at the vertex of the cone. Find the area of the part that lies inside the cone. Divide by the total area, which is $ 4\pi$.


function
quadratics
limit
algebra unsolved
algebra

Find all functions $f: \mathbb R \setminus \{\pm 1\} \to \mathbb R$ satisfying
\[ f(1-2x) = 2 f\left(\frac 1x \right) + 1, \quad \forall x \in \mathbb R, x \neq 0,\pm 1.\]

Ok, I consider that the question is :
Find all functions $ f$ : $ \mathbb R\backslash\{ - 1, + 1\}\to\mathbb R$ such that $ f(1 - 2x) = 2f(\frac {1}{x}) + 1$ $ \forall x\in\mathbb R\backslash\{ - 1,0, + 1\}$

$ f(1 - 2x) = 2f(\frac {1}{x}) + 1$ $ \forall x\in\mathbb R\backslash\{ - 1,0, + 1\}$ $ \Leftrightarrow$ $ f(x) = 2f(\frac {2}{1 - x}) + 1$ $ \forall x\in\mathbb R\backslash\{ - 1, + 1, + 3\}$

Let $ g(x) = f(x) + 1$ and $ h(x) = \frac {2}{1 - x}$

The problem is now : Find all functions $ g$ : $ \mathbb R\backslash\{ - 1, + 1\}\to\mathbb R$ such that $ g(x) = 2g(h(x))$  $ \forall x\in\mathbb R\backslash\{ - 1, + 1, + 3\}$

We'll have (with good conditions over $ x$) : $ g(x) = 2^ng(h(h(h( ...$n times$ ...(x)))) ... )$. And clearly the $ n^{th}$ composit of $ h(x)$ is an homographic function.

Let then the sequence $ a_n$ defined $ \forall n\in\mathbb Z$ as :
$ a_0 = 0$
$ a_1 = 2$
$ \forall n > 2$ : $ a_n = a_{n - 1} - 2a_{n - 2}$

$ \forall n < 0$ : $ a_n = \frac {a_{n + 1} - a_{n + 2}}{2}$

It's easy to check that $ a_n = \frac {\sqrt 2}{2}(b^n - (\frac { - 1}{b})^n$ (where $ b = 1 + \sqrt 2$) $ \forall n\in\mathbb Z$ and that $ a_n\neq 0$ $ \forall n\in\mathbb Z^*$.

Let then the family of functions defined for any ${ k\in\mathbb Z}$ as : $ h_k(x) = 2\frac {a_{k - 1}x - a_k}{a_kx - a_{k + 1}}$ ($ h_0$ : $ \mathbb R\to \mathbb R$; $ h_k$ : $ \mathbb R\backslash\{\frac {a_{k + 1}}{a_k}\}\to\mathbb R$)

It's easy to check that $ h_0(x) = x$ and $ h_{k + 1} = h\circ h_k$ $ \forall k\in\mathbb Z$. So, $ g(x) = 2^kg(h_k(x))$
It's also easy to check that, $ \forall k\neq 0$, $ h_k(x)$ has no fixed point (the equation $ h_k(x) = x$ implies a quadratic whose discriminant is $ - \frac {7a_k^2}{4} < 0$

Let's then the relation $ \sim$ defined as $ x\sim y$ $ \iff$ $ \exists$ a unique $ k\in \mathbb Z$ such that $ y = h_k(x)$
This relation is an equivalence one :
$ x\sim x$ : $ h_0(x) = x$ and no other $ k$ is available since $ \forall k\neq 0$, $ h_k(x)$ has no fixed point.
$ x\sim y$ $ \implies$ $ y\sim x$ : $ y = h_k(x)$ $ \iff$ $ x = h_{ - k}(x)$ and unicity is obtained thru the no_fixed_points property.
$ x\sim y$ and $ y\sim z$ $ \implies$ $ x\sim z$ :  $ y = h_i(x)$ and $ z = h_j(z)$ imply $ z = h_{i + j}(x)$ and unicity is obtained thru the no_fixed_points property.

So we can define a function $ k(x,y)$ defined in each equivalence class of $ \sim$ as $ y = h_{k(x,y}(x)$
Let $ C(x)$ the equivalence class of $ x$ and $ A = C( - 1)$

We have $ - 1 = h(3)$ and $ 1 = h( - 1)$ and so $ \{ - 1, + 1, + 3\}\subset A$

Let $ r(x)$ any choice function which associates to each real $ x$ a representative of its equivalence class (unique per class).

Then :
$ \forall x\notin A$, we have $ x = h_{k(r(x),x)}(r(x))$ and so $ g(x) = 2^{k(x,r(x))}g(r(x))$ and so we just have to define $ g(r(x))$ as we want.

The situation for $ A$ is a little bit different :
We have $ 0 = h_4(3)$ but we dont have $ g(0) = 2^4g(h_4(3))$ since we dont have $ g(1) = 2g(h(3))$
So we can choose g(x) freely for $ x = 3$ and for $ x = 0$.

So here is the general solution :
================================
Let $ U = \{r(x)$ $ \forall x\notin A = C( - 1)\}\cup\{0,3\}$
Let $ m(x)$ any function from $ U\to\mathbb R$

$ \forall x\notin C( - 1)$ : $ f(x) = 2^{k(x,r(x))}m(r(x)) - 1$
$ \forall x\in C( - 1)\backslash\{ - 1, + 1\}$ such that $ k(0,x)\geq 0$ : $ f(x) = 2^{k(x,0)}m(0) - 1$
$ \forall x\in C( - 1)\backslash\{ - 1, + 1\}$ such that $ k(0,x) < 0$ : $ f(x) = 2^{k(x,3)}m(3) - 1$

And obviously this is a general solution.

If we want a continuous solution :
================================

If we suppose $ g(x)$ continuous in $ \mathbb R\backslash\{ - 1, + 1\}$, and since $ \lim_{k\to - \infty}h_k(x) = -2b$, we have, using the continuity and $ k\to - \infty$, $ g(x) = 0$ $ \forall x\in\mathbb R\backslash A$. Then, since $ A$ is a countable set, we can always find, in any neighborhood of an element of A, elements of $ \mathbb R\backslash A$. And so $ g(x) = 0$ $ \forall x\in R\backslash\{ - 1, + 1\}$ (remember $ g(x)$ is not defined in $ \{ - 1, + 1\}$ And so the only continuous function solution of the initial problem is $ f(x) = - 1$


Very nice and good inequality! Congratulations, Kim.

[quote="Sung-yoon Kim"]For positive reals $ a,b,c$, prove that
\[ (a^{2}\plus{}ab\plus{}b^{2})(b^{2}\plus{}bc\plus{}c^{2})(c^{2}\plus{}ca\plus{}a^{2})\leq (((b\plus{}c)(c\plus{}a)(a\plus{}b))^{\frac{2}{3}}\minus{}(abc)^{\frac{2}{3}})^{3}.\]
[/quote]
Rewrite the inequality as
$ \left( 1\plus{}\sqrt[3]{\frac{(a^{2}\plus{}ab\plus{}b^{2})(b^{2}\plus{}bc\plus{}c^{2})(c^{2}\plus{}ca\plus{}a^{2})}{a^{2}b^{2}c^{2}}}\right)^{3}\le\frac{(a\plus{}b)^{2}(b\plus{}c)^{2}(c\plus{}a)^{2}}{a^{2}b^{2}c^{2}}$
By Holder inequality,
$ \left( 1\plus{}\sqrt[3]{\prod_{cycl}\frac{a^{2}\plus{}ab\plus{}b^{2}}{ab}}\right)^{3}\le\prod_{cycl}\left( 1\plus{}\frac{a^{2}\plus{}ab\plus{}b^{2}}{ab}\right) \equal{}\prod_{cycl}\frac{(a\plus{}b)^{2}}{ab}$

Nice. In fact, my solution uses the similar method.

$ \prod (a^{2}\plus{}ab\plus{}b^{2}) \equal{}\prod ((a\plus{}b)^{2}\minus{}ab)\leq (RHS)$, since $ \prod (p_{i}\minus{}q_{i})\leq ((\prod p_{i})^{\frac{1}{3}}\minus{}(\prod q_{i})^{\frac{1}{3}})^{3}$ holds by Holder.


number theory
prime factorization
number theory unsolved

For each natural $ n$,we define $ f(n)$ as the exponent of number $ 2$ in the prime factorization of $ n!$.Show that for any natural
$ a$, the equation $ n\minus{}f(n)\equal{}a$ has infinitely many solutions.

We have $ (2n)!\equal{}2^n \cdot n!\cdot1\cdot3\cdot5\cdot...\cdot(2n\minus{}1)$ so $ 2n\minus{}f(2n)\equal{}2n\minus{}(n\plus{}f(n))\equal{}n\minus{}f(n)$.
But $ 1\minus{}f(1)\equal{}1 \Rightarrow 2^a\minus{}f(2^a)\equal{}1$ so $ f(2^a\minus{}1)\equal{}(2^a\minus{}1)\minus{}a$ and $ 2^k(2^a\minus{}1)\minus{}f(2^k(2^a\minus{}1))\equal{}a$ for every nonnegative integer $ k$.

[quote="mazur89"]We have $ (2n)! \equal{} 2^n \cdot n!\cdot1\cdot3\cdot5\cdot...\cdot(2n \minus{} 1)$ so $ 2n \minus{} f(2n) \equal{} 2n \minus{} (n \plus{} f(n)) \equal{} n \minus{} f(n)$.
But $ 1 \minus{} f(1) \equal{} 1 \Rightarrow 2^a \minus{} f(2^a) \equal{} 1$ so $ f(2^a \minus{} 1) \equal{} (2^a \minus{} 1) \minus{} a$ and $ 2^k(2^a \minus{} 1) \minus{} f(2^k(2^a \minus{} 1)) \equal{} a$ for every nonnegative integer $ k$.[/quote]
Note that $ S_p(n)$ is some  digits  of n in base p.
Then $ v_p(n)\equal{}\frac{n\minus{}S_p(n)}{p\minus{}1}$
Apply this for $ p\equal{}2$ then
$ f(n)\equal{}n\minus{}S_2(n)$
So $ n\minus{}f(n)\equal{}S_2(n)$
Obvious the equation $ S_2(n)\equal{}a$ have infinite solution on N.


Prove that the centroid and orthocenter of an equilateral triangle are concurrent.

[hide]
The median of an equilateral triangle is perpendicular to the opposite side, and so is collinear to the altitude. Hence, the intersection of the medians is the same as the intersection of the altitudes, and so the centroid and the orthocenter are concurrent.
[/hide]


function
abstract algebra
complex numbers
linear algebra
geometry unsolved
geometry

how do we prove that an isometry in the complex plane is always a function of the form az+b or a(conjugate of z)+b, where modulus of a = 1, a and b are complex numbers
Also, anyone pls explain me what is meant by 'fixed points' of an isometry

First question: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/isometrycpx.pdf

(knowing the types of isometric transformations and the way to represent them in complex plane makes your question simple, and the process is given in the paper.)

Second: If $ f$ is a transformation, point $ z_0$ for which $ f(z_0)\equal{}z_0$ is called fixed point of the transformation-it's simply the point which remains intact during the transform.

thanks for the help hsiljak. Will u pls tell me what is meant by 'kernel'?

Haven't had much time to study the paper, but I assume that the typical definition of linear operator's kernel and image is what you need.

And this is how it was presented to me in my Linear algebra course:

[quote]The set of all vectors from vector space $ X$ being mapped in vector space's $ Y$ null-vector is denoted with $ \text{Ker} A$ and called kernel or zero-subspace of linear operator $ A$.

The set of all vectors in $ Y$ which are mapping images of elements in $ X$ is denoted with $ \text{Im} A$ or $ A(X)$ and called image of operator $ A$.

In other words:

$ \text{Ker} A : \equal{} \{ x \in X|A(x) \equal{} 0\}$
$ \text{Im} A : \equal{} \{y \in Y|y \equal{} A(x) \text{for some } x \in X\}$.[/quote]


Let $O_1,O_2,O_3$ be the centers of ciclers $(C_1),(C_2),(C_3)$ , respectively.
The circles $(C_1),(C_2),(C_3)$ intersect at point $O$. Let $A_1,A_2,A_3 $ be point on $(C_1),(C_2),(C_3)$, respectively, such that: $ OA_1||O_2O_3;OA_2||O_3O_1;
OA_3||O_1O_2$.
Prove that quadrilateral $OA_1A_2A_3$ ic cyclic. (maybe it was posted before)

[color=red][Moderator edit: Typo corrected.][/color]

Grobber, that one is for you  :D

  darij

You're too kind.. :D:D

Do you know how hard I tried not to use inversion? I failed :)

After inversion wrt $O$ (any power you want) you get the following problem (I'll leave out the details of how you get to it: it's straightforward): Given a triangle $ABC$ and a point $O$, assume the perpendicular through $O$ to $OA$ cuts $BC$ in $A_1$. In the same way obtain $B_1,C_1$. Show that $A_1,B_1,C_1$ are collinear.

This has been discussed before: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1036&highlight=]here[/url]'s the link.

Indeed, I don't have a non-inversive solution either...

Just a remark:

[quote="grobber"]you get the following problem (I'll leave out the details of how you get to it: it's straightforward): Given a triangle $ABC$ and a point $O$, assume the perpendicular through $O$ to $OA$ cuts $BC$ in $A_1$. In the same way obtain $B_1,C_1$. Show that $A_1,B_1,C_1$ are collinear.[/quote]

There is a famous solution of this problem in Jacques Hadamard's geometry book. You can find it in the note "Problem: The ortho-intercepts line" on [url=http://de.geocities.com/darij_grinberg]my website[/url].

  Darij

[quote="pestich"]In triangle O1O2O3   we have point  O and lines thru O parallel
       to the sides.  Parallels intersect the 3rd circle in a point symmetrical to O
      wrt  perpendicular  thru  center of 3rd circle.   So points are concyclic
     with the  center of the  resulting circle being orthocenter of O1O2O3.[/quote]

This is correct and very beautiful except of the last line, where "orthocenter" has to be replaced by "circumcenter". Thank you!

  Darij


Let $ x,y,z>0$ . Show that :

\[ \frac{x^2(y\plus{}z)}{(x\plus{}y)(z\plus{}x)}\plus{}\frac{y^2(z\plus{}x)}{(y\plus{}z)(x\plus{}y)}\plus{}\frac{z^2(x\plus{}y)}{(z\plus{}x)(y\plus{}z)}\le \frac{x\plus{}y\plus{}z}{2}\]

[quote="Pain rinnegan"]Let $ x,y,z > 0$ . Show that :
\[ \frac {x^2(y \plus{} z)}{(x \plus{} y)(z \plus{} x)} \plus{} \frac {y^2(z \plus{} x)}{(y \plus{} z)(x \plus{} y)} \plus{} \frac {z^2(x \plus{} y)}{(z \plus{} x)(y \plus{} z)}\le \frac {x \plus{} y \plus{} z}{2}\]
[/quote]

[quote="Pain rinnegan"]Let $ x,y,z > 0$ . Show that :
\[ \frac {x^2(y \plus{} z)}{(x \plus{} y)(z \plus{} x)} \plus{} \frac {y^2(z \plus{} x)}{(y \plus{} z)(x \plus{} y)} \plus{} \frac {z^2(x \plus{} y)}{(z \plus{} x)(y \plus{} z)}\le \frac {x \plus{} y \plus{} z}{2}\]
[/quote]
without loss of generality, we may assume that $ x\ge y\ge z$.
\[ \frac {x^2(y \plus{} z)}{(x \plus{} y)(z \plus{} x)} \plus{} \frac {y^2(z \plus{} x)}{(y \plus{} z)(x \plus{} y)} \plus{} \frac {z^2(x \plus{} y)}{(z \plus{} x)(y \plus{} z)}\le \frac {x \plus{} y \plus{} z}{2}\]
\[ \Longleftrightarrow \sum{\frac {x (x \minus{} y) (x \minus{} z)}{2 (x \plus{} y) (x \plus{} z)}} \ge 0\]
\[ \Longleftrightarrow \frac{{x\left( {x \minus{} y} \right)^2 }}{{2\left( {x \plus{} y} \right)\left( {x \plus{} z} \right)}} \plus{} \frac{{x\left( {x \minus{} y} \right)\left( {y \minus{} z} \right)z}}{{\left( {x \plus{} y} \right)\left( {y \plus{} z} \right)\left( {z \plus{} x} \right)}} \plus{} \frac{{z\left( {y \minus{} z} \right)^2 }}{{2\left( {z \plus{} x} \right)\left( {z \plus{} y} \right)}} \ge 0\]
Clearly.


probability
graph theory
combinatorics theorems
combinatorics

[b]I have a few recommendations:[/b]

- Bodendiek, Rainer and Burosch, Gustav: [url=http://www.amazon.de/Streifz%C3%BCge-durch-Kombinatorik-Rainer-Bodendiek/dp/386025393X]Streifzüge durch die Kombinatorik:[/url]  An excellent book in German using many problems from old shorlists and Russia written for the German IMO preparation. [url=http://www.baltic-college.de/kontakt/42-campus-guestrow/18-prof-dr-gustav-burosch-.html]Gustav Burosch[/url] was team leader of the German Democratic Republic from 1970 to 1990 except for 1976, 1971 and 1981 due to political problems. For other political reasons he could not continue his work after the reunification of the two Germanys after 1990. He specifically discusses how many problems can be phrased in the language of graph theory. Burosch was a world expert on combinatorics and graph theory.

- László Lovász, [url=http://www.amazon.com/Combinatorial-Problems-Exercises-Chelsea-Publishing/dp/0821842625/ref=sr_1_1?ie=UTF8&s=books&qid=1247754784&sr=8-1]Combinatorial problems and exercises:[/url] IMO math and more advanced research problem collection

- Ioan Tomescu, R.A. Melter: [url=http://www.amazon.com/Problems-Combinatorics-Interscience-Discrete-Mathematics/dp/0471801550/ref=sr_1_1?ie=UTF8&s=books&qid=1247754876&sr=1-1]Problems in Combinatorics and Graph Theory, 1985:[/url] Romanian IMO Training from the 80s and 90s

- Diestel, Reinhard: [url=http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/]Graph Theory:[/url] Good reference book and can can be downloaded from the linked URL.

- Alon, Noga and Spencer, Joel: [url=http://www.amazon.com/Probabilistic-Wiley-Interscience-Discrete-Mathematics-Optimization/dp/0470170204/ref=sr_1_1?ie=UTF8&s=books&qid=1247754925&sr=1-1]The Probabilistic Method:[/url] This book uses graph theory and also teaches a useful technique which can be seen from the book title. A few sections can be downloaded from [url=http://cs.nyu.edu/cs/faculty/spencer/nogabook/nogabook.html]here.[/url]

- Engel, Konrad: [url=http://opt.math.uni-rostock.de/engel/sperner.html]Sperner Theory:[/url] This book is also quite specialized but teaches you thinking in combinatorics and graph theory

[quote="orl"]

- Diestel, Reinhard: [url=http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/]Graph Theory:[/url] Good reference book and can can be downloaded from the linked URL.
[/quote]

Thank you, and can you tell me how good is the quoted book because I already have it?

Well, I think it is an excellent book. But it is a book written for university students and not necessarily targeted at olympiad preparing students only.

What about

Douglas B. West's

Introduction to Graph Theory

http://www.amazon.com/Introduction-Graph-Theory-Douglas-West/dp/0130144002

But it is too expensive in English. Nevertheless, we can download by searching in Google Book.

Also try Andreescu's 102 Combinatorial Problems and A Path to Combinatorics for Undergraduates.. Sure, they are really good books for your training.

There is a new book on graph theory, specifically aimed towards mathematical olympiads. I don't know how much theory is dicussed in this book, but I browsed the first few pages. The problems discussed are very insightful. I haven't read it, so I can't recommend that you buy it, but it is certainly worth a look.

Graph Theory by Xiong Bin and Zheng Zhongyi, vol.3 of the Mathematical Olympiad Series, published by World Scientific. [url=http://www.amazon.com/Graph-Theory-Mathematical-Olympiad-Xiong/dp/9814271128]Here[/url] is a link.

[quote="fallinlovewithmaths"]graph theory is the most important topic in combinatorics. [/quote]
Surely not!
Anyway, Douglas West's book is good for theory and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=45&t=414427]here[/url] are a few problems  :)

[url=http://www.amazon.com/Introduction-Graph-Theory-Douglas-West/dp/0130144002]Introduction to Graph Theory by Douglas B. West[/url] or [url=http://www.amazon.com/Graph-Theory-Applications-Adrian-Bondy/dp/0444194517]Graph Theory With Applications by John Bondy and U.S.R Murty[/url] are 2 books that I have and am studying for Iran National Olympiad in Informatics.

[url]http://www.amazon.com/Walk-Through-Combinatorics-Miklos-Bona/dp/9810249012[/url]

This book seems to have some chapters in graph theory.
Are they suitable for Olympiad problems?

I am searching about graph theory too. You have given really important suggestions about some graph theory books. I just want to know the applications of graph theory. Can you make me get some information about it?

[quote="aestik"][quote="fallinlovewithmaths"]Hello everybody. I'm preparing for mathematical olympiads. I want to learn graph theory. Can you suggest me good graph theory books? As far as I know , graph theory is the most important topic in combinatorics. I found some but i decided to ask you for help. thx.[/quote]
Introduction to Graph Theory By Gary Chartrand
1.　Introduction .
1.1.　Graphs and Graph Models 1
1.2.　Connected Graphs 9
1.3.　Common Classes of Graphs 19
1.4.　Multigraphs and Digraphs 26
2.　Degrees
2.1.　The Degree of a Vertex 31
2.2.　Regular Graphs 38
2.3.　Degree Sequences 43
2.4.　Excursion:Graphs and Matrices 48
2.5.　Exploration:Irregular Graphs 50
3.　Isomorphic Graphs
3.1.　The Definiition of Isomorphism 55
3.2.　Isomorphism as a Relation 63
3.3.　Excursion:Graphs and Groups 66
3.4.　Excursion:Reconstruction and Solvability 76
4.　Trees
4.1.　Bridges 85
.
..
..
.[/quote]

Wow! This looks like an excellent resource! However, AoPS Intermediate Counting and Probability only uses one chapter to introduce graph theory to the reader. Is there anything that I should do before I start reading this book or should I go dive into it?


[b][size=100][color=DarkBlue]ΠΡΟΤΑΣΗ. - Δίδεται ισοσκελές τραπέζιο $ ABCD$ με $ AB\parallel CD$ και έστω $ P\equiv AC\cap BD$ και $ Q\equiv AD\cap BC.$ Με διαμέτρους τις βάσεις του $ AB,$ $ CD,$ γράφουμε τους κύκλους $ (O_{1}),$ $ (O_{2})$ αντιστοίχως, οι οποίοι τέμνονται στα σημεία έστω $ E,$ $ F,$ $ ($ $ E,$ $ B,$ $ C,$ προς το αυτό μέρος της $ O_{1}O_{2}$ $ ).$ Αποδείξτε ότι $ \angle KEM \equal{} \angle NEQ$ και $ PE\perp EQ,$ όπου $ K\equiv O_{1}O_{2}\cap EF$ και $ M\equiv O_{1}O_{2}\cap (O_{1})$ και $ N\equiv O_{1}O_{2}\cap (O_{2})$ $ ,($ $ M,$ μεταξύ των $ O_{1},$ $ Q$ και $ N,$ μεταξύ των $ O_{2},$ $ Q$ $ ).$[/color][/size][/b]

Κώστας Βήττας.

[b][size=100][color=DarkBlue]PROBLEM. - An isosceles trapezium $ ABCD$ is given, with $ AB\parallel CD$ and let be the points $ P\equiv AC\cap BD$ and $ Q\equiv AD\cap BC.$ We draw the circles $ (O_{1}),$ $ (O_{2}),$ with diameters its bases $ AB,$ $ CD$ respectively, which intersect each other at points $ E,$ $ F,$ $ ($ $ E,$ $ B,$ $ C,$ in to the same place with respect to the line segment $ O_{1}O_{2}$ $ ).$ Prove that $ \angle KEM \equal{} \angle NEQ$ and $ PE\perp EQ,$ where $ K\equiv O_{1}O_{2}\cap EF$ and $ M\equiv O_{1}O_{2}\cap (O_{1})$ and $ N\equiv O_{1}O_{2}\cap (O_{2}),$ $ ($ $ M,$ between $ O_{1},$ $ Q$ and $ N,$ between $ O_{2},$ $ Q$ $ ).$[/color][/size][/b]

Kostas Vittas.

Το πρώτο ερώτημα φαίνεται δύσκολο (βασικά είναι αρκετά μπερδεμένο το σχήμα) αλλά το δεύτερο νομίζω ότι ισχύει χωρίς καν το $ ABCD$ να είναι τραπέζιο!
Σύμφωνα με το θεώρημα Gauss-Bodenmiller σε ένα πλήρες τετράπλευρο οι κύκλοι με διαμέτρους τις διαγωνίους του έχουν κοινό ριζικό άξονα ο οποίος μάλιστα διέρχεται κι απ' τα ορθόκετρα των τριγώνων του πλήρους τετραπλεύρου! Εδώ φανερά οι διαγώνιοι είναι οι $ AB, CD, PQ$ και ο ριζικός άξονας η $ EF$ άρα ο κύκλος με διάμετρο $ PQ$ διέρχεται απ' τα $ E,F$ άρα το ζητούμενο :)

Για το πρώτο ερώτημα αρκεί να είναι:
$ ENM \equal{}$ ισοσκελές γιατί τότε $ EF // AB$ άρα $ EF$ κάθετη στην $ MN$ και άρα $ EK$ = ύψος $ \Leftrightarrow EK$ = διχοτόμος

πράγματι:
[hide]αν $ O, P$ τα μέσα των $ AB, CD$ είναι $ K \equal{}$ μέσο $ OP$ άρα $ EOP \equal{}$ ισοσκελές,
αφού είναι ακόμα $ EK \equal{}$ κάθετη στην $ OP$

άρα $ \widehat{KOE} \equal{} \widehat{KPE} \Leftrightarrow$
(επειδή τα τρίγωνα $ EPN, EOM$ είναι ισοσκελή αφού έχουν δύο πλευρές τους ίσες ως ακτίνες των κύκλων $ (O2), (O1)$ αντίστοιχα)
$ \widehat{OME} \equal{} \widehat{PNE}$

άρα $ ENM \equal{}$ ισοσκελές :D
[/hide]

Έγραψα τις σκέψεις μου πολύ συμπηκνωμένες γιατί είναι σπάσιμο να τις γράφω σωστά. :blush: Και παίρνοντας υπόψιν ότι δεν έκανα καν σχήμα ίσως να έχω λάθη. Πείτε αν βρείτε κάποιο.

Δυστυχώς, ξαναβλέποντας την άσκηση, συμπεραίνω ότι το πρώτο υποερώτημα δεν ισχύει γιατί
σε όλες τις σχέσεις που χρησιμοποίησα ισχύει διπλή συνεπαγωγή και
για να είναι $ K \equal{}$ μέσο του $ OP$ πρέπει $ EO \equal{} EP$ αφού $ EK \equal{}$ κάθετο στο $ OP$
δηλαδή πρέπει οι ακτίνες των κύκλων να είναι ίσες, δηλαδή οι βάσεις να είναι ίσες, δηλαδή να έχουμε ορθογώνιο παραλληλόγραμμο, οπότε δεν τέμνονται οι πλευρές
ακόμα, πρέπει να δοθεί η διευκρίνηση ότι το ύψος πρέπει να είναι μικρότερο του ημιαθροίσματος των βάσεων για να τέμνονται οι κύκλοι,


inequalities
geometry
search
inequalities unsolved

Prove or disprove: $a, b, c > 0 \rightarrow a^4 + b^4 + c^4 + d^4 + 8abcd \geq a^2bc + ab^2c + abc^2 + a^2bd + ab^2d + abd^2 + a^2cd + ac^2d + acd^2 + b^2cd + bc^2d + bcd^2$

Are there any other inequalities with a weak $abcd$ term on the wrong side (which is no schurable)? Edit: I added this line just as harazi replied.

For positive real numbers, take a look at a more general problem in the open questions area, caller "Schur for n numbers", posted by Vasc. I still remember the nice solution I gave to it. Your inequality is the case n=4.

To answer your second question, the strongest inequality I know with the product in the suffering part of the inequality is Suranyi's inequality.

Can you write that inequality for me ? Ok.

---------------
Reyes.

Just use search button and not only you'll find it, but alsoi its proof.

I have read it .It's hard and beautiful.Thanks!


As the subject says, the problem below may seem ugly at first, but there is an excellent solution to it.

Find the largest integer N such that $N < (\sqrt{33+ \sqrt{128}} + \sqrt{2} - 8)^{-1}$

FWIW - A "fun" way is to notice, as you have shown $5\sqrt2 - 7$ is less than 1(being reciprocal of $5\sqrt 2 +7 )$)  and we  see
$(5\sqrt 2 +7 ) - (5 \sqrt 2 - 7 )$ is  $14$  (so the requred answer is 14 without sort of  pluging in  the value of $\sqrt 2$  :)


function
real analysis
real analysis unsolved

Prove or supply a counterexample: If the function $f$ from $\mathbb{R}$ to $\mathbb{R}$ has both a left limit and a right limit at each point of $\mathbb{R}$, then the set of discontinuities of $f$ is,at most,countable.

It's a very well-known problem, and I think it's been discussed before (I remember Myth posting  very similar one, used in a math contest from Russia this year).

The set $A_n$ consisting of points $x$ for which $|f_+(x)-f_-(x)|\ge\frac 1n$ is discrete and thus countable ($f_+,f_-$ are the right and left limits, respectively), and since the set of discontinuities is the union of the $A_n$'s, that's that.

[quote="grobber"]The set $A_n$ consisting of points $x$ for which $|f_+(x)-f_-(x)|\ge\frac 1n$ is discrete and thus countable ($f_+,f_-$ are the right and left limits, respectively), and since the set of discontinuities is the union of the $A_n$'s, that's that.[/quote]
It doesn't include all the cases, not $|f_+(x)-f_-(x)|\ge\frac 1n$. Use $\max\{|f_+(x)-f(x)|,|f_-(x)-f(x)|\}\ge\frac 1n$ instead.

Right. Sorry, I was thinking about monotonic functions :).

For monotonic functions, we don't need to do it in that way...


What is the sum of all the perfect cubes under 100?

How do you solve this? Please help. :?:

This seems too easy so maybe I'm wrong: $ (1)^{3}\plus{}2^{3}\plus{}3^{3}\plus{}4^{3}\equal{}1\plus{}8\plus{}27\plus{}64\equal{}100$.

Well, we know that

$ 1^3\plus{}2^3\plus{}3^3\plus{}...\plus{}n^3\equal{}(1\plus{}2\plus{}3\plus{}...\plus{}n)^2$. So, since the biggest

perfect cube less then $ 100$ is $ 64\equal{}4^3$, $ n\equal{}4$.

So, $ (1\plus{}2\plus{}3\plus{}4)^2\equal{}10^2\equal{}100$.


Jean is paving a 2 by 10 rectangle with 1 by 2 paving stones. In how many different ways can she do this?

super008 wrote:[hide]Let f(n) be the number of ways to pave a 2xn rectangle. By induction, we found that f(n)=f(n-1)+f(n-2), where f(1)=1, f(2)=2. So f(10)=89.[/hide]

super, to find your pattern did you calculate some of the numbers and notice the pattern or do you have a reason why it would follow that recurrence? If you have a reason, you should probably explain it.

I have done this problem before. I first saw the pattern then did the induction.

[quote="zabelman"]lol. Did you put this up because you like the question or the pun?[/quote]

Just the pun, sorry.

Though Super008 should show his induction.

What I meant is that you should explain why the pattern is true so other people can follow your solution.

[hide]We observed that f(1)=1, f(2)=2, f(3)=3, f(4)=5, f(5)=8. And they look like consecutive Fibonacci numbers.

Claim: f(n) = F_(n-1), where F_n is the nth Fibonacci number.

Proof: We will prove the claim by induction on n.

We have verify that it is true for n=1.

Assume that f(n)=F_(n-1) is true for n=k, we need to prove that it is also true for n=k+1.

Now in order to pave a 2x(k+1) rectangle, we could either put a 2x1 rectangle down like "|", then pave the rest, which is a  2xk rectangle; or we could put down two rectangles across like "=", then pave the rest, which is a 2x(k-1) rectangle. Since thses are the only starting position to pave the rectangle, we have f(k+1) = f(k) + f(k-1).

By assumption, f(k) = F_(k-1) and f(k-1) = F_(k-2). Substitute them in, we have f(k+1) = F_(k-1) + F_(k-2) = F_k. Thus we have proved that the assumption is true for n=k+1. By induction, we know that f(n)=F_(n-1) is true for positive integer n.[/hide]

That should actually be F_(n + 1), but good job otherwise.


suppose a,b,c are side length of a triangle, proof
(1)$ a^3\plus{}b^3\plus{}c^3\plus{}3abc\minus{}2b^2a\minus{}2c^2b\minus{}2a^2c\ge0$
(2)$ 3a^2b\plus{}3b^2c\plus{}3c^2a\minus{}3abc\minus{}2b^2a\minus{}2c^2b\minus{}2a^2c\ge0$ :D

I've also been wondering whether I was really the first one to find them. None of them is hard anyway.
http://www.mathlinks.ro/Forum/viewtopic.php?t=35295

  darij

[quote="greentreeroad"]Easy :?: .. not at all to me

why $ x^{3} \plus{} 2xy^{2} \plus{} y^{3} \plus{} 2yz^{2} \plus{} z^{3} \plus{} 2zx^{2}\geq 3(x^{2}y \plus{} y^{2}z \plus{} z^{2}x)$? (not necessary sides of triangle any more)[/quote]
 its easy by EMV theorem :)

[quote="greentreeroad"]Easy :?: .. not at all to me

why $ x^{3} \plus{} 2xy^{2} \plus{} y^{3} \plus{} 2yz^{2} \plus{} z^{3} \plus{} 2zx^{2}\geq 3(x^{2}y \plus{} y^{2}z \plus{} z^{2}x)$? (not necessary sides of triangle any more)[/quote]
If $ x\equal{}\min\{x,y,z\},$ $ y\equal{}x\plus{}u$ and $ z\equal{}x\plus{}v$ then $ \sum_{cyc}(x^3\minus{}3x^2y\plus{}2xy^2)\geq0\Leftrightarrow$
$ \Leftrightarrow2(u^2\minus{}uv\plus{}v^2)x\plus{}u^3\minus{}3u^2v\plus{}2uv^2\plus{}v^3\geq0,$ which obviously true.

[quote="M.A"][quote="greentreeroad"][quote="M.A"]
 its easy by EMV theorem :)[/quote]

What is EMV...  :blush:[/quote]
 its entirely mixing variables method :wink:[/quote]
what is entirly mixing variables please post that theory

[quote="Javkhaa"][/quote][quote="M.A"][quote="greentreeroad"][quote="M.A"]
 its easy by EMV theorem :)[/quote]

What is EMV...  :blush:[/quote]
 its entirely mixing variables method :wink:[/quote][quote="Javkhaa"]
what is entirly mixing variables please post that theory[/quote]
Download file: http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf

[quote="greentreeroad"]suppose a,b,c are side length of a triangle, proof
(1)$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \minus{} 2b^2a \minus{} 2c^2b \minus{} 2a^2c\ge0$
[/quote]
Suppose that $ c\equal{}min\{a;b;c\}$
This ineq can rewrite:
$ (a\plus{}b\minus{}c)(a\minus{}b)^2\plus{}(a\plus{}c\minus{}b)(a\minus{}c)(b\minus{}c) \geq 0$
Note that $ a,b,c$ are side length of a triangle,so we have $ a\plus{}b\minus{}c>0;a\plus{}c\minus{}b>0$
Done!

[quote="greentreeroad"]suppose a,b,c are side length of a triangle, proof
(2)$ 3a^2b \plus{} 3b^2c \plus{} 3c^2a \minus{} 3abc \minus{} 2b^2a \minus{} 2c^2b \minus{} 2a^2c\ge0$ :D[/quote]
Suppose that $ c\equal{}min\{a;b;c\}$
We can rewrite this ineq:
$ c(a\minus{}b)^2\plus{}(2a\minus{}b)(a\minus{}c)(b\minus{}c) \geq 0$
We have $ 2a\minus{}b \geq a\plus{}c\minus{}b >0$
Done!


Ramsey Theory
combinatorics unsolved
combinatorics

Find the minimum number of point in plane in which three arbitrary point aren't collinear satisfy for all ways of painting all segment over two point always exit a triangle in which three sides are painted by a same colour!

Well...it depends on the number of colours... :wink:
For $ n$ colours, the desired result is the Ramsey number $ R(3,\cdots,3)$.
In particular, for two colours, the desired minimum is $ 6$.

Pierre.


Columbia
function
integration
algebra
polynomial
real analysis
real analysis unsolved

Let $ n$ a positive integer, and $ f: [0,1]\to\mathbb{R}$ a continuos function such that \[ \int_{0}^{1}x^{k}f(x) dx\equal{}1\] for $ k\equal{}0,1,2,\ldots,n\minus{}1$. Prove that \[ \int_{0}^{1}f^{2}(x) dx\geq n^{2}\]

See http://www.mathlinks.ro/Forum/viewtopic.php?&t=146373 or http://www.mathlinks.ro/viewtopic.php?p=443980#443980
(The approaches are similar, but I don't understand why $ \int_{0}^{1}\left( f(x)\minus{}q(x)\right) q(x)\, dx \equal{} 0$ in jmerry's solution.)

It's about orthogonality. Let $ p_{0},p_{1},\dots,p_{n\minus{}1}$ be an orthogonal basis for polynomials of degree $ \le n\minus{}1$; the hypothesis is equivalent to the system $ \langle f,p_{i}\rangle\equal{}c_{i}$ for some fixed constants $ c_{i}$. When we subtract the solutions, we get $ \langle f\minus{}q,p_{i}\rangle\equal{}0$ for each $ i$. Since the $ p_{i}$ span the polynomials of degree $ \le n\minus{}1$, $ f\minus{}q$ is orthogonal to all polynomials of degree $ n\minus{}1$, including $ q$ in particular.


I had breadsticks and Marinara.(breakfast, not lunch.)

wtf is up wit tis

whatev, ill do it

i had poha

guz wat tat is

I had banana bread, 2 bagels, 1 sports drink, 1 water, and two  small pieces of sandwich...lunch time!

Right now I'm having lunch....chocolate and bananas :wink:

a bowl of cinnamon toast crunch with 1% milk.

Rice with butter and a fried egg - I leave for college after that.

[quote="star99"]poha is good!

i had puri aloo kudu and hulwa, because its janmashtmi[/quote]

GARBA

[quote="funcia"]I'm eating waffles!!!!! Ego my Lego! Or wtf.
I always thought that meant I have my I say b/c of greek.[/quote]
Lol! :rotfl:

Sorry, but I find this very funny.

The first thing I do when I wake up is eat.  I've tried to wait until I'm out of my room to eat since I think it bothers my roommate when I rummage through my cabinet while he's still trying to sleep :P (I mean, I try really hard to be quiet :( ).

Anyway, today I had an energy bar, a couple pieces of bread with peanut butter, and a banana.

My two favorite things for breakfast are muffins and breakfast burritos (from Anna's Taqueria)...mmm.

I had a chocolate chip cookie dough poptart and strawberries. Now I get to have a bowl of cereal as an almost-1am snack. :jump:  :rotfl:

I usually have something random: The probability that I'll have oatmeal is definately greater tha 1/2  :(

Ah, I need to update this with what I ate today:

two bagels (with nutella), 2 pieces of french toast, 1 pancake, a handful of fruits+nuts, 2 glasses of smoothies, 1 slice of ham, 1/2 nutri-grain bar

So full....

Two fried eggs, a pickle, and green tea. :D

Two boiled eggs, a piece of toast with peanut butter, a bowl of cereal, three glasses of orange juice, a bagel with cream cheese, and half a blueberry muffin.

Me too, that's why I ate so much yesterday.


Go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209860]here[/url] for the rules

One difference: there will be no quests.

14 people to start. (1 king, 1 queen, 2 known knights, 1 unknown knight, 9 citizens)

Signups:

1. meewhee009
2. Scamper
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.

1. meewhee009
2. Scamper
3. jjfun1
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.

Sounds fun.
[b]Join.[/b]
W00t, I got my lucky number. :)

Then why aren't you jjfun3?  This has been a public shaming of jjfun1.  Thank you. [yet another revival]

1. meewhee009
2. Scamper
3. jjfun1
4. ernie
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.

[quote="ernie"]Then why aren't you jjfun3?  This has been a public shaming of jjfun1.[/quote]
Maybe in ernie-land, three and fun rhyme. But not in jjfun1-land. :P

1. meewhee009
2. Scamper
3. jjfun1
4. ernie
5. 15!!!
6. westiepaw
7. EggyLv.999
8. spaceguy524
9. FlyAgaric
10.
11.
12.
13.
14.

Sure, I'll play.

1. meewhee009
2. Scamper
3. jjfun1
4. ernie
5. 15!!!
6. westiepaw
7. EggyLv.999
8. spaceguy524
9. FlyAgaric
10. cf249
11.
12.
13.
14.

join!
 :nhl:
I will beat everyone I want to beat using this hockey stick.
Also, this reminds me of "The once and future king"!
*Read, Read, Journal, Read...*

Mwahahaha :).

meewhee009-20 pts
Scamper-21 pts
jjfun1-9 pts
westiepaw-5 pts
EggyLv.999-10 pts
spaceguy524-10 pts
FlyAgaric-10 pts
Bramblestar-7 pts
mathling235-18 pts
[b]Scamper for 1[/b]

meewhee009-19 pts
Scamper-21 pts
jjfun1-9 pts
westiepaw-5 pts
EggyLv.999-9 pts
spaceguy524-10 pts
FlyAgaric-10 pts
Bramblestar-7 pts
mathling235-18 pts

Day 8 ends.

Day 9 starts. Everyone gets their salaries:

meewhee009-22 pts
Scamper-23 pts
jjfun1-10 pts
westiepaw-6 pts
EggyLv.999-10 pts
spaceguy524-11 pts
FlyAgaric-11 pts
Bramblestar-8 pts
mathling235-19 pts

Random Event: King Execution Day

The king may pay 15 points to instantly kill any one person. Under the terms of the new law, EggyLv.999 can be executed for free.

If we attack the king today, can the king execute us for free or does she have to wait for the next execution day? Or is the law no longer being imposed?

EggyLv.999 should attack the king.

meewhee009-22 pts
Scamper-23 pts
jjfun1-10 pts
westiepaw-6 pts
[b]EggyLv.999 haz been killedzzz![/b]
spaceguy524-11 pts
FlyAgaric-11 pts
Bramblestar-8 pts
mathling235-19 pts
 :P
I have a new aim...
hehehehe...

The king now has to wait until day 12 until another execution can be done.

meewhee009-22 pts
Scamper-23 pts
jjfun1-10 pts
westiepaw-6 pts
spaceguy524-11 pts
FlyAgaric-11 pts
Bramblestar-8 pts
mathling235-19 pts

Day 9 ends.

Day 10 starts. Everyone gets their salaries

meewhee009-25 pts
Scamper-25 pts
jjfun1-11 pts
westiepaw-7 pts
spaceguy524-12 pts
FlyAgaric-12 pts
Bramblestar-9 pts
mathling235-20 pts

Random Event: Small Money Flow

So money is becoming scarce in the country, and very little is getting to the king.

[b]Starting tomorrow, the king only gets 2 points a day in salaries.[/b]

Does the law passed on Day 9 still apply?

It does, but meewhee can only execute one of you at a time, and there are 7 of you left...


How do I go about doing this?

I know how to prove the distance formula by using the pythagorean theorem, however, I  don't know how to prove the mid-point formula by using the distance formula.

I can prove the mid-point formula by using a graph and relating it to the distance formula, but how do I do this by solely using the distance formula?

Please enlighten me.

Remember:

D = sqrt((Xsub2 - Xsub1)squared + (Ysub2 - Ysub1)squared)

and

M = ((Xsub2 + Xsub1)/2 , (Ysub2 + Ysub1)/2)

Thanks,
dfunk

So we have the points $A \ (x_1,y_1)$ and $B \ (x_2,y_2)$ and want to prove that $M \ \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ is the midpoint.  It is sufficient to show that the distance $AM=BM=\frac 12 AB$.  So we have:

\begin{eqnarray*}
AB &=& \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\
AM &=& \sqrt{\left(x_1-\frac{x_1+x_2}{2}\right)^2+\left(y_1-\frac{y_1+y_2}{2}\right)^2}\\
BM &=& \sqrt{\left(x_2-\frac{x_1+x_2}{2}\right)^2+\left(y_2-\frac{y_1+y_2}{2}\right)^2}
\end{eqnarray*}

So $AM=\sqrt{\left(\frac{x_1-x_2}{2}\right)^2+\left(\frac{y_1-y_2}{2}\right)^2}$.  Doing the algebra we easiliy see this is equivalent to $BM$.  Now we only need to prove that this is $\frac 12 AB$ as follows:

\begin{eqnarray*}
AM &=& \sqrt{\left(\frac{x_1-x_2}{2}\right)^2+\left(\frac{y_1-y_2}{2}\right)^2}\\
      &=& \sqrt{\frac 14\left(\left(\frac{x_1-x_2}{2}\right)^2+\left(\frac{y_1-y_2}{2}\right)^2\right)}\\
      &=&\frac 12 \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\
      &=& \frac 12 AB
\end{eqnarray*}


geometry
calculus
topology
analytic geometry
graphing lines
slope
geometric transformation

What sorts of topics and questions belong on this forum?

To help you decide what the use of this forum is, I have compiled some example questions that might be asked here.  These are not strict guidelines by any means.  Generally, you can post anything here that is math related but does not fall into one of the three other math forums.  If you are unsure, please use the other posts on this forum and the example questions below as a guide.

(1)  Resources for problems and problem solving.
�	Where can I find a book full of past AMC 12-level geometry problems?
�	What is a good textbook on Number Theory?
�	Would you recommend Mathematical Reasoning: Writing and Proof by Ted Sundstrom for beginners learning how to write proofs?

(2) Problems from topics that are not generally tested on high school mathematics competitions, like a difficult problem from a textbook.
�	I don't understand this calculus question; can you give me a hint?
�	I would like to suggest this math-oriented Physics/Chemistry problem as a challenge.
�	Can someone help me with this beginning topology question?

(3) Questions about theory (not a problem).
�	I do not understand mathematical induction.  Can you explain the difference between inductive and deductive reasoning?
�	What is the proof for the Binomial Theorem?
�	How many *Fundamental Theorem of XYZ* are there and what are they?

(4) Discussion of current events in mathematics.

(5) Methods of teaching mathematics.

(6) Studying mathematics.  These should be questions about studying mathematics that do not belong in the �College� forum.
�	What is the calculus sequence you learn in college?
�	What are the other types of mathematics you can learn in college that are not part of the calculus sequence, such as combinatorics, statistics, number theory, and discrete math?

(7) Discussion of other topics in mathematics which are similar to threads on the �Round Table� but are mathematics-oriented.
�	Should I take AP Calculus AB, AP Calculus BC, or AP Statistics?
�	Why is the letter m used to represent the slope of a line?

Please remember that these are not strict guidelines.

Thanks, and please use the forum responsibly!

That's a great list of topics, and well reflects how this forum has been used over the last school year. I would only add that if the topic has to do with math, it really belongs [b]here[/b], on Other Problem Solving Topics, and [b]not[/b] on the Round Table, which at its best is more for issue-oriented political such as several of the highest-volume threads posted there recently.

I'm glad to see that you've come on board as a co-moderator here.

Yes, I think that that is a wonderful guide.  Welcome to the elite corps of moderators! ;)

Thanks!  I feel elite already  :D !  I will edit it w/Tokenadult's suggestion.


ceiling function
logarithms
number theory
greatest common divisor
relatively prime
number theory proposed

Given $a_0 > 1$, the sequence $a_0, a_1, a_2, ...$ is such that for all $k > 0$, $a_k$ is the smallest integer greater than $a_{k-1}$ which is relatively prime to all the earlier terms in the sequence.
Find all $a_0$ for which all terms of the sequence are primes or prime powers.

No odd numbers larger than 3 can be included unless the number following is a power of two, and the number following that is a power of three. 2 and 3 work in this set, as 4 is a power of a prime, 2, and all following numbers will be prime (as they are relatively prime to 3 and 4, and every number before that relatively prime to 3 and 4 etc.). 7 also works, because directly above 7 is a power of 2, and above that is a power of 3, and a factor of 5 will be included at 25, also a power of a prime.

I don't think any other numbers exist, as there it must go in the order (prime, power of 2, power of 3, power of 5, etc) all the way up to a power of the prime below the starting prime, which can be shown (I think) cannot happen.

A hint: show that if n is sufficiently large then $a_n$ is prime.

Does anyone have a full solution to the problem?

All solutins are $a_{0}=2,3,4,7,8.$ It is easy to check that if $a_{0}<10$ only these solutions. For prove, that it had not another solutions sufficiently consider equations $2^{k}\pm1=3^{l}$, $2^{k}\pm 1=5^{l}$ (k>3) and $3^{k}\pm 2=5^{l}$(k>1). This equations had not solutions. It give if $a_{0}>9$ had not another solutions.

Suppose $a_1,a_2,...,$ is a sequence such that the $a_i$ are all prime powers. (I'm starting at $a_1$ because starting with $a_0$ is annoying)

[b]Lemma 1:[/b] Every prime divides a term in the sequence.
[b]Proof:[/b] Take any prime $p$. Suppose $k$ is the least integer such that $a_1 \le p^k$. Now let $z$ be the least such integer such that $a_z \ge p^k$. If $a_z \neq p^k$, then we have $p$ divides one of $a_1,a_2,...,a_{z-1}$. But then we're done.

[b]Corollary:[/b] If $p$ is prime, then $p^{\lceil \log_p{a_1} \rceil}$ is one of the $a_i$.
It immediately follows for sufficiently large $n$, we have the $a_n$ are just a bunch of consecutive primes. Note that by the corollary we have that if $a_n > a_0^2$ then $a_n$ is prime.

I claim we can pick two primes $p,q$ such that $pq \in [a_1,a_1^2]$ but $p^{\lceil \log_p{a_1} \rceil},q^{\lceil \log_q{a_1} \rceil} > pq$

By Bertrand's Postulate we can find a prime $p$ such that $\frac{\sqrt{a_1}}{2} - 1 < p < \sqrt{a_1}$ when $a_1 \ge 16$. Let $\frac{\sqrt{a_1}}{p} = z$. Then we can also find a prime $q$ such that $z\sqrt{a_1} < q < 2z\sqrt{a_1}$.
Now, clearly $pq  > a_1$. However, $p^{\lceil \log_p{a_1} \rceil} \ge p^3 > pq$. We also have $q^{\lceil \log_q{a_1} \rceil} = q^2 > pq$. Thus it easily follows $pq$ must be one of the $a_i$, contradiction and thus we have $a_1 < 16$. Checking which of these work is just stupid casework.

EDIT : So I just realized $16$ is not big enough as then in strange cases we can have $q \ge p^2$. This method actually yields something like $a_1 \le 256$ which is pretty hard to check up to... though working with [b]proglote[/b] a bit on this we deduced $a_0$ must be of the form $2^{2^n}$ except in some exceptional cases so I guess its not too bad.

[quote=proglote]Here's another solution with some casework:

If $a_0$ is odd, the next terms in the sequence are clearly $a_0 + 1$ and $a_0 + 2$. Since $a_0 +1$ is even, it must be a power of two, and either $a_0$ or $a_0 + 2$ is a power of three. So it remains to solve $2^{a} \pm 1 = 3^{b}$ which is easy, giving $a = 2,3 \implies a_0 \in \{3,7\}.$
If $a_0$ is even, then the next terms in the sequence are clearly $a_0 +1, a_0 +3, a_0 +5$ (though we can't say anything further since $a_0 + 7$ and $a_0 + 1$ may both be multiples of 3). Also note that $a_0 = 2^a$ for some positive integer $a$.
If $a$ is 1 mod 3, then $7|2^{a} +5$, i.e. $a_0 + 5$ is a power of $7.$ So $2^a + 5= 7^b$, which implies $a=1$, so $a_0 = 2.$
If $a$ is 2 mod 3, then similarly it remains to solve $2^a + 3 = 7^b$, and the only solution is clearly $a =  2$, so $a_0 = 4.$
If $3|a$, then $a_0 + 1= 2^{3a'} + 1$ for some positive integer $a'.$ Note that $x^3 + 1 = (x+1)(x^2 - x + 1)$ and $(x+1, x^2 - x + 1) = (x+1, 3).$ Since $2^{3a'} + 1$ is a prime power, it cannot be factored into two coprime integers greater than 1, so the gcd of the two factors is 3 and we get $2^a + 1 = 3^b$ for some $b.$ So the only solution is $a=3$, i.e. $a_0 = 8.$
Therefore, $a_0 \in \{2,3,4,7,8\}.$[/quote]

macetei


Could anybody prove the following theorem

[color=green][b]Theorem.[/b] Let P and Q be two convex polygons such that P is on the non-externally region of Q, then prove that perimeter(P)=< perimeter(Q).[/color]

Using this lema (I proved this lema)

[color=green][b]Lema. [/b]Let Q be a convex polygons and let P be a triangle such that P is on the non-externally region of Q, then prove that perimeter(P)=< perimeter(Q).[/color]

Thanks a lot..

José Carlos

In [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=197437#p197437]http://www.mathlinks.ro/Forum/viewtopic.php?t=31419 post #3[/url], this was proven for the case when the external figure Q is a circular disc rather than a polygon, but for a polygon the proof is exactly the same.

  Darij


limit
trigonometry
inequalities
real analysis
ratio
quadratics
continued fraction

Is it true that
\begin{eqnarray*} \lim_{n \rightarrow \infty} \frac{1}{n \sin n} & = & 0 \end{eqnarray*}
What about
\begin{eqnarray*} \underset{n \longrightarrow \infty}{\lim \sup} \frac{1}{n \sin n} & = & ? \end{eqnarray*}

I should think not. We can find some $m$ (infinitely many in fact) so that $|\pi - \frac{n}{m}| < \frac{1}{m^2}$. Then $\frac{1}{n\sin(n)}$ is about $\frac{1}{n(\frac{1}{m^2})} \approx \frac{m}{\pi}$, which is arbitrarily large.

(for rigor you should probably take two terms of the taylor series for sine so that the "about" is an inequality in the direction we want, and then bound from each side by a constant times $m$)

I bet it does converge in some weaker sense, like Cesaro.

The first fact is a corollary of Hurwitz theorem, as far as I know.
Otherwise, I don't think you really need to mess up with Taylor series. Some basic inequalities should suffice.

For any $c$, $\limsup \left|\frac1{n\sin cn}\ge\frac1{\pi}\right|$. This limsup is $\infty$ when the continued fraction for $\frac c{\pi}$ has arbitrarily large quotients or $\frac c{\pi}$ is rational.

For any integer $m$, $|\sin cn|=|\sin(cn-m\pi)|< |cn-m\pi|=n\pi\left|\frac{c}{\pi}-\frac mn\right|$. If $\frac pq$ is a convergent (in lowest terms) of the continued fraction for $\frac c{\pi}$, $\left| \frac{c}{\pi}-\frac pq\right|<\frac1{q^2}$. There are infinitely many convergents $\frac pq$; choosing these as our $m,n$, we have $n\pi\left|\frac{c}{\pi}-\frac mn\right|<\frac{\pi}n$. Taking the reciprocal and dividing by $n$, we have $|\frac1{n\sin cn}|> \frac1{\pi}$.

Xevarion's argument misses one of these factors of $n$; whether the $\limsup$ is finite in any particular case is an interesting question. It is always nonzero.

If we drop the absolute value signs, we can show that $\limsup \frac1{n\sin cn}\ge\frac1{4\pi}$ and $\liminf \frac1{n\sin cn}\le-\frac1{4\pi}$. The convergents $\frac pq$ alternate between being too large and too small; if we then let $n=2q$, $\sin 2cn$ will be positive when $\frac pq$ is low and negative when $\frac pq$ is high. In both cases, we lose a factor of 4 by doubling.

:blush:
Thanks for correcting my error jmerry. Can you please give some more detail on the case where the continued fraction for $\frac{c}{\pi}$ has arbitrarily large quotients? I'm afraid I don't see immediately why it's so ...  :oops:

If the quotient after the convergent $\frac pq$ is $r$, we have $\left|\frac c{\pi}-\frac pq\right|\le\frac1{rq^2}$.

I think so, anyway; it's been a while since I worked out the details. The two quantities are comparable in any case.
As a corollary, large quotients indicate unusually good rational approximations. As a famous example, here's the first few terms of the continued fraction for $\pi$:
3;7,15,1,292
If we cut off just before the 292, we get $\frac{355}{113}$ as a convergent.

Oh, I get it, that's really cool! Thanks  :D

Are there general methods for knowing whether the continued fraction of some irrational has arbitrarily large quotients (other than actually knowing the general term, like in case of the golden ratio)?

[quote="Xevarion"]... like in case of the golden ratio)?[/quote]
Theorem: the quotients of the continued fraction of a number are periodic if and only if the number is a quadratic irrational - that is an irrational number that can be written in the form $\frac{a+\sqrt{b}}c$ or $\frac{a-\sqrt{b}}c$ for integers $a,b,c.$ (If the number is rational, then the continued fraction representation terminates.) Of course, the golden ratio is a quadratic irrational.

So quadratic irrationals cannot have arbitrarily large quotients, because the quotients are periodic. But we can't say anything about the converse.


If $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}3$, then:

$ (a\plus{}\frac1{b^2\plus{}c^2})(b\plus{}\frac1{a^2\plus{}c^2})(c\plus{}\frac1{b^2\plus{}a^2})\le\frac{27}8$

[quote="akai"]If $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$, then:

$ (a \plus{} \frac1{b^2 \plus{} c^2})(b \plus{} \frac1{a^2 \plus{} c^2})(c \plus{} \frac1{b^2 \plus{} a^2})\le\frac {27}8$[/quote]
Try $ a\equal{}b\equal{}0.001,c\equal{}2.998$.
If you meant that
\[ (a \plus{} \frac1{b^2 \plus{} c^2})(b \plus{} \frac1{a^2 \plus{} c^2})(c \plus{} \frac1{b^2 \plus{} a^2})\ge\frac {27}8\]
Then try $ a\equal{}b\equal{}\frac{1}{2},c\equal{}2$.


if f(z)=z+1/z-1, then find f to the 1991 power (2+i).

[hide]
Try the first two values to find that f^2(2+i) = 2+i, so f^1991(2+i) = f^1(2+i) = (3+i)/(1+i).[/hide]


$ O$ is incircle of a $ \triangle ABC$. An another circle $ O'$ tangent to rays $ AB,AC$.If center of $ O'$ is lie circle $ O$, prove that $ \triangle ABC$ is isosceles.

                                                                          H.İBRAHİM AYANA

i don't see why this needs to be true:
[asy]size(200); pair A,B,C,D,X,Y,Z; C=origin; B=(50,20); A=(30,70); draw(A--B--C--cycle); draw(incircle(A,B,C)); D=incenter(A,B,C); dot(D); X=(13.5,32); Y=(43,37); draw(incircle(A,X,Y)); Z=incenter(A,X,Y); dot(Z); label("A",A,N); label("B",B,E); label("C",C,W); label("O",D,S); label("O'",Z,N);[/asy]


[quote="chess64"]1. (AoPS) Factor 141, 1441, and 14441.

2. (AoPS) Find the base 8, 9, and 16 representations of $47_{10}$.

3. (AoPS) Find the base 10 equivalents of $BEE_{16}$, $DEF_{16}$, and $A1_{16}$.

4. (MATHCOUNTS 1986) Find the value of digit $A$ if the five-digit number 12A3B is divisible by both 4 and 9, and $A\ne B$.

5. (MATHCOUNTS 1988) In how many ways can a debt of 69 dollars be paid exactly using only 5 dollar bills and 2 dollar bills?

6. (MATHCOUNTS 1988) Find the sum of all four digit natural numbers of the form $4AB8$ which are divisible by 2, 3, 4, 6, 8, and 9.

7. (Mu Alpha Theta 1990) If $a\ne 1$ and $\sqrt[a]{10000_{a}}= 10_{a}$, find $a$.

Good luck![/quote]

[hide="1."]
$141 = 47 \cdot 3$
$1441 = 11 \cdot 131$
$14441 = 7 \cdot 2063$
[/hide]

[hide="2."]
$48_{10}= 57_{8}$

$48_{10}= 52_{9}$

$48_{10}= 2F_{16}$

i want my ranking to be fixed...
[/hide]

About time somebody did them...

Both correct. :)

I'm not in middle school but I'll try these anyway.  :)

[hide="#4"]
To be divisible by 4, the last two digits must form a number divisible by 4, so B=2 or B=6.  To be divisible by 9, the sum of the digits must be a multiple of 9.  1+2+3+6=12, so if B=6, A=6, which is not allowed.  1+2+3+2=8, so if B=2, A=1.  This is the only solution, so A=1.
[/hide]

uhh.. I hope you don't have to be from Maryland to answer these but here's #7.
[hide="7"]Changinge everything to base 10, we get $\sqrt[a]{a^{4}}=a\rightarrow a^{4}=a^{a}\rightarrow a=4$[/hide]

well actually you have to be from cockeysville middle school, but since they aren't answering them you can :lol:

Hmm I understand the divisibility stuff but like number 5 how can you use those rules and apply them to real life based problems?

[hide="5"]
We have $5a+2b=69$ for some positive integers $a,b$. Modulo 5, the equation becomes $2b\equiv 4\pmod{5}$ so $b\equiv 2\pmod{5}$. Therefore, the smallest possible value of $b$ is 2, and when $b=2$, $a=13$. Notice that decreasing $a$ by 2 and increasing $b$ by 5 will give us another solution. Therefore, the solutions are (13, 2), (11, 7), (9, 12), (7, 17), (5, 22), (3, 27), (1, 32). So the answer is $\boxed{7}$.
[/hide]


calculus
integration
function
real analysis
calculus computations

How can one prove that
\[ \int f(x) \, \mbox{sgn} \left( f(x) \right)  \, \mbox{d}x \equal{} \mbox{sgn} \left( f(x) \right) \int f(x) \, \mbox{d}x\]
It is trival, but I don't have slightest idea how to prove it in a nice way :(

Is it enough when we consider intervals when f>0 and f<0  :?:

What is sgn? Does it stand for the sign (+ or - symbol) of a function? And if I'm interpreting that correctly, the proof is indeed trivial, write the LHS as a Riemann sum and factor out the sign of the function.

sgn stands for sign function.
Ok, thanks.

If you look carefully the equation doesn't make sense (at least without some conditions). How do you 'factor out' the sign function while keeping the variable the same?


AoPSwiki AMC AIME LaTeX articles USA(J)MO USAMO

If you haven't checked out the [url=http://www.artofproblemsolving.com/Wiki/]AoPSWiki[/url] yet, you really ought to!

One of the ongoing projects in the AoPSWiki is the development of the [url=http://www.artofproblemsolving.com/Wiki/index.php/AMC_Problems_and_Solutions]AMC Problems/Solutions[/url].  This is not a small task, so we will need a lot of help.  Anyone can help.  You just use your AoPS account for the AoPSWiki.

I've already started with the [url=http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I]2006 AIME I[/url].  If you look around, you should get a feel for it.

I haven't put the solutions in for the 2006 AIME I.  Also, if you want to put links to message board posts with the problems, that would be a good idea.  Just make sure to put the links in the ''See also'' section of the article.

The biggest advantage of the wiki over the contest directory is that anyone can edit it at any time.

Thanks.

Another note for entering problems into the wiki:

Look at the [url=http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems]2006 AIME 1 Problems[/url].  It has all the problems listed as their own section. After the problem statement it has a solution link which brings you to a page dedicated solely for that problem.  When making these solution pages you MUST name them like the following: 2006 AIME I Problems/Problem 1.   What this does is it creates a subarticle for that problem (instead of making a whole new article).

Thanks.

An additional note:

Look at the way I entered the [url=http://www.artofproblemsolving.com/Wiki/index.php/University_of_South_Carolina_High_School_Math_Contest/1993_Exam]USC exam[/url] into the wiki.  I think the AMC's should follow this format (it's open to improvement though if you have suggestions).

Also, note that on the individual problem pages ( e.g. http://www.artofproblemsolving.com/Wiki/index.php/University_of_South_Carolina_High_School_Math_Contest/1993_Exam/Problem_2 ) I have links to the previous problem, next problem, and back the exam page.  I also added the [[Category:...] tags for all of them.

Sorry to bump....

but the AIME's certainly still need more work!!!

Today I fixed around a couple things about the AIME's (preliminaries for like 3 or 4 tests, one completely added)
For many, the Latex is up but it just need to have <math> tags added and split into problems.

When you work on an AIME, still try to post here :)

for editing the AIME pages, is it ok to copy and paste the "official AoPS solution", which is linked to the right of the problem statement in the Contests section? and if so should the original solver's name be posted?

[quote="me@home"]Seeing as those aren't truly "official AoPS solutions" but solutions that members posted and mods edited, yes, its okay, and yes the solver's name should be up there.
Also if you have a different solution you can post it as a second solution.[/quote]
You think so?  When I come up with my own solutions I usually don't give myself credit—in fact, I usually don't give anybody credit at all.  This is probably okay because it is quite possible for several people to come up with practically identical solutions to the same problem.  In any case, rather than copying and pasting, you would probably do better to learn the solution and rewrite it yourself, especially since the proofs posted on the forums are sometimes poorly written (and sometimes written in an odd sort of slang).  Plus, you might as well link to the problem's discussion on the forum, where anybody interested can see who posted what solutions.

Can someone start putting up solutions to this year's AIME I?  I typed up the problems [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems]here[/url] but don't have time to put up all my solutions.

[quote="joml88"]The biggest advantage of the wiki over the contest directory is that anyone can edit it at any time.

Thanks.[/quote]That's also a disadvantage in my view but it doesn't matter :P

A pessimist is just a realist who's not high...

But w/e.  Do we have MAA's permission to break copyright? I know that the Kalva archive has to avoid that.

On second thought, if the problems are split up it should be legal...at least I think that's what the pamphlet said.

Why do we have a wiki and contest resources section?
We should combine those as some problems aren't entered in the one but is in the other and it gets annoying to filp back and forth.
Also alot of pictures and solutions are missing in the contest resources and typos are present.

Time and energy...you're more than welcome to do it :)

Aren't only certain moderators able to edit the resources section?

[quote="ceegers"]So obviously both the wiki and the resources sections are incomplete. Now I've got some extra time in here for a couple months, so I quite possibly will go ahead and add some stuff to the wiki.[/quote]

That is great!  :)

[quote="ceegers"]So... is AoPS still concerned with updating the competition resources section? If so, should I maybe post problems that aren't in the resources section in the appropriate forum so they can be linked? Anything else I should do? If you are still updating both, which has priority to be updated, the wiki or the resources?[/quote]

The resources section is more frequently updated and is of greater priority. If you have problems that don't already appear in the resources section, of course do go ahead and post them. That being said, many of the American contests have already been uploaded to the resources section; the wiki by contrast is missing many pages, including both problems and articles, and it is much easier to directly make changes in the wiki. Chances are you'll find that much more work needs to be done with the wiki.

And post 1000.

Is this the place to report mistakes in the posted problems, because either I am reading something wrong, or Problem 15 on the 2005 AIME II has an error in it.  I think the question is asking for the smallest possible positive value.

I'd better put why I think it's wrong inside of spoiler space, in case somebody wants to work on the problem without a hint:
[hide]The ellipse of possible centers for the circle goes through the Y-axis, which means that if you choose a point where X = a very small negative value, a < -1,000,000, so the answer > 1000.  The answer makes it look like they are looking for a positive slope.[/hide]

-- Don

You post this in the discussion section of that particular thread.

Or you could just edit it yourself.

If it's an accident, people will probably change it back to what it's supposed to be.  If it's on purpose, then that's a whole other thing.

What's difference between AoPS wiki and the solutions found on the "contest section" with all the different questions?

The questions in the contest section link you back to the original thread from whence it came (where users have posted solutions).  AoPSWiki contains solutions in which users type in there, not necessarily the same as the solution in the original thread.

[quote="Herp"]http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems

Like how all the links to the solutions are to Friday?[/quote]
What?  I don't get it...the links to the solutions are fine.
[quote="1=2"]Who is this HyperSet guy???[/quote]
Where did that come from?

[quote="JSGandora"][quote="Herp"]http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems

Like how all the links to the solutions are to Friday?[/quote]
What?  I don't get it...the links to the solutions are fine.[/quote]

As of 12:30 PM Eastern Time, the links to the solutions of all the problems were links to the "Friday" video.

[quote="JSGandora"][quote="1=2"]Who is this HyperSet guy???[/quote]
Where did that come from?[/quote]

He last edited the article before the actual solutions came up.

[quote="AoPS wiki"]# (cur) (prev)  17:23, 13 May 2011 Ksun48 (Talk | contribs) (6,902 bytes) (Undo revision 38515 by Xantos C. Guin (Talk)) (undo) [/quote]

Nice try, ksun48.

[quote="1=2"]Who is this HyperSet guy???[/quote]

http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=102351

He/She has a history of putting youtube links in to pages, as well as putting nonsense in pages (http://www.artofproblemsolving.com/Wiki/index.php?title=Demonic_Possession ).  I will block him/her from editing the wiki and will revert the vandalism.


Find the least value of the expression

x20y-11 + y20z-11 + z20x-11

where x, y, z are positive real numbers satisfying x + y + z = 2001.

We have x^20/y^11+y^20/z^11+z^20/x^11=( x^2/y)^10/y+... \geq (  \sum _ciclyc (x^2/y)^5)^2/(sum x). But  \sum  x^2/y \geq  \sum x. So, the expression is at least (5*(  \sum x^2/y /5)^5)^2/sum (x^2/y). Since  \sum x^2/y is at least 2001, the minimum is attained when x=y=z