induction
logarithms
inequalities
number theory proposed
number theory

Solve in prime numbers $ 2p^q \minus{} q^p \equal{} 7$.

obviusly $ q$ is odd.
case(1):$ p \equal{} 2 \Rightarrow2^{q \plus{} 1} \equal{} q^2 \plus{} 7$ but for $ k\geq4$ we have $ 2^{k \plus{} 1} > k^2 \plus{} 7$(one can prove it by induction) therefore $ q < 4$ and $ q$ is odd prime,therefore $ q \equal{} 3$.obviusly $ (p,q) \equal{} (2,3)$ is a solution.
case(2):$ p\geq3$
lemma.if $ a\geq{b}\geq{e}$ then $ b^a\geq{a^b}$.
proof:let $ f(x) \equal{} \frac {x}{\ln{x}}$ then $ f^{\prime}(x) \equal{} \frac {\ln{x} \minus{} 1}{(\ln{x})^2}$,therefore on interval $ [e,\infty)$ we have $ f^{\prime}(x)\geq0$ and therefore in this interval $ f$ is increasing....
$ a\geq{b}\geq{e}\Rightarrow\frac {a}{\ln{a}}\geq\frac {b}{\ln{b}}\Rightarrow a\ln{b}\geq{b\ln{a}}\Rightarrow\ln{b^a}\geq\ln{a^b}\Rightarrow{b^a\geq{a^b}}$.
from original equation by fermat little theorem we get that $ p|q \plus{} 7$ but $ p \equal{} q \plus{} 7$ is impossible, because $ p,q$ are odd.therefore$ p\leq\frac {q \plus{} 7}{2}$
if $ q\geq{p}$
then $ q\geq{p}\geq3 > e$ therefore $ p^q \minus{} q^p\geq0$ and now from original equation we get that $ p^q\leq7$which is impossible ,because $ p,q\geq3\Rightarrow{p^q\geq27}$
therefore $ p > q$ and also we have $ p\leq\frac {q \plus{} 7}{2}$ therefore $ q\leq\frac {q \plus{} 7}{2}\Rightarrow{q < 7}$
if $ q \equal{} 5$ because $ q < p\leq\frac {q \plus{} 7}{2}$ we have $ 5 < p\leq6$ which is impossible...
if $ q \equal{} 3$ because $ q < p\leq\frac {q \plus{} 7}{2}$ we have $ 3 < p\leq5$ therefore $ p \equal{} 5$
and $ (p,q) \equal{} (5,3)$ satisfy original equation...therefore all solution are $ (p,q) \equal{} (2.3),(3,5)$

[quote="ali666"]
lemma.if $ a\geq{b}\geq{e}$ then $ b^a\geq{a^b}$.
[/quote]
is there any solution,without using this inequality?

Yes it can be done by an simple solution as following .
The first easy to check that $ p,q$ are different from each other .
Apply Fermat's Theory we have
$ 2p^q\equiv 2p (\mod q)$
$ p^q\equiv p (\mod q)$
Therefore :$ q|2p \minus{} 7$ and $ p|q \plus{} 7$
It is can be solve by inequalities .
The solution of this problem are $ (p,q) \equal{} (2,3),(3,5)$

[quote="TTsphn"]
$ p^q\equiv p (\mod q)$
Therefore :$ q|2p \minus{} 7$ and $ p|q \plus{} 7$
It is can be solve by inequalities .
[/quote]
by inequalities,we obtain $ p \equal{} 2,q \equal{} 3$ or $ q \equal{} 2p \minus{} 7$...then?!how we can find solution of the form $ (p,2p \minus{} 7)$ ?

Anyone has an elementary proof?  :blush:


Crestview's school colors are purple and gold. The students are designing a flag using three solid-colored horizontal stripes, as shown. Using one or both of the school colors, how many different flags are possible if adjacent stripes may be the same color?

[asy]size(75);
draw((0,0)--(0,12));
dot((0,12));
draw((0,12)--(8,12)--(8,11)--(0,11));
draw((8,11)--(8,9)--(0,9));
draw((8,9)--(8,6)--(0,6));[/asy]

There are $ 2$ choices, purple or gold, for each stripe. Thus $ 2^3 \equal{} 8$ flags can be made.


AMC
AIME
summer program
Mathcamp
USA(J)MO
USAMO

I am a student in Singapore.

My mathematics level as below...
AMC 12: 136
AIME: 10

I'm wondering whether my mathematical ability has met the average standard of Mathcamp paticipants? Was it good enough for Mathcamp?

I'm going to apply for Mathcamp this year... but I have little confidence, the campers look all very fantastic who make me feel rather inferior.

Anyway, I've just started the qualifying quiz... the questions seem fairly OK for me!

Good luck to all Mathcamp 2006 applicants!

i've gone to mathcamp for the past two years and ive never done that well on either exam.  at mathcamp, there are of course people who are amazing who will make you feel stupid but if you like math have some skill (i mean even i got in :lol: ) and are motivated you will probably do fine and fit right in.  I really advise you to just apply and dont worry.  If you are really worried you may want to aply somewhere else as well or be ready to apply elsewhere but dont worry that everyone who goes is not a usamo qualifier.  Much more important is how many quiz questions you answer and how motivated you are.


hi, i am completely an outsider in this thing...may i ask a simple question?

what are top 25% certificates of the fermat, cayley....etc?

i assume they are papers sent to individual students, but how come i never got any....i should b in top 25% ah....130 for fermat....i guess?

thanks!

I think you should ask your teacher about the certificates because they are the ones that hand the certificates to us.

The top 25% certificates are Certificate of Distinction (i think) and basically, it's just to tell you that you made it to the top 25% of Canada.

I think the top 25% varies but usually it is around 128/150. The ones that make the Honour Roll are the top 25%.

[quote="Lucky707"]Nope.

If you're talking about Pascal/Cayley/Fermat, the typical cutoff for a certificate (ie. top 25%) is around 100 with cutoffs ranging from 90 to 110 depending on the difficulty in any given year.

The ones who make Honour Roll are a lot higher than top 25%.[/quote]

That's interesting to know... This year's Honour Roll found on the CEMC website shows a wider range of contestants than previous year's Honour Rolls.

[quote="kool_dudy"]The top 25% certificates are Certificate of Distinction (i think) and basically, it's just to tell you that you made it to the top 25% of Canada.[/quote]
what is a Certificate of Distinction? is it something real, like a piece of paper?

The Cutoffs for the Pascal, Cayley, and Fermat contests canbe found on the CEMC website, [url=http://www.cemc.uwaterloo.ca/english/contests/past_result/2005/pcf_avg.shtml]here[/url]


Points on a square with side length $ c$ are either painted blue or red. Find the smallest possible value of $ c$ such that how the points are painted, there exist two points with same color having a distance not less than $ \sqrt {5}$.

$\textbf{(A)}\  \frac {\sqrt {10} }{2} \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\  \sqrt {5} \qquad\textbf{(D)}\  2\sqrt {2} \qquad\textbf{(E)}\ \text{None}$

[b]Edit:[/b] [i]Ignore this post![/i]

----------------------

[hide]$ c \geq \sqrt {5} / \sqrt {2} \equal{} \sqrt {10}/2$, or else we can't even have two points a distance $ \sqrt {5}$ apart.

If $ c \equal{} \sqrt {10} / 2$, we can construct an equilateral triangle using two vertices and one point in the interior of the square. At least two must be the same color by the Pigeonhole principle, and thus this $ c$ is large enough.

But we know $ c$ cannot be less than this value, thus the minimum possible $ c$ is just $ \boxed{\frac {\sqrt {10}}{2}}$[/hide]

[quote="TZF"][hide]$ c \geq \sqrt {5} / \sqrt {2} \equal{} \sqrt {10}/2$, or else we can't even have two points a distance $ \sqrt {5}$ apart.

If $ c \equal{} \sqrt {10} / 2$, we can construct an equilateral triangle using two vertices and one point in the interior of the square. At least two must be the same color by the Pigeonhole principle, and thus this $ c$ is large enough.

But we know $ c$ cannot be less than this value, thus the minimum possible $ c$ is just $ \boxed{\frac {\sqrt {10}}{2}}$[/hide][/quote]

[hide="I do not follow"]If I understand correctly, you construct a triangle with the edge $ c\equal{}{\sqrt{10}\over 2}$. But $ {\sqrt{10}\over 2}<\sqrt{5}$...?[/hide]

i think, the question is about the points on the square not inside.

[quote="TZF"]$ c \geq \sqrt {5} / \sqrt {2} \equal{} \sqrt {10}/2$, or else we can't even have two points a distance $ \sqrt {5}$ apart.

If $ c \equal{} \sqrt {10} / 2$, we can construct an equilateral triangle using two vertices and one point in the interior of the square. At least two must be the same color by the Pigeonhole principle, and thus this $ c$ is large enough.

But we know $ c$ cannot be less than this value, thus the minimum possible $ c$ is just $ \boxed{\frac {\sqrt {10}}{2}}$[/quote]

Paint the left side of the square with red, and right side with blue.

[code]KKMM
K  M
K  M
KKMM[/code]

If $ c \equal{} \frac {\sqrt{10}}{2}$, then we can find two points having a distance at most $ \frac {\sqrt{10}}{2}\frac {\sqrt{5}}{2} \equal{} \frac {5\sqrt{2}}{4} < \sqrt{5}$. So $ min(c) \not \equal{} \frac {\sqrt{10}}{2}$.

[quote="Farenhajt"]If I understand correctly, you construct a triangle with the edge $ c \equal{} {\sqrt {10}\over 2}$. But $ {\sqrt {10}\over 2} < \sqrt {5}$...?][/quote]

This is true.

[hide="Solution"]If $ s < 2$, let all points a distance $ < s/2$ from one side be red, and the rest be blue. We've found a counterexample.

If $ s \equal{} 2$, let such a square be ACEG, and let the midpoints of the four sides be B, D, F, H, so that, starting at one corner and reading clockwise, the points go A,B,C,D,E,F,G,H.

Let A be [b], wlog., and attempt to avoid a similarly colored pair of sufficient distance. Then, D and F, both a distance of $ \sqrt {5}$ from A, will both be [r]. Then, G and C, each a distance $ \sqrt {5}$ from D and F, respectively, must both be [b]. But C and G are a distance $ 2 \sqrt {2} > \sqrt {5}$ apart, so they cannot have the same color -- contradiction.

We therefore have $ \boxed{\textbf{(B)} \; 2}$[/hide]


Given a line segment of length L, how to find a line segment of length 1/L.

[quote="lotrgreengrapes7926"][quote="pkerichang"][hide]
draw a straight line and put the segment with length $L$ on it.  On  one of the endpoints of $L$, put a perpendicular line through it.  Then construct a segment with length 1 on it (I think this is necessary...).  Then connect the other endpoint of $L$ to the endpoint of the segemnt of length 1.  Then construct a perpendicular line through that endpoint of 1 again.  This line's intersection with the one line drawn will form the segment desired with the endpoint of $L$
[/hide][/quote]
In a construction, you can't construct a segment of length 1.

And this is a bad question, just like constructing the square root of a length. It all depends on the units of measure. For example, if we have a segment of $1$ foot or $\frac{1}{5280}$ miles, then the reciprocal of that would be $1$ foot or $5280$ miles, which are obviously not equal.[/quote]

that's why I say that you have to know unit length in order to do multiplication, division, and square root.  The unit length has to be given in order to do the operation.


What are the names of high school books in Canada?
In Physics , which subjects students are  studying?
How about chemistry and mathematic?

Thats aa very broad question. Unfortuantely i don't know the names of any of the high school books. I do, though, remember that they were by a publisher by the name of Pearson. :)

[quote="Elyot"]Yeah... mine were by the publisher Nelson Thomson Learning... I think they were just called Chemistry 11, Chemistry 12, physics 11, physics 12.  They were Ontario editions... probably came out around 2001-2003 or so, when the curriculums changed.[/quote]
I had nelson too. I heard the curriculum changed this year too.

[quote="samangharib"]no i didn't mean that i mean are you studying specific book in high school that you have to study it because of the high school rules?
my question is in general , in all around canada.[/quote]

No, generally there are no specific books that we have to study from.  Each school can individually select the textbooks it wants to use in teaching various courses.  However, the curriculum is standardized across the province (at least it is in Ontario), and generally there are 2 or 3 good textbooks brought out by different publishers from which schools will choose one to use in teaching the course.  Some teachers will teach math and science classes without ever using a textbook.

And in Canada, the curriculum varies from province to province anyway... so you're not gonna get a good general answer to your question.

[quote="Elyot"]...and generally there are 2 or 3 good textbooks brought out by different publishers from which schools will choose one to use in teaching the course.  Some teachers will teach math and science classes without ever using a textbook.[/quote]

What "good textbooks" are you talking about? I've never seen a good textbook in my life! In my experience, people who are actually really talented in teaching in a specific subject area have no intention of publishing a textbook in that area...Which means crummy textbooks...

I wish more of my math and science classes had been taught without a textbook...

In grade 11 enriched math at my high school, we got to use this book:
http://www.harcourtcanada.com/school/math/functions.htm

I think it's the best math textbook I've ever seen for high school kids.  (You can tell Waterloo minds were behind it.  Anyone recognize the logo on the cover!?  :lol: )  It was light, had tons of relevant material, proofs if you were interested, and the exercises were always split up into parts A, B and C.  I was just a really big fan of that book lol.  The books for MCB 4U and MGA 4U seemed good too, but I've only skimmed through them.  Oh!  And I loved how they were so colourful!   :D

hehe, i used the calc and geometry texts made my the UW people too!

[quote="Elyot"][quote="WindingFunction"]I've never seen a good textbook in my life![/quote]

Well... even if all textbooks you've ever seen are SHlTE, some of them must be better than the others, yes?  These would then be the "good" textbooks... unless every textbook in the world is of equal crappiness.  What you're saying is rhetorically equivalent to saying something like "I have never seen a big person" or "I have never seen a sharp knife".  Qualitative attributes like good and bad, big and small, or sharp and dull only exist in one's mind when they can compare different objects with respect to the given attribute.  So if you've never seen a good textbook, then you don't even know what one is, and therefore don't know whether or not you've ever seen a good textbook![/quote]

Haha...well by "bad textbooks," I mean books that don't explain things very clearly, or use bad form, or annoyed me in some other way.  And they have lots of useless pictures and articles of people who have math-related careers or something.

[quote="lotus87"]I think it's the best math textbook I've ever seen for high school kids. [/quote]

But you had a much better experience in the course with Caldwell...  I didn't like that book very much.  I think Romiens ruined it for me.

[quote="WindingFunction"]

Haha...well by "bad textbooks," I mean books that don't explain things very clearly, or use bad form, or annoyed me in some other way.  And they have lots of useless pictures and articles of people who have math-related careers or something.[/quote]

Heheh yeah.  They have to make those kind of textbooks appeal to everyone; not just us super-bright enriched-math people.  I love the comment about the stupid pictures and articles of people who have math-related careers.  I guess they want to prove that math IS useful for something, after all.


find all integers like a,b and c so that :
a|bc ,b|ac, c|ab.

This problem should be posted in number theory's forum .

[quote="rem"]Let $gcd(a,b)=m, gcd(a,c)=k, gcd(b,c)=n$. Then we can write:
$a=mka', b=mnb', c=nkc'$ and it is pretty obvious that $a',b',c'$ are pairwaise relatively prime. Now going back to the fact that $a|bc$, we get:
$mka'|mkn^{2}b'c'\Leftrightarrow a'|n^{2}$ However if $a'>1$, we will get $gcd(b,a)>m$ because some factor of $a$ that is greater than $1$ divides $n$ and therefore that factor divides both $b$ and $a$ and is not included in $m$. So we get a contradiction, therefore $a'=1$. By analogy, we get $c'=1,b'=1$.
So the solution must be numbers $a,b,c$ of form $mk,mn,nk$ where $m,n,k$ are positive integers.[/quote]

P.H post the problem in a wrong way,He meant:

$b+c|bc\ ,\ a+c|ac\ ,\ a+b|ab$

[quote="P.H"]I edited the problem yesterday.But I dont know why there is nothing here  :blush:
any way amir.s is right .the corect form of the problem is what he said.[/quote]

no You posted problem twice here is another one:
http://www.mathlinks.ro/Forum/viewtopic.php?t=104620


Determine whether or not there exists a positive integer $n$ such that $n$ is divisible by exactly $2000$ different prime numbers, and $2^{n}+1$ is divisible by $n$.

This topic (and a couple of others) have been moved from the AMC forum.

Please limit discussion in the AMC forum to AMC-related topics.


logarithms
limit
calculus
calculus computations

On the surface, this seems simple but I can't figure it out.

Find $ \lim_{x\rightarrow\infty}(\ln(x))^{\minus{}\ln(x)}$

Here's my first step, then I'm stuck

$ \lim_{x\rightarrow\infty}(\ln(x))^{\frac{1}{\ln(x)}}$

Now if we plug in, we get the indeterminate form $ \infty^{0}$. I cannot factor the highest power of x, use the squeeze theorem, use conjugates or seemingly simplify in any way. Does anyone have any helpful hints?

That first step is wrong. Look at the original- $ \infty^{\minus{}\infty}$ is not an indeterminate form.

If you do get an indeterminate exponential form, taking logarithms is usually the best way to work with it.


find all $ x \in R$ such that :
\[ 3^x \plus{} 4^x\equal{}5^x\]

Divide both sides by $ 5^x$ (which is never zero) to get $ \left(\frac35\right)^x \plus{} \left(\frac45\right)^x \equal{} 1$, which has its only root at $ x \equal{} \boxed2$, since the LHS is strictly decreasing.

Edit: Sorry, fixed.


MATHCOUNTS
number theory
prime factorization

MD's highest score was 45.  I was wondering what other states had, since everyone seems to care for cutoffs.  No one seemed to have done this

yea.  pretty different.  i made it that way on purpose, cuz i wanna know as well.  it was supposed to be.

i know, i was sort of kidding about that :P

mario...do you know who i'm talking about when i say "elevator kid"?

jorian

is the elevator kid dan tobias?
or
is it someone else?

why would dan be the "elevator kid"?

well...i asked choi as he was at nats last year...

jorian

i dont know who the elevator kid is.  all i can remember is that there was this annoying girl that asked me for a pin.

[quote="Fanatic"]I remember hearing a story about one kid who kept on screwing with the elevator buttons?

I remember when I first met the Utah kids:

"Hey are you guys from Utah?"
"Yes. WE'RE NOT MORMONS"
"O... k..."

They were all 7th graders, and I think at least 2 returned, maybe 3, not 4.[/quote]
Dan Tobias from CT was the kid who kept on screwing with the elevator buttons, it was annoying, I just used the stairs because of it.

[quote="jhredsox"]he claims he only did that once...

@dioph:what state did you go to nats w/ last year?

jorian[/quote]
Maybe the elevators were just slow, either way, I used the stairs.

I went with FL

Way to get off topic. I got 33 which was the highest score for Alaska

Top 4: 33,30,27,24 then 3 or 4 people got 23

[quote="jhredsox"]well, isn't this sorta different from cut-offs...

anyways

the elevator kid got a 36 (i should've gotten a 39, w/e)

jorian[/quote]

You are kidding about the part you should have gotten a 39 right?  :D

nope, read my target complaints

and i should've easily gotten a 23 on sprint

jorian

Well really, that's pretty egregious to say something like that.

If you got x wrong, and they were stupid mistakes, then you got x wrong, and you should have gotten x wrong, as punishment for stupid mistakes or whyever you got them wrong. Whether it's good or bad, Mathcounts tests reading questions, and you don't get points for careless mistakes, even if you solved it correctly. If you don't like that, tough, do another competition that doesn't stress that as much.

But for you to say you should have gotten X instead of Y is bad. To say you [i]could[/i] have gotten X instead of Y, I guess, could be acceptable. I'm sure I could have gotten 30, 32, 36, 40, 46 instead of my 25 at Nationals last year, but I'm completely at fault. I didn't get enough sleep, I wasn't prepared with the small paper, and I should have avoided getting a cold, and I should have read better. But I should not have gotten anything more or less than a 25 with the specific situation that I put myself in.

to get it back on topic...
texas high = 46 if no one figured out already  :lol:
yeah and um...
I WISH I LIVED IN ALASKA!!! XD

Michigan's was 42, and I wished I lived in Alaska too :mad:

You wouldn't once you froze to death.  :rotfl:

42 is good. I checked, and it's a lot better than SOME others.  :roll:

And me in my awesome 22nd place finish with a 28 :P

My scatch paper was so crowded that on the last target problem, I forgot about 1 2 in the prime factorization, and that messed me up. :o
2 easy mistakes on sprint, 2 easy targets wrong. I still blame myself though for the silly mistakes.
There's always next year, and the year afer that!

CA's top score was 41, the top 5 people or so got it  :o

yes, but we are not allowed to tell anyone.

No, b/c a more diffcult state means better perfromance at nats.

Assuming you can make the team...

KS 32, tie i think

I know-it's more difficult to be there, but it's also more rewarding.

I'd have to agree.
I feel lucky for anyone on the Texas team....

i would actually agree that living in texas is a LOT better: ever seen be talk about how GA will NOT win? look around, and you'll c it....
but in case you can't tell, i was really responding 2 the person who wanted 2 live in alaska!
i barely trained this year, buti still have another year anyway, so i hope to meet some of you at nats next year!

[quote="footballrocks41237"]yeah kstan won WI with 36.  i think we have the lowest cutoff besides for states like North Dakota and Alaska...[/quote] Heh right... apparently GA along with KS have lower high scores.

to people like me, i just want to make Nats. How well you do at  Nats matters, but it's not as important as getting there.(duh?)

Eh, I'd rather get 5th at states than 198th at nats.

well, most AoPsers can get more than 5 on a Nats test. 198th at Nats is better than not even REACHING NAts at all. You're saying that you would rather quit than perform badly at the next level?

whose teams got a 46 as a high score?   are all of them texan votes?

since this topic seems to consist of mainly high performers at mathcounts...when was the last time that someone made like coundown or so from a REALLY bad state?

jorian

NY's score(high) was 37. The scores were like this:

37, 36, 35, 34, 33, 33, 32, so on. It consisted of almost all positive integers less than 38.

Wisconsin's were similar (towards the end)--
1st-36
2nd-36
3rd-32
4th-32
5th-32
6th-31
7th-31
8th-30
9th-30
10th-29

Hey it's not my fault.

[quote="academicsrule"]heh i was 22nd 2. about 2-3 points separated me from sitan. (argh) for GA the high's between 25 and 35, but i doubt its that high. bet u guys wish u live here![/quote]

me too... :mad: gaah! i made a stupid mistake on a target round question, and that ruined my chances!

It seems most people don't pay enough attention to the importance of reading carefully in mathcounts, even when it means sacrificing speed.

...bus trips cost more than $\$$ 4.50 altogether.  >_< I even messed up the ABCD square due to lack of...psychological preparation for the pressure of my state's competition.

Well, it's a good thing MATHCOUNTS isn't the only competition- having 9 hours on 6 problems would be nice.

I'd rather bomb nats then not make it- At least you get the rare experience of travel for a math competition and you don't worry "If only I'd made the nationals...then I might have/would have/could have etc."


I was reading this chapter and at the top of page 2 i read that the surface is a complicated 4th degree curve.  Could someone dervive this curve for me, i've been having trouble figuring out the correct steps.  Thanks

I first draw a coordinate system, it's origin is V (in fig 27-2). The curve is, say, y(x). I call the x coordinate of $O$ $a$, and of $O'$ $b$. And x-coordinate of point $P$ is $x$. The length of the path is, then
$|OP|+|PO'|$
$=\sqrt{(x-a)^{2}+y(x)^{2}}+\sqrt{(b-x)^{2}+y(x)^{2}}= Constant$. Solving for $y(x)$ gives something pretty lengthy, but it should be 4th degree.

The second constraint is, two lines, $|OP|$ and $|PO'|$ should obey Snell's law. Just draw the tangent line to the curve at x, (so tangent is $y'(x)$), and after a bit geometrical play, it should be OK.


[quote="The Zuton Force"]For any value of $m$, there exists infinitely many values of $x$ to make that statement false as it stands right now.

You might want to fix your problem.

[hide="Some simplification"]$t=x^{2}-2x+m$

$log_{2}\sqrt{x^{2}-2x+m}+4\sqrt{log_{4}(x^{2}-2x+m)}= \frac{1}{2}\log_{2}(t)+4 \sqrt{\frac{1}{2}\log_{2}(t)}$

$p = \frac{1}{2}\log_{2}(t)$

$(\sqrt{p})^{2}+4 \sqrt{p}-5 \leq 0$
$(\sqrt{p}+5)(\sqrt{p}-1) \leq 0$
$-5 \leq \sqrt{p}\leq 1$
$0 \leq \sqrt{p}\leq 1$
$0 \leq p \leq 1$
$0 \leq \log_{2}(t) \leq 2$
$1 \leq t \leq 4$
$1 \leq x^{2}-2x+m \leq 4$
$2 \leq (x-1)^{2}+m \leq 5$
$\boxed{2-(x-1)^{2}\leq m \leq 5-(x-1)^{2}}$[/hide][/quote]
i'm sorry,complete ,Find m to satisfy $log_{2}\sqrt{x^{2}-2x+m}+4\sqrt{log_{4}(x^{2}-2x+m)}\leq 5$ with x in [0,2]
 :)  :)

Under the condition of anti logarithm, $x^{2}-2x+m>0\Longleftrightarrow m>-x(x-2).$ Thus for $0\leq x\leq 2$, we have $0\leq-x(x-2)\leq 1$, yielding $m>1.$ Afterwards as The Zuton Force  derived, $-(x-1)^{2}+2\leqq m\leqq-(x-1)^{2}+5.$ For $0\leq x\leq 2$, $Max_{0\leq x\leq 2}\{-(x-1)^{2}+2\}\leq m\leq min_{0\leq x\leq 2}\{-(x-1)^{2}+5\}.$ The desired answer is $2\leq m\leq 4$, which satisfies the above condition $m>1.$

[quote="kunny"]Thus for $0\leq x\leq 2$, we have $0\leq-x(x-2)\leq 1$, yielding $m>1$[/quote]
i dont think so


Ασκησεις σαν αυτες ειναι πολυ χρησιμες και οχι ιδιαιτερα δυσκολες...Αποδειξτε οτι υπαρχουν απειροι φυσικοι $ n$ ωστε $ n/2^n\minus{}2$... :)

[quote="Αρχιμήδης 6"]Ασκησεις σαν αυτες ειναι πολυ χρησιμες και οχι ιδιαιτερα δυσκολες...Αποδειξτε οτι υπαρχουν απειροι φυσικοι $ n$ ωστε $ n/2^n \minus{} 2$... :)[/quote]

Ισχύει για κάθε n πρώτο από θεώρημα Euler : $ a^n\equiv a(\mod p)$ gia $ (a,p)\equal{}1$ κι άρα είναι πράγματι άπειροι

Για n συνθετο? :)

Οι αριθμοί που έχουν την παραπάνω ιδιότητα λέγονται ψευδοπρώτοι .
Όσο για την λύση επιλέγουμε $ n\equal{}F_k\equal{}2^{2^k}\plus{}1$   :)

Να και ένα παρομοιο problem  . Νδο ότι υπάρχουν άπειροι n ώστε $ n|2^n\plus{}2$

Σωστα Σιλουανε...ευκολακι ε? :) Και η δικη σου ειναι γνωστη.. Ομως δεν απεδειξες οτι υπαρχουν απειροι συνθετοι $ F_k$.. :lol:

[quote="Αρχιμήδης 6"]Ομως δεν απεδειξες οτι υπαρχουν απειροι συνθετοι $ F_k$.. :lol:[/quote]

Αν είχα απόδειξη γιαυτό δεν θα έγραφα αυτή τη στιγμή στο φόρουμ μάλλον θα έγραφα ιστόρια  :D
Θα το προσπαθήσω πάντως  :P  (που ξέρεις καμια φορά .(άμα το λύσω τα λεφτα μισά μισά) )


geometry
parallelogram
trapezoid
geometry proposed

Let $ A_{1}A_{2}A_{3}A_{4}A_{5}$ be a convex pentagon, such that
\[ [A_{1}A_{2}A_{3}] \equal{} [A_{2}A_{3}A_{4}] \equal{} [A_{3}A_{4}A_{5}] \equal{} [A_{4}A_{5}A_{1}] \equal{} [A_{5}A_{1}A_{2}].\]
Prove that there exists a point  $ M$ in the plane of the pentagon such that
\[ [A_{1}MA_{2}] \equal{} [A_{2}MA_{3}] \equal{} [A_{3}MA_{4}] \equal{} [A_{4}MA_{5}] \equal{} [A_{5}MA_{1}].\]
Here $ [XYZ]$ stands for the area of the triangle $ \Delta XYZ$.

Denote $ B_{i}\in A_{i \plus{} 1}A_{i \plus{} 3}\cap A_{i \plus{} 2}A_{i \plus{} 4}$, where the indices are taken modulo $ 5$.

The basic thing is that the equality of the areas imply the following parallelism: $ A_{i}A_{i \plus{} 1}\| A_{i \plus{} 2}A_{i \plus{} 4}$.

It is straightforward to see that $ M \equiv X \in A_{1}B_{1}\cap A_{2}B_{2}\cap A_{3}B_{3}\cap A_{4}B_{4}\cap A_{5}B_{5}$, based on the above parallelism.

Instead, another solution, more nicer I think, is by observing that there exists an affine transformations that maps the initial pentagon in a regular one, and the problem is now more like trivial.

Remark: A very similar problem, related to this one, is the following I posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=149421[/url]

I did not expect to see it here. This is was a problem
that I gave to my team leader to propose somewhere
and he did that for Romanian TST. So just to say
that this is my problem, even so it may appear too easy.

[quote="pohoatza"]

Now a straightforward solution is by proving that $ M \equiv X \in A_{1}B_{1}\cap A_{2}B_{2}\cap A_{3}B_{3}\cap A_{4}B_{4}\cap A_{5}B_{5}$, based on the above parallelism.
[/quote]

Well, why they intersect at one point, and why the point $ M$ satisfies the condition?

Since $[A_1A_2A_3 ] = [A_2A_3A_4]$, we have $A_1A_4||A_2A_3$. Similarly are proven the relations
$A_{i-1}A_{i+2}||A_iA_{i+1}$' where all indices are taken modulo $5$.
Denote by $U$ the intersection of $A_1A_3$ and $A_2A_4$.
Notice that $UA_4||A_1A_5$ and $U A_1||A_4A_5$'
Thus $A_1A_5A_4U$ is a parallelogram, and $A_5 U$ passes through
the midpoint $H$ of $A_1A_4$. Denote by $V$ the midpoint of $A_2A_3$' Since $A_1A_2A_3A_4$
is a trapezoid, it is well-known that $HU$ (i.e. $A_5 U$) passes through $V$. Let $G$ be
the centroid of the pentagon. Notice that $G$ lies on $A_5 V$, since the centroids of
$A_2A_3, A_1A_4$, and $A 5$ , all lie on $A_5 V$. Similarly, the segments joining the vertices
$A_i$ with the midpoints of $A_{i+2}A_{i+3}$ will all pass through $G$.
We shall prove that $G$ satisfies the requirement.
Notice that $[A_5 GA_2 ] = [A_5GA_3]$ (from $A_2 V = A_3 V$).
Similarly, it is proven that $[A_2 GA_5 ]=[A_2 GA_4 ] = [A_1 GA_4 ] = [A_1 GA_3 ] = [A_3 GA_5 ]$.
Denote by $S_2$ this common value, by $S_1$ the common value of
$[A_1 A_2 A_3 ] = ...,$  and by $S_T$ the area of the pentagon.
It follows that $[A_2 GA_3 ] = S_T -[A_5 A_1 A_2 ]-[A_5A_4A_3] - [A_5 GA_2 ] - [A_5GA_3] = S_T - 2S_1 - 2S_2 $,
therefore all $[A_iGA_{i+1}]$ are equal, and thus $G$ satisfies the desired requirement. So done.


Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.

We will prove the result using mathematical induction.
[b]Base Step[/b] For $ n \equal{} 1$ we can take the $ 1$-digit number $ 5$ and $ 5$ is odd, which is divisble by $ 5^{1}$, so the result is true for $ n \equal{} 1$
[b]Induction Step[/b] Assume the result is true for $ n \equal{} k$ and let $ q$ be the $ k$-digit number composed only of odd digits and divisible by $ 5^{k}$. Then, $ q \equal{} 5^{k}*t$ where $ t$ is a positive integer.
Of the numbers $ 1*2^{k}\plus{}t, 3*2^{k}\plus{}t, 5*2^{k}\plus{}t, 7*2^{k}\plus{}t, 9*2^{k}\plus{}t$ each gives a different residue $ \pmod 5$, because if we assume two numbers $ a*2^{k}\plus{}t, b*2^{k}\plus{}t$ have the same residue $ \pmod 5$ where $ a,b\in\{1,3,5,7,9\}$ and $ a$ and $ b$ are not equal, then $ a*2^{k}\plus{}t\minus{}( b*2^{k}\plus{}t)\equiv 0\pmod 5$ so $ (a\minus{}b)2^{k}\equiv 0\pmod 5$, but $ 2^{k}$ and $ 5$ are coprime as $ 2,5$ are coprime; also $ (a\minus{}b)\in\{\minus{}8,\minus{}6,\minus{}4,\minus{}2,2,4,6,8\}$ so $ 5$ and $ a\minus{}b$ are coprime. Hence, we cannot have $ a*2^{k}\plus{}t\minus{}( b*2^{k}\plus{}t)\equiv 0\pmod 5$ so of the numbers $ 1*2^{k}\plus{}t, 3*2^{k}\plus{}t, 5*2^{k}\plus{}t, 7*2^{k}\plus{}t, 9*2^{k}\plus{}t$ each gives a different residue $ \pmod 5$. But then, because there are $ 5$ different residues $ \pmod 5$ then one of the numbers $ 1*2^{k}\plus{}t, 3*2^{k}\plus{}t, 5*2^{k}\plus{}t, 7*2^{k}\plus{}t, 9*2^{k}\plus{}t$ gives a residue of $ 0\pmod 5$, let this number be $ x*2^{k}\plus{}t$, where $ x\in\{1,3,5,7,9\}$, then the number $ (x*2^{k}\plus{}t)5^{k}$ is divisible by $ 5^{k\plus{}1}$.
But $ (x*2^{k}\plus{}t)5^{k}\equal{} x*10^{k}\plus{}5^{k}*t \equal{} x*10^{k}\plus{}q$, which is the number with the first digit $ x$ followed by the decimal representation of $ q$, so we get a $ k\plus{}1$-digit number $ r$ composed only of odd digits (as $ x$ is odd and all digits of $ q$ are odd) and divisible by $ 5^{k\plus{}1}$, so the result is true for $ k\plus{}1$,
Hence the result of the problem follows by mathematical induction.


Diagonals of convex quadrangle $ ABCD$ intersect in point $ E$. Triangles $ ABE$, $ CDE$ are of the same area. The AC diagonal is alsow bisector of $ \angle DAB$. Knowing that $ AB\equal{}4$ find length of $ BC$.

[hide]From the area condition it follows $ AE\cdot BE\equal{}CE\cdot DE\implies{AE\over DE}\equal{}{CE\over BE}$. Since $ \angle CEB\equal{}\angle DEA$, we get $ \triangle AED\sim\triangle CEB$, and that gives $ \angle BCE\equal{}\angle EAD$. However, that implies $ \angle BCE\equal{}\angle EAB$, and hence $ \triangle ABC$ is isosceles, giving $ BC\equal{}AB\equal{}4$.[/hide]

Oh, nice :) But could you explain me something?

I agree with $ \triangle AED\sim\triangle CEB$ but why $ \angle BCE \equal{}\angle EAD$ not $ \angle BCE \equal{}\angle ADE$?
Does it take from the proportion $ \frac{AE}{DE}\equal{}\frac{CE}{BE}$?

Exactly. $ DE$ corresponds to $ BE$ - cf. the proportion - therefore the angle opposite $ DE$ corresponds to the angle opposite $ BE$. (The correspondance of vertices is also incorporated into notation: $ \triangle AED\sim\triangle CEB$ - and it's usually a good habit to observe that when writing similarity relations.)

[quote="Farenhajt"][hide]From the area condition it follows $ AE\cdot BE \equal{} CE\cdot DE\implies{AE\over DE}\equal{}{CE\over BE}$. Since $ \angle CEB \equal{}\angle DEA$, we get $ \triangle AED\sim\triangle CEB$, and that gives $ \angle BCE \equal{}\angle EAD$. However, that implies $ \angle BCE \equal{}\angle EAB$, and hence $ \triangle ABC$ is isosceles, giving $ BC \equal{} AB \equal{} 4$.[/hide][/quote]

wait, how do u know that $ \triangle AED\sim\triangle CEB$?

[quote="mihail911"]wait, how do u know that $ \triangle AED\sim\triangle CEB$?[/quote]

Two pairs of proportional sides and equal angle between them...?

[quote="Farenhajt"][hide]From the area condition it follows $ AE\cdot BE \equal{} CE\cdot DE\implies{AE\over DE}\equal{}{CE\over BE}$. Since $ \angle CEB \equal{}\angle DEA$, we get $ \triangle AED\sim\triangle CEB$, and that gives $ \angle BCE \equal{}\angle EAD$. However, that implies $ \angle BCE \equal{}\angle EAB$, and hence $ \triangle ABC$ is isosceles, giving $ BC \equal{} AB \equal{} 4$.[/hide][/quote]

how does the area condition imply $ AE\cdot BE \equal{} CE\cdot DE$ ?

[quote="Ish"]how does the area condition imply $ AE\cdot BE \equal{} CE\cdot DE$ ?[/quote]

$ [AEB]\equal{}{AE\cdot BE\over 2}\sin\angle AEB$

Or you can avoid trigonometry to prove it: See the attached diagram.

Assume $ [AOB]\equal{}[COD]\implies{AO\cdot BN\over 2}\equal{}{CO\cdot DM\over 2}\implies{AO\over CO}\equal{}{DM\over BN}$

But $ {DM\over BN}\equal{}{DO\over BO}$, hence $ {AO\over CO}\equal{}{DO\over BO}\implies AO\cdot BO\equal{}CO\cdot DO$


inequalities
AMC
AIME
triangle inequality

[quote="simulacrum"]Do you think geometric inequalities; i.e the triangle inequality,etc is heavily tested on the AMC/AIME? or will there just be at most like one problem on it?[/quote]
Well you want to make sure you know them just in case, though I doubt there will be over 2 problems on the AMCs and AIMEs that require them. The only ones you'll need are triangle inequality and pythagorean inequality for that anyways.

the triangle inequality comes up often...[note two aime problems posted recently]...but the triangle inequality is pretty straightforward...the applications are not always though...


geometry
number theory
modular arithmetic

Hi, I was thinking of studying the aops volume 2 "And beyond", but i dont feel like studying book 1 also. For those of you that might have studied both, do u guys think i can get throught volume 2 without knowing some concepts from volume 1??  :help:

It depends how much you know already. Look at the table of contents of volume 1 - if there are multiple things you don't know well, look at that too.

Volume 1 has a lot of standard mathematics knowledge, such as simple geometry and algebra. However, there are some things you should know before continuing to Volume 2 (such has modular arithmetic and combinatorics). In fact, I had to reference Volume 1 many times while studying Volume 2 in order to refresh my memory and knowledge.

Like the previous poster said, it depends on how much of the topics you know now.

Hmm ok thanks! :lol:   :D
I guess ill just have to study both books.


we can remark that $y^{4}\equiv 0\left( %Error. "func" is a bad command.
{mod}4\right) $
which means that $y\equiv 0\left( %Error. "func" is a bad command.
{mod}4\right) $ or $y\equiv 1\left( %Error. "func" is a bad command.
{mod}4\right) $
can anyone go on ???  :D

I don't know if that has anything to do with it. But I have an independent solution: if $x$ is odd then we have $3|y^4+y+2$, which is impossible. On the other hand, if $x$ is even then $y^4+4y+1$ is a square. This is larger than $(y^2)^2$ and the next square is $y^4+2y^2+1$. So we get $y \leq 2$. Checking yields $x=2,y=2$.

[quote="Bojan Basic"]Find all pairs $(x,y)$ of integer satisfying $5^x=1+4y+y^4$.[/quote]
Don't forget $(0,0)$. ;)


I would like to know which is the difference between serie A and B and why only GAZETA MATEMATIKA serie B will be in future an electronic journal on the web.
Does anyone know when it will be active?
I have send an e-mail  to

contact@gazetamatematica.ro

but nobody sent me a reply.

Thank you.

The A series is for teachers and college students.
The B series is for highschool students or younger.
The B series is based on problems, and A series is based on research articles.

The PR problems are something ususal in Romania.

at least that is the way it should be :). unfortunately I don't think that the GM site will be online soon.


geometry
incenter
circumcircle
geometry proposed

Let $ ABC$ be an acute triangle with $ I$ and $ H$ its incenter and orthocenter, respectively. Let $ B_1$ and $ C_1$ be the midpoints of segments $ AC$ and $ AB$ respectively. Ray $ B_1I$ intersects segment $ AB$ at $ B_2 \neq B$. Ray $ C_1I$ intersects segment $ AC$ at $ C_2$ with $ C_2A > CA$. Let $ K$ be the intersection of segments $ BC$ and $ B_2C_2$. Prove that triangles $ BKB_2$ and $ CKC_2$ have the same area if and only if $ A, I, A_1$ are collinear, where $ A_1$ is the circumcenter of triangle $ BHC$.

$ a \equal{} BC, b \equal{} CA, c \equal{} AB.$ $ AI$ cuts $ BC$ at $ X,$ $ IB_1$ cuts $ BC$ at $ P.$ By Menelaus for $ \triangle AXC$ cut by $ PIB_1,$ $ \minus{}\frac{\overline{IX}}{\overline{IA}} \equal{} \frac{\overline{PX}}{\overline{PC}} \equal{} 1 \minus{} \frac{\overline{XC}}{\overline{PC}}.$
By Menelaus for $ \triangle ABC$ cut by $ PB_1B_2,$
$ \frac{\overline{B_2B}}{\overline{B_2A}} \equal{} \minus{}\frac{\overline{PB}}{\overline{PC}} \equal{} \frac{\overline{BC}}{\overline{PC}} \minus{} 1 \equal{} \frac{\overline{BC}}{\overline{XC}} \left(1 \minus{} \frac{\overline{IX}}{\overline{AI}}\right) \minus{} 1 \equal{}$ $ \frac{b \plus{} c}{c} \left(1 \minus{} \frac{a}{b \plus{} c}\right) \minus{} 1 \equal{} \frac{c \minus{} a}{b}$
In exactly the same way, $ \frac{\overline{C_2C}}{\overline{C_2A}} \equal{} \frac{b \minus{} a}{c}$ $ \Longrightarrow$ $ \frac{\overline{CC_2}}{\overline{CA}} \equal{} \frac{1}{1 \plus{} \frac{\overline{C_2A}}{\overline{CC_2}}} \equal{} \frac{1}{1 \plus{} \frac{c}{a \minus{} b}} \equal{} \frac{a \minus{} b}{c \plus{} a \minus{} b}.$

$ [\triangle KBB_2] \equal{} [\triangle KCC_2] \Longleftrightarrow [\triangle AB_2C_2] \equal{} [\triangle ABC] \Longleftrightarrow$ $ \overline{B_2A} \cdot  \overline{C_2A} \equal{} \overline{BA} \cdot \overline{CA} \Longleftrightarrow$
$ B_2C \parallel BC_2 \Longleftrightarrow \frac{\overline{B_2B}}{\overline{B_2A}} \equal{} \frac{\overline{CC_2}}{\overline{CA}} \Longleftrightarrow  \frac{c \minus{} a}{b} \equal{} \frac{a \minus{} b}{c \plus{} a \minus{} b} \Longleftrightarrow  a^2 \equal{} b^2 \plus{} c^2 \minus{} bc  \Longleftrightarrow$
$ \angle A \equal{} 60^\circ  \Longleftrightarrow  A, I, A_1$ are collinear.

it's incredible... the same ugly problems are haunting us over and over, just because somebody proposed them for a china MO...
http://www.mathlinks.ro/Forum/viewtopic.php?t=660

  darij


A regular hexagon is inscribed in a circle. Given that the circumference of the circle is 12$ \pi$ inches, how many inches are in the perimeter of the hexagon?

Circumference is 12pi, so radius is 6, side length of hexagon is 6, perimeter is 36.

Just a little side-note. A hexagon can be split into six equilateral triangles. That's why the radius of the circle is equal to the side lengths of the hexagon.


You are a pilot on an aircraft carrier. You must fly to another aircraft carrier, now 1450 km at 45 degree of your position, moving at 56 km/hr due east. The wind is blowing from the south at 72 km/hr. Calculate the heading and airspeed needed to reach the carrier 2.5 hour after you take off.

i drew a picture but i dont even know if i have the picture right or what to do from there on please help. http://img127.imageshack.us/my.php?image=physicsvo6.jpg

Basically, you want to add all of the velocity vectors for the plane and your final resultant is the speed of the plane.  You can use trig to get the angle at which it is flying.  After, do the same for the boat and then it is easy from there.

i did not understand that. my teacher does not really teach so its like a self study class so i really did not get that. by the way is my picture correct?

For the most part it seems correct.  I can't put up my own diagram so it's rather difficult for me to explain it well.

than if i droped the perpendicular than i would have one triangle with 1450 as hypotenuse and 45 degree so the x and y would both be 1025.3 but the other line going east would only have a horizontal component of 140 so the total x would be 1165.3 and y would be 1025.3 right? and than i draw a new triangle and find the hypotenuse and the angle.
do i have it right or am i completely off?
thx for the help!


function
LaTeX
inequalities
real analysis
real analysis solved

How would you show that if f is a continuous function on interval [a,b], and c in (a,b) is a local maximum, then

D-(f(c))=<0<=D+(f(c))

where D-(f(x))=limsup(h to 0+) of (f(x)-f(x-h))/h    and

 D+(f(x))=liminf(h to 0+) of (f(x+h)-f(x))/h.

(These are derivates, not derivatives.)

When I try this, I seem to get the opposite inequalities.

Also what if the maximum occures at the endpoints a or b?

Any suggestions?

Translating your post into $\LaTeX$ (which was very easy in this case):
[quote="mark1234"]
How would you show that if $f$ is a continuous function on interval $[a,b]$, and $c \in (a,b)$ is a local maximum, then

$D^-f(c)\le0\le D^+f(c)$

where $D^-f(x)=\limsup_{h \to 0^+} \frac{f(x)-f(x-h)}{h}$ and

$D^+f(x)=\liminf_{h \to 0^+} \frac{f(x+h)-f(x)}{h}$.

(These are derivates, not derivatives.)

When I try this, I seem to get the opposite inequalities.

Also what if the maximum occures at the endpoints $a$ or $b$?

Any suggestions?[/quote]

You're right: those inequalities should be reversed. If all of those difference quotients are nonnegative, their $\limsup$ must also be nonegative. A similar argument applies on the other side.
At an endpoint, one of these inequalities is meaningless, and the other works the same way as inside the interval.


A number N consisits of r digits chosen from 5,6,7,8,9 any of which may be repeated any number of times

Let S=sum of digits in the number N

Let D=sum of digits in the number 2N

Prove D =2S-9r

. 6785
. . . . 2
 x-----
12
 .14
 . .16
 . . .10
--------
13570

This example shows well the principle. for each number in 5-9 we have the property that its double starts with 1, but the last digit can at most be 8 so we can't have a 10-jump again.

So we just do the thing individually and we see that the property is valid for for each of the numbers 5-9.

So valid for each digit, and composable digit by digit, means it's proven :)

Hi,

Slight different explanation, I guess.. (I was working on it, and saw that my solution is almost similar to Peter's)

[hide]Lets analyse the situation from a summing point of view.

    - - - - - ........ - - - (r digits)

 + - - - - - ........ - - - (r digits)

Consider 5. 5+5 = 10. Final sum of digits = 1+0 = 1. So compared to the case where we didnt have carry over in addition (assume), we lose 10 - 1 = 9 units.

Similarly, 6+6 = 12. and 12 - (1+2) = 9.

The other cases yield this too. Thus, loss of 9 for each digit pair sum.

Now, note that adding two digits (out of 5,6,7,8, or 9) results in 1 carry over, and since the maximum unit's digit of sum will be 8 (in case of 9+9), so carry over 1 will be added to this place. Thus, for each digit place, 1 gets carried out, and the 1 carried into this place is added. So, each digit place contributes exactly 9 to the loss.

Thus, twice the sum of digits will be 9*(no. of digits) greater than sum of the digits of double the number;

i.e., 2S = 9r + D

[/hide]

************************************************************************

Sambit

Sambit - please don't put that huge string of asterisks at the end of your posts - it makes the screen format too wide and forces admins and mods to have to edit (or delete) your posts.


Let $ n$ be a positive integer. Show that there is a positive integer $ N$ such that:
If $ A$ is a subset of $ \left\{1,2,...,N\right\}$ satisfying $ \left|A\right|\geq\frac {N}{2}$, then there is a positive integer $ m$ with $ m\leq N \minus{} n$ such that $ \left|A\cap\left\{m \plus{} 1,m \plus{} 2,...,m \plus{} k\right\}\right|\geq\frac {k}{2}$ for all $ k\in\left\{1,2,...,n\right\}$.

Let us call [i]black numbers[/i] numbers that belong to A and [i]white[/i] - the others. Suppose that there exist A fulfilling the conditions for which the thesis is false. We will divide the sequence $ (2, 3,...,N\minus{}n)$ into subsequences using the following rules:
(1) We take elements until we have more white elements than black ones or we have reached the end of the sequence. If we didn't we must have more white than black before we take $ n\plus{}1$th element because the thesis is false.
(2) We make a subsequence out of the taken elements if we didn't reached the end of the sequence.
Say we have $ k$ subsequences. In $ i$th subsequence the number of while ($ w_i$) is bigger than the number of black ($ b_i$): $ w_i \ge b_i\plus{}1$. But in whole sequence the number of black is greater or equal to the number of white, so $ \sum_{i\equal{}1}^{k}b_i \plus{}n\ge\sum_{i\equal{}1}^{k}w_i$. From $ w_i \ge b_i\plus{}1$, transforming, we get $ n \ge k$. But $ k\ge\frac{N\minus{}n}{n}$, because each of the subsequences has at most $ n$ elements and we only leave less than $ n$ elements undivided. From here we see that $ k$ grows to infinity with $ N$, so for sufficiently big $ N$ $ k$ must be bigger than $ n$. Contradiction.


calculus
integration
calculus computations

$ \int\frac{1}{t^3 \sqrt{t^2 \minus{}36}}{dt}$
[hide]Use a triangle[/hide]

Put $ t \equal{} 6\sec\theta$
What do you mean by using a triangle?

[quote="kabi"]we can also do this by partial fraction.I will post the solution later if u need it by partial fraction.[/quote]

You could try $ t^2 \minus{} 36 \equal{} a^2$. Then maybe you could use the partial fractions.


inequalities
calculus
inequalities proposed

Find the best constant $k$ for the inequality:

$a^k+b^k+c^k \ge ab+bc+ca$

If $a,b,c$ are nonegative real and $a+b+c=3$.

I think it deals with Schur's inequality..

[quote="Namdung"]Dear Hung, can you explain the meaning of "best constant" in your problem?[/quote]
Dear teacher Nam Dung, I think Hung want to find the smallest $k$ so that if $a,b,c$ is positive real and $a+b+c=3$  the inequa is true:
$a^k+b^k+c^k \geq (ab+bc+ca)$
I think he want to know Vasc's magic number : $\frac{2\ln3-3\ln2}{\ln3-\ln2} \approx 0.29$ (Look at http://www.mathlinks.ro/Forum/viewtopic.php?t=56732).
I don't know if Schur's inequa can  handle this one because it involved "ln". So probably, caculus is the simple way to approach it.
  I just ask one question:
I remembered there exist one inequa that can be solved easily by Bernuli . It like:
If $x_i \geq 0$ and $\sum_{i=1}^{n} x_i \geq n$ and $r \geq s$ then
$\sum_{i=1}^{n} x_i^r \geq \sum_{i=1}^{n} x_i^s$
   If this inequa is true then after finding the best constant $k$ all of the number greater than k is also true too.
  I hope s.on can prove my asking inequa here. I think it is not difficult .

Ok. $k=\frac{2ln3-3ln2}{ln3-ln2}$.  But It's not the key for this problem.

We must say: Best positive constant

Be cause all k<0 work

Namdung

OK. For detail, find the minimum positive real $k$. Easy if $k \le 0$ or $k \ge 1$.


Remember the cyst I told you about in the "mods delete this topic" thread? Well here is a curious piece of conversation I had with shadysaysurspammed today:

shady: hey u really have that cyst thing?
Ritam Bhaumik: yeah
Ritam Bhaumik: sad case
shady: its curable right?
Ritam Bhaumik: anyway we all must die one day
Ritam Bhaumik: i kust wished i lived a bit longer
Ritam Bhaumik: 18 is too young .... too early ....
shady: lol...i'm not dumb,that can be cured
Ritam Bhaumik: it works this way
Ritam Bhaumik: after the operation
Ritam Bhaumik: they take the cyst for biopsy
Ritam Bhaumik: if its cancerous i am a gonner
Ritam Bhaumik: my only hope to live is if its not cancerous
Ritam Bhaumik: which is a slim chance
shady: oh sorry ..all the best
Ritam Bhaumik: i hope that helps
Ritam Bhaumik: i really do
Ritam Bhaumik: this is only the beginning of my life
Ritam Bhaumik: to be cut off so early ...
Ritam Bhaumik: can't we switch topics?
shady: ya sure
shady: so i'm leavin india in a week
shady: :((
Ritam Bhaumik: at least you don't have to worry about leaving the world soon
shady: aw man
shady: don't make me cry
Ritam Bhaumik: let us go to some very different topic
shady: yeah ..hey i'll be right back
shady: 2 mins
shady: m back
Ritam Bhaumik: cantemplating life?
Ritam Bhaumik: well that's waht it's like
Ritam Bhaumik: you build up dreams
Ritam Bhaumik: carefully and lovingly
Ritam Bhaumik: and suddenly
Ritam Bhaumik: it all goes wham!
shady: life's unpredictable
Ritam Bhaumik: it is mate
Ritam Bhaumik: live it up
shady: hey,ur not joking around with me right,cuz thats ritam
Ritam Bhaumik: while you can
shady is running a version of Yahoo! Messenger with Voice (BETA) that does not support plugins.
shady: ya thats true dude
shady: k gotta go
shady: cya
Ritam Bhaumik: if i live, yes
shady: i will cya
Ritam Bhaumik: let me quote something from julius caesar
shady: ya
Ritam Bhaumik: if we meet again, we shall smile indeed,
Ritam Bhaumik: if not, well then, this parting was well made.
shady: nice...bye
shady: hey does that yashvardhan thing work?
Ritam Bhaumik: of course not
Ritam Bhaumik: even if it did
Ritam Bhaumik: i would hardly have time to enjoy it
shady has signed out. (7/14/2006 1:46 PM)

it looks weird because I directly got it off the yahoo!messeenger window, but you can still make it out.

What's your opinion on this conversation?

lolzzz  u could have suckered me up a li'l bit more if u didn't exaggerate ur dying too much  :D  i got bored n i thought maybe he's dying so better off to leave without asking him whether he's lyin n stuff  and sympathise with him :lol:  i was still online,just offline to u  :rotfl: yeah,bcame a bit too obvious hehe

lol
he indeed exaggerated too much, it was funny though.


arithmetic sequence
combinatorics proposed
combinatorics

Suppose that $S$ is a finite set of real numbers with the property that any two distinct elements of $S$ form an arithmetic progression with another element in $S$. Give an example of such a set with 5 elements and show that no such set exists with more than $5$ elements.

I  tried  this  for  a half  an  hour  and  got nothing.  of  course  this is  normal (im really  weak  at  combi)  but  im  trying  to  get  better  so  could  someone  please post  a  solution??? thanks :)

Cool problem :).

Assume we can have such a set with a finite but larger than $5$ number of elements. Also assume the smallest and largest elements are $0$ and $1$, for simplicity. Then $\frac 12$ is also an element, and there must be at least one element $a_0\ne 0,1,\frac 12,\frac 13,\frac 23$. More precisely, we asume WLOG that $a_0\in\left(\frac 13,\frac 12\right)$. Then there will be an element of our set, $b_0\in\left(\frac 23,1\right)$, the midpoint of the segment $(a_0,1)$. After that, we can find an element of our set, $a_1\in\left(\frac 13,a_0\right)$, the midpoint of $(0,b_0)$. Then we find $b_1\in\left(\frac 23,b_0\right)$, the midpoint of $(a_1,1)$, and so on. This sequence of steps cannot stop (we get a sequence $a_n$ converging to $\frac 13$ whose terms are always $>\frac 13$), so we're done.

Sorry for reviving this, but no one posted a 5-element set example. Grobber's solution is very nice, anyway..

[hide]A 5-element set is $ \{0,2,3,4,6\}$. We can easily verify that this set satisfies the given conditions.

Now assume that there exists a 6-element set $ \{0,v,w,x,y,z\}$ which satisfies the conditions. WLOG, assume 0 and $ v$ makes an arithmetic progression with $ w$, so assume that $ w\equal{}2v$. Similarly, assume that $ y\equal{}2x$. So the set becomes $ \{0,v,2v,x,2x,z\}$. Now, $ 0,z$ form an arithmetic progression with (WLOG) $ v$ or $ 2v$. It is easy to check that for both cases, that set is of the form $ \{0,a,2a,4a,b,2b\}$, with $ a,b\in\mathbb{R}^\plus{}$.

$ a,4a$ make an arithmetic progression with $ 7a$. So $ b\equal{}7a$ or $ 2b\equal{}7a$. Easy to check that both cases are impossible. Contradiction, Q.E.D.[/hide]

[quote="Johan Gunardi"]Q.E.D.[/quote]  No: deriving a contradiction for sets of size 6 does not automatically rule out sets of larger size.  (Also, grobber did construct the same 5-element example you did (though he didn't say this was what he was doing) when he mentioned 0, 1/3, 1/2, 2/3, 1.)


geometry
geometric transformation
reflection
incenter
perimeter
projective geometry
geometry proposed

Let $ABC$ be a triangle and $P$ be any point. Denote by $X, Y, Z$ respectively the reflections of $P$ on the sides $BC, CA, AB$ of triangle $ABC$. Show that the triangle $XYZ$ is always perspective with the orthic triangle $DEF$ of triangle $ABC$.

[hide="Hint"]  What is a special property of the orthic triangle?[/hide]

The orthocentre of triangle ABC is the incentre of the orthic triangle?  (Actually, sometimes its an excentre.)  Minimal perimeter?

my way: using Desargues theorem but it is quite long,
if the intersection of $XY,DE;YZ,EF;ZX,FE$ is colinear then two triangles are perspective.
We can check it by assume $P(x,y,z)$ and write equation of $XY,DE$ and calculate intersection point, and check the condition of three points !

Here is armpist's solution which is the same as mine

[hide="Solution"] As nsato pointed out the prominent property that the orthocenter $H$ is the incenter of the orthic triangle $DEF$ and since the problem gives you reflections (i.e. some equal angles), you may predict that probably the perspector $P'$ has something to do with bisectors of the orthic triangle and the point $P$. So, one good guess is that $P'$ is the isogonal conjugate of $P$ wrt $DEF$; and fortunately, it turns out to be true.  :)  [/hide]


Prove that
There are infinitely prime numbers :)

Assume that there are only finite number of primes $ p_{1} ,p_{2} ,p_{3} ....p_{k}$ where $ p_{1} < p_{2} < .......<p_{k}$ ,note that if there are finite number of primes we have $ p_{k}$ to be the largest prime . Consider the number $ n \equal{} p_{1}p_{2}....p_{k} \plus{} 1$ .Clearly $ n > p_{i}$ where $ i\equal{}1,2,3,.....k$.By assumption this number is composite so by fundamental theorem of arithmetic ,this number should have prime factors which will be from the set of prime numbers considered  . But it is easily observed that none of the primes $ p_{1} ,p_{2},....p_{k}$ divide $ n$ ,which implies $ n$ itself has to be prime ,contradicting the maximality of $ p_{k}$ .
This completes the proof .


email
probability
calculus
summer program
PROMYS
number theory
IMO Shortlist

Does anyone have any recommendations about the summer programs offered by the Johns Hopkins University Center for Talented Youth? There look to be several interesting math and computer science courses offered for summer of 2004 through that program, and I was wondering if any of you have tried those out in previous years. Any been-there, done-that descriptive information you could provide would be much appreciated.

I have actually attended CTY and now currently work there as an instructor in the distance learning program.

If you have any specific questions, feel free to email me.

jpo1@math.jhu.edu

Some of the classes at CTY can be wonderful because of exceptional instructors. However, some of the other classes can be mediocre as well. Overall, CTY is a good experience though.

CTY was pretty good. Most of the classes are excellent, but most of the math classes are probably too easy for the people on this forum. I went two summers ago.

The CTY [url=http://www.jhu.edu/gifted/summer/osopencourse.html]Open Course List[/url] for 2004 (the complement of the closed course list) is now posted. Did you get into the course you wanted this year?

I notice that some of the better-looking courses this year at CTY seemed to fill instantly. The math courses that took the longest to fill, dismayingly, were mostly the hard-but-not-hardest courses with algebra prerequisites, which made me wonder how many program participants come on board with a background in beginning algebra.

Are any of you AoPSers planning to be at CTY this summer? So far A+MATH is in like flint for the second session Probability and Game Theory class at Lancaster, and wait-listed for the Number Theory class, which filled up right away. He is a first-time student, so he didn't have the highest enrollment priority even by meeting the early application deadline.

Hmmm, it's pretty easy to get into the course of your choice if your scores are good enough.  Anyone here go to Chestertown 03-04 or St. Mary's City 05?

I was a member of CTY for 5 years and it is excellent! I would recommend most of all the Number Theory course at Lancaster. It is both the hardest class, the most interesting, and the teachers are by far the best I have had in my life.

I took the Game theory and probability course at lancaster session 2 a few years ago. Overall, I'd say it was a good experience. However, I did not feel my teacher was very good at explaining things. There was hardly any math or even probability taught, much to my surprise. Most of it was going over Nash equilibrium and prisoner's dilemna, etc. We did learn the basic calculus derivation laws though, like quotient, product, chain, etc. If I had the chance to do CTY over again, I would not take the course. Instead, I would have taken physics or chemistry, so I could have skipped physics/chemistry honors at my school and went directly to the AP courses in my freshman or sophomore year.

I would second the vote for the number theory course at Lancaster -- three of the most fascinating and mathematically exciting weeks of my life!!  A really wonderful and fun class.

Who are you mathilde? Maybe you were in my class.

CTY was pretty fun... but i always had ppl who were really stupid and stuckup in my sessions...i went to the site director every session lol...had to sit out on some socials (cruddy as they were)
im attending intro to lab or sumthing this year, any1 else?

How old were you, and how many years of "baby CTY" attendance or CTY attendance had you had before, when you took the number theory class? Have you taken any more number theory since then?

Yes, I was Chestertown 03, with Mr. Rozakis and the Writing class.

I have not yet done a math class through CTY. The humanities are awesome, but I signed up for Prob Game Theory this year. I wonder how well it's gonna be...

I wanted Crypto, but I didn't want Number Theory because I'm gonna take that next year at PROMYS. Would you say PROMYS-theo is better than CTY-Theo?

Interesting story about the number theory class, actually. No, I didn't go to baby CTY. But I got an 800 on the Math SAT before my last year and they still wouldn't let me in. So I had to yell at them a little bit before they finally did. I read a couple books about Number Theory and I read nsato's notes since then. I can now solve some of the easier IMO shortlist number theory problems.


Two different prime numbers between $ 4$ and $ 18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

$ \textbf{(A)}\  21 \qquad \textbf{(B)}\  60\qquad \textbf{(C)}\ 119 \qquad \textbf{(D)}\  180\qquad \textbf{(E)}\ 231$

[hide="Solution"]I get C.

Let the primes be $m$ and $n$, so that we're looking for $mn-(m+n)$. This can be factored as $m(n-1)-1(n-1)-1=(m-1)(m-1)-1$.

This obviously equals 119 when m=13 and n=11.  :)  [/hide]

Here's an easier solution:
[hide]
It is clear that the product is odd and the sum is even, so the difference will be odd. That rules out B and D.
The minimum value of our expression is 5*7-5-7=23, so that rules out A.
The maximum value of our expression is 13*17-13-17=221, so that rules out E.
Therefore the only answer choice left is C.
(Of course, while m and n increase, the expression mn-m-n increases too, because mn increases faster than m+n.)
[/hide]

[hide=Solution]Denote the numbers be $x,y$. Then we have $xy-x-y=(x-1)(y-1)-1$. Clearly the prime numbers are both odd, so $x-1, y-1$ are each even and the whole expression is odd. This leaves only $21, 119, 231$. If it's $21$, then $(x-1)(y-1)=22$, which does not work. If it's $119$, then $(x-1)(y-1)=120$, which works with $x=11, y=13 \implies 119 \implies \boxed{\text{(C)}}$.[/hide]


function
logarithms
algebra theorems
algebra

Is it true $f(x+y)=f(x)f(y)$ is given, taking $ln$ from both sides, then we will get
$lnf(x+y)=lnf(x)+lnf(y)$. After taking $ln$ does the finction will have any property?
 Abdurashid

If you do that, you're going to obtain all the solutions that are strictly positive for every $x$ (you assume $f(x)>0$ for every $x$), but you won't know anything about wether there are or not any more solutions, with some zero or negative values.

If you assume $f(x)>0$ for every $x$, the function $g(x)=\ln f(x)$ will have to be a solution to Cauchy's equation $g(x+y)=g(x)+g(y)$, continuous or not (the only continuous solutions are $g(x)=Cx$, but there are infinite non-continuous solutions).
The solutions with $f(x)>0$ for every $x$ are $f(x)=e^{g(x)}$, including the continuous $f(x)=B^x$, $B=e^C>0$.
For instance, you've already lost here the solution $f(x)=0$. And maybe other solutions with some zero or negative values.

If there exists an $a$ so that $f(a)=0$
then $f(x+a)=f(x)\cdot f(a)=0$
so $f(x)=0$  $\forall x \in \mathbb{R}$

What about $f(x)<0$  ?

$f(x)<0$ $\forall x \in \mathbb{R}$ cannot be true because then
$f(x+y)<0$ , $f(x)<0$ and $f(y)<0$ but this contradicts to the equation:
$f(x+y)=f(x)\cdot f(y)$

What about the mixed case ($f(x)<>0$)?

Let´s assume f(x) to be continous then we solved the problem or not?

By the way a question:
What would you call continous?
Does it mean that the function is continous everywhere? or only where the function is defined?

 If continous would mean the first thing then we really finished the problem.
Because then $f(x)$ could have no zeros otherwise $f(x)=0$ and so $f(x)<0$ or $f(x)>0$ but $f(x)<0$ is not possible so $f(x)>0$ and one could use $ln$.

If it would mean the second thing then I don´t know...

sqrt4


inequalities
geometry
parallelogram
perpendicular bisector
geometry proposed

Let A, B, C, D be four points in the Euclidean plane. Prove that the following two assertions are equivalent:

[b](a)[/b] For every point P in the plane, we have $AP+BP\geq CP+DP$.
[b](b)[/b] The points A, B, C, D lie on one line, the segment CD lies inside the segment AB, and the midpoints of the segments AB and CD coincide.

  Darij

If we make $P$ move freely on the perpendicular bisector of $AC$ we find that the condition can only be satisfied if the perpendicular bisectors of $AC,BD$ are parallel, i.e. $AC\|BD$. In the same way we prove $AD\|BC$. We now know that $A,B,C,D$ are the vertices of a parallelogram with $AB,CD$ being the diagonals. To show that they are in fact collinear, put $P=A$. This proves [b]a)[/b]$\Rightarrow$[b]b)[/b].

The other implication can be proved in lots of ways, I guess. For instance, we can translate the triangle $PAC$ until it reaches the position $P'DB$. Then our nequality is reduced to something like "in a convex quadrilateral, the sum of the diagonals is at least as large as the sum of two opposite edges", which is pretty obvious.


[quote="Victory.US"]Consider $ 15$ polynomial $ x^2 \minus{} p_ix \plus{} q_i$ , $ i \equal{} 1,2,...,15$
$ p_i,q_i \in \left\{ {1,2,...,15} \right\}$, $ p_i \ne q_i$ ,$ \forall i$

At most, how many roots of the polynomials $ > 20$ ?[/quote]

Let $ f_i(x)\equal{}x^2 \minus{} p_ix \plus{} q_i$ : $ f_i(20)\equal{}400\minus{}20p_i\plus{}q_i> 400\minus{}20\times 15 >0$ and $ f_i'(20)\equal{}40\minus{}p_i>0$.

So no root $ >20$


Homologation of alcohols.

Devise methods for converting an alcohol to its next higher homolog,

$ROH \longrightarrow RCH_{2}OH$,

and for converting an alcohol to its next lower homolog,

$RCH_{2}OH \longrightarrow ROH$.

i don't know whether this procedure is a classic one or not.

still
1.treat $ROH$ with $PX_{3}$ treating with this and not $HX$ reduces chances of rearrangement.
2.treat the obtained compound with $C_{2}H_{5}ONa$/$C_{2}H_{5}OH$
for 1,2 elimnation to occur.
3.follow this up by alkoxymercuration-demercuration.

for the 2nd part let me think....

Regarding your steps 2 and 3, what happens if there is no beta hydrogen available?
Let's also consider the special case $R = C_{6}H_{5}$.

i thought by convention $R$ was an Alkyl group

R can still be an alkyl group with no beta hydrogens.

Sorry saw the thread only now.
$ \ce{ROH}$ ---> $ \ce{RCH2OH}$
1. Convert ROH into RCl using $ \ce{SOCl2}$
2.Convert RCl into RCN using KCN
3. Convert RCN into RCOOH by acidic hydrolysis .
4. Use LAH

2. RCH2OH oxidise to RCOOH .
RCOOH ---> NH3 heat --->RCONH2
RCONH2 --->Hoffmann Bromamide rn (Br2,KOH) --->RNH2
RNH2 ---->HNO2 --->ROH

Of coures Phenol is a Special Case .

ROH->RCOOH
RCOOH+SOCl2->RCOCl
RCOCl+CH2N2->RCOCHN2
RCOCHN2+Ag2O->RCH=C=O
RCH=C=O+H2O->RCH2COOH
WHICH ON REDUCTION GIVES CORR ALCOHOL

Basically what you did is almost the Ardnt Eistert Synthesis
 :)

for the second part first oxidize RCH2OH->RCOOH Foll by HVZ and hydrolysis which on decarb gives corr prim alcohol

Why should it have an $ \alpha$ hydrogen?

how abt this. carry out dehydration ( preferably using p2o5 so that no rearrangement occurs) which gives alkene foll by chlorination and successive elimination to give a terminal alkyne foll by hydration wid hgso4 to give methyl ketone foll by haloform reaction and acidification and red to alcohol.

i know this involves a lot of steps but at least alpha h is req

Ya but still why should it have an $ \alpha$ hydrogen?

i meant alpha h is not req in the above steps

I mean how do you do it for tertbutylmethanol i.e. OH attached to primary C which is inturn attached to tert carbon

sorry i meant beta not alpha. got confused. i know this method cant be used for tert butyl alcohol but youll be degrading it to what. isopropyl alcohol????

you can dehydrate tert butyl alcohol to give alkene foll by ozonolysis and red to alcohol

No! I'm asking you in $ \ce{RCH2OH}$ what if R is tert butyl group?????

[quote="Madness"]1. Convert ROH into RCl using
2.Convert RCl into RCN using KCN [/quote]

How about if R is a tertiary group?

[quote="Madness"]RNH2 ---->HNO2 --->ROH[/quote]

Wrong.

[quote="VARUNARASU"]ROH->RCOOH[/quote]

How?

[quote="VARUNARASU"]WHICH ON REDUCTION GIVES CORR ALCOHOL[/quote]

Wrong: your alcohol have one extra carbon.

[quote="VARUNARASU"]Foll by HVZ and hydrolysis which on decarb gives corr prim alcohol[/quote]

Wrong.


group theory
abstract algebra
quadratics
Galois Theory
superior algebra
superior algebra unsolved

Let $ p$ be an odd prime and let $ \xi$ be a primitive root of unity of order $ p$ in $ \mathbb{C}$. Let $ E \equal{} \mathbb{Q} (\xi)$. Then $ E/Q$ is a Galois extension with Galois group $ G \cong (\mathbb{Z} / p \mathbb{Z})^{\times}$. Let $ H$ be the subgroup of index 2 in $ G$ and let $ \alpha : \equal{} \sum_{i \in H} \xi^{i}$ and $ \beta : \equal{} \sum_{i \in G \backslash H} \xi^{i}$ (where we identify $ H$ with the appropriate subgroup in $ (\mathbb{Z} / p \mathbb{Z})^{\times}$).

Prove that $ \alpha \beta \equal{} \frac {1 \plus{} p}{4}$ if $ p \equiv 3 (\mathrm{mod } 4)$ and $ \alpha \beta \equal{} \frac {1 \minus{} p}{4}$ if $ p \equiv 1 (\mathrm{mod } 4)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199531&search_id=1659080872]Here[/url] I proved $ \alpha\plus{}\beta\equal{}\minus{}1$ and $ (\alpha\minus{}\beta)^2\equal{}(\minus{}1)^{p\minus{}1\over 2}p$.
So $ \alpha\beta\equal{}{(\alpha\plus{}\beta)^2\minus{}(\alpha\minus{}\beta)^2\over 4}\equal{}{1\minus{}(\minus{}1)^{p\minus{}1\over 2}p\over 4}$.

Here is an outline of a more direct (though uglier) approach.

The first thing to notice is that $ \alpha\beta \equal{} \sum_{n\equal{}0}^{p\minus{}1} f(n) \zeta^n$, where $ f(n)$ is the number of ways in which $ n$ can be written as the sum of a quadratic residue and nonresidue $ (mod\ p)$.

You can show that $ f(n)\equal{}\frac{p\minus{}2\plus{}\left(\frac{\minus{}1}{p}\right)}{4}$ when $ p\not | n$, and $ f(n)\equal{}\frac{p\minus{}1}{4} \left(1\minus{}\left(\frac{\minus{}1}{p}\right)\right)$ when $ p | n$. (I did this by considering the sum $ \sum_{y} \left(\frac{x^2\minus{}y^2}{p} \right)$, for example)

It follows that $ \alpha\beta \equal{} \left(\frac{p\minus{}1}{4} \left(1\minus{}\left(\frac{\minus{}1}{p}\right)\right)\right) \minus{} \left(\frac{p\minus{}2\plus{}\left(\frac{\minus{}1}{p}\right)}{4}\right) \equal{} \frac{1\minus{}\left(\frac{\minus{}1}{p} \right)}{4}$.


Say you received a chain letter that read as follows:

Title: BECOME A MILLIONEARE!!!
Cc: (List of all the email addresses that end with @comcast.net)

HOW WOULD YOU LIKE TO BECOME A MILLIONEARE!? It's simple as 1, 2, 3!!!
1.) Just write your name at the bottom of this list
(List of 10 people with their addresses)
2.) Send $1.00 to each of these people
3.) Wait for others to send you $1.00!!!

It's so simple why didn't others think of this up! You'll become a millioneare in a matter of minutes! Join Now!
(CAUTION: IF YOU SEND THIS MESSAGE TO MORE THAN ONE PERSON YOU WILL BE SUBJECT TO WEST NORTH CAROLINE LAW 102-53/a AND VIOLATORS WILL BE PURSECUTED)

Why is this an idiotic chain letter? (no, I did not get this in real life I made this up, there is no "West North Carolina" and I spelled "millionare" wrong :lol: )

What does this have to do with Mathcounts??

Well this does require you to do some thinking, and I think these "real world" trickery will help people to get into the right mind set of solving difficult problems and coming up with creative solutions for them. I have changed the naming nonetheless.

P.S. It also gives someone time to relax after solving math problems all day, like that lollipop in Family/Consumer Science after a hard day :lol:

[hide]is it becausde the letter is sent by e-mail and to send the dollar, you have to have their addresses?[/hide]

Hehe, you're thinking too much real-worldy.  Think about what happens if ten people get sent the letter, and then each person sends it to ten more, and then . . .


calculus
integration
trigonometry
calculus computations

Evaluate:

1)  $ \int\frac{\tan x}{1\plus{}x^{2}}\text{d}x$

2)  $ \int\frac{\text{d}x}{\left(1\plus{}x^{2}\right)\tan x}$

[quote="borislav_mirchev"]If we try to move (1/(x.x+1)) under the differential sign we may solve them.[/quote]

Any more details?

[quote="vinskman"][quote="Kent Merryfield"]If those were arctangents rather than tangents, then this would be a straightforward substitution exercise from an ordinary calculus course.

With tangents there, it's just a mess. There is no good reason to believe that these antiderivatives are elementary functions.[/quote]

Yes, I think so.[/quote]

Wolfram Integrator says that it is very probably that there are no elementary functions.
You can check  here http://integrals.wolfram.com/index.jsp?expr=tg(x)%2F(1%2Bx^2)&random=false

Sinajackson, did you create this integrals by yourself or did you toke them from some source?

Hi Carcul! :) no i didnt! one of my friends was solving some defferential equation and the answer needed to evaluate the above integrals... i tried many ways but i coundnt do anything!
anyway thanks everyone! Yeah as [b]Svejk[/b] said, wolfram says its unsolvable!
strange...


let a,b,c be positive numbers
prove that $ \frac{a^2(b\plus{}c)}{b^2\plus{}c^2}\plus{}\frac{b^2(a\plus{}c)}{a^2\plus{}c^2}\plus{}\frac{c^2(b\plus{}a)}{b^2\plus{}a^2}\geq a\plus{}b\plus{}c$  :)

Sorry, Silouan, but your application of Chebyshev is not more correct than the spelling of my name in the first post...

Harazi's proof at http://www.mathlinks.ro/Forum/viewtopic.php?t=14577 is still unbeaten in simplicity.

  Darij


search Columbia USAMTS email articles MATHCOUNTS AMC

Just for fun, how far down the list of results when googling your name do you have to go, before you find a page talking about yourself??  When you search for your name, search with you first name first, and then your last name.  For example, John Doe would search John Doe on google.  So no just using your first name, or just your last name or anything.  That we, we can compare more easily.

Wow, according to google,
[code]Results 1 - 10 of about 739,000 for David Jimenez[/code]
Actually, I have found (and I have been found) people with my name around the world. There is a Doctor in the west coast of the US, there is a guy who is a HSchooler in Colombia, who also like Math Competitions, there is a journalist in Spain, there is a violin concertist in Austria, and I know that there has been a few more, but I can't recall right now!!

Not a lot of people have the name: 'calderbank' However, I found some stuff about a British socialist leader, and some Scottish stuff (town named Calderbank), then stuff about my dad, and then a quote from AoPS!!!!!!!!

1st for both my username (drspyder) and my name (Daesun Yim).

EDIT: 1st for Treething too.

First result: USAMTS Leaders page
There are 323,000 results, but most of the pages have one person with my first name and one with my last. If I put my name in quotes, there are only 238 results.

Then again, the number of results isn't always accurate. One time I searched for something and got about 500 results, then I searched again, excluding some word, and got 1000 results.

I couldn't even find myself on Google.
I googled my username, though, And I was the first one :D

I have 11 things...from honor rolls from schools, tourneys, ect.

I searched my name without quotes, got 712,000 results.   I got too bored looking for one about me.  Searched for name in quotes, got 5690 results.

I searched my username, apparantly "eryaman" is a german word or something like that.  6020 results.  I found one search result on the AOPS website on result number 281.

Hehe, it seems that even if I put it on quotes, my name is pretty popular

[code]Results 1 - 10 of about 14,200 for "David Jimenez". [/code]

hmm my real name (vincent dang) when googled i am the 4th one down and my two sn's dangvh and dragonvd24 are both 1st lol

[quote]Résultats 1 - 10 sur un total d'environ 28 pour "Igor Kortchemski"[/quote]

[quote="h_s_potter2002"]When you search for your name, search with you first name first, and then your last name.  For example, John Doe would search John Doe on google.[/quote]This is not a correct search for your name. It searches through pages that contain both "John" and "Doe" inside. There are, for some names, zillions of pages like that. The correct search is [code]"John Doe"[/code] With the " in place it will look for an exact phrase match ;)

there is a doctor, a secretary of the navy, and and executive in liverpool with my name (i'm staying annonomous), so i never can find the true me without adding something like math in the search.

There's a mathamatician supossedly with my name, and my dads email adress  :D

The first result is an AoPS quote. Because my first name's so rare, I guess...

Oh, yes, my name is Billy Dorminy. Look at all the results for me now!

I am all of those people.

If you search my name without quotes... it's the 10th one.
If you search my name with quotes... it's the 4th one.
If you search my name with quotes and then click on "Search for English results only"... it's the 2nd one. xD

:D

My name, which I will not disclose here, comes up as the 2nd through 12th results, when in quotes.  The first result is for some person with my name geting married to someone from Waldorf, wherever that is!  (I see why the US is dead last in Geograhy literacy!)

[quote="solafidefarms"]Jiucy, did you place? I was 4th in JV. One of the least of my accomplishments.[/quote]

no i didn't do well(second tier).  the questions was too hard

[quote="ExterAY"]My name closely resembles an international action movie star.
[hide]
Which means that i am never gonna to find me....... :( [/hide][/quote]

Umm... I found you. :P

If I search for "Annika Heckel", I get 41 pages, and they're all about me. *lol*

If I enter Annika Heckel without quotes, I get 1140 pages, and the first 16 ones are about me.

Doesn't seem like I have a very common name.  :lol:

Oh yeah? Well
Results 1 - 10 of about 7,400,000 on google.

mine didnt appear until something like in 400s. All the ones before were Doctors, PhDs, Lawyers, PHysicists, Pianists, and other people who shared my name. By the Way, that first  one under mine was from the 2005 USAMO qualifier's list.

First five results: all me, three different sites. I'm about half of the ~110 with my name in quotes.

If I leave my first name off, I have to go down almost 300 out of 68000 to find myself. Before then, I pick up a British actor (more than 50 of him), several towns and farms in England, mayors of another British town who have been dead 500 years, buildings at Oregon State, my father- actually some of those are AOPS/Mathlinks pages that have me by pseudonym, and many others.
I've even got a London pub on page 10.

It's a pretty rare name in the US, but obviously more common in England.

Still, that's only 68000 results.

Peter VDD -> the first (15?) are almost entirely mine, and far over 100 in total that's for sure :)
when using my full name I get more results but they're not mine since I hardly ever use my name. (I guess there I'd get 20 out of 100 first results)

Wow! I guess my name's rather unique.  I've never thought of it that way before because it sounds entirely American, but all 12 results of the search are about me.  Not even any concerning my dad who has the same name.

The first result is my mathlinks post.(My name is David Rhee)
But my name seems to be common. Most of the results are not me.

With quotes I found 13, all me.  Mostly ski race and honor roll results.  There is one with pictures, so no name here!


Solve

$ x^{2}\plus{}y^{2} \plus{}6x\plus{}2y\equal{}0$

$ x\plus{}y\plus{}8\equal{}0$.

[hide="Solution"]From the second equation, we get that
\[ x \plus{} y \plus{} 8 \equal{} 0\implies x \equal{} \minus{} y \minus{} 8\]
Substituting that into the first equation yields
\[ ( \minus{} y \minus{} 8)^2 \plus{} y^2 \plus{} 6( \minus{} y \minus{} 8) \plus{} 2y \equal{} 0\]
Simplifying...
\[ y^2 \plus{} 16y \plus{} 64 \plus{} y^2 \minus{} 6y \minus{} 48 \plus{} 2y \equal{} 0 \\
2y^2 \plus{} 12y \plus{} 16 \equal{} 0 \\
y^2 \plus{} 6 \plus{} 8 \equal{} 0 \\
(y \plus{} 4)(y \plus{} 2) \equal{} 0\]
Thus
\[ y \equal{} \minus{} 4 \text{ or }y \equal{} \minus{} 2\]
When $ y \equal{} \minus{} 4$, (substituting into the equation $ x \plus{} y \plus{} 8 \equal{} 0$)
\[ x \minus{} 4 \plus{} 8 \equal{} 0\implies x \equal{} \minus{} 4\]
When $ y \equal{} \minus{} 2$, (substituting into the equation $ x \plus{} y \plus{} 8 \equal{} 0$)
\[ x \minus{} 2 \plus{} 8 \equal{} 0\implies x \equal{} \minus{} 6\]
Thus, the solution set to this system of equations is $ \boxed{\{( \minus{} 4, \minus{} 4),( \minus{} 6, \minus{} 2)\}}$[/hide]


some one asked me this question but i cannot solve it.
abcd is a square with ab=8.

Let $ m = BX$, $ n = BY$. We are given $ m + n + \sqrt {m^2 + n^2} = 16$, or
\begin{align*}\sqrt {m^2 + n^2} = (8 - m) + (8 - n) & \implies m^2 + n^2 = (8 - m)^2 + (8 - n)^2 + 2(8 - m)(8 - n) \\
& \implies(8 - m)(8 - n) = 8m + 8n - 64.\end{align*}
But
\begin{align*}\tan(90^\circ - \angle XDY) & = \tan(\angle ADX + \angle CDY) \\
& = \frac {\tan\angle ADX + \tan\angle CDY}{1 - \tan\angle ADX\tan\angle CDY} \\
& = \frac {\frac {8 - m}{8} + \frac {8 - n}{8}}{1 - \frac {8 - m}{8}\frac {8 - n}{8}} \\
& = \frac {8(16 - m - n)}{64 - (8 - m)(8 - n)} \\
& = \frac {8(16 - m - n)}{64 - (8m + 8n - 64)} \\
& = 1,\end{align*}
so $ 90^\circ - \angle XDY = 45^\circ\implies\angle XDY = \boxed{45^\circ}$. Yeah sorry, I'm probably missing a much nicer synthetic solution or something.

Circle $ (d)$ with center $ d$ and radius $ da\equal{}dc$ is b-excircle of $ \triangle xby,$ tangent to $ xy$ at $ t.$
$ \angle xdy \equal{} \angle xdt \plus{} \angle tdy \equal{} \frac{_1}{^2}(\angle adt \plus{} \angle tdc) \equal{} \frac{_1}{^2}\angle adc \equal{} 45^\circ.$

Let $ Z$ be a point on ray $ XA$ such that $ CY\equal{}AZ$ then $ \Delta DAZ\equal{}\Delta DCY \Rightarrow DY\equal{}DZ.$
$ XZ\equal{}AX\plus{}AZ\equal{}AX\plus{}CY\equal{}16\minus{}YB\minus{}XB\equal{}XY$ then $ \Delta DXZ\equal{}\Delta DXY \Rightarrow \angle YDX\equal{}1/2 \angle YDZ\equal{}45^o.$


I don't know if this problem is possible or not, if so please help of you know how to do it. thanks :)

Question: Say you have two pieces of carpet: a 1x8- foot strip and a 10-foot square. How can you make a [u]single continuous cut in just one of the pieces[/u] so that you can combine the resulting [b]three[/b] pieces into a 9x12- foot carpet?

I havent found a possible way either. I looks like by the areas it's possible (but that's not the whole story), and you must cut up the square.

I do not think that this is possible. I will move it to the high school forum to make sure though.

I did a search :D

[url=http://www.bonus.com/bonus/nav/scooter/scooterize.htmp?loc=http%3A%2F%2Fwww%2Ebonus%2Ecom%2Fcontour%2Faskdrmath%2Fhttp%40%40%2Fmathforum%2Eorg%2Flibrary%2Fdrmath%2Fview%2F55365%2Ehtml&useragent=Mozilla%2F4%2E0+%28compatible%3B+MSIE+6%2E0%3B+Windows+NT+5%2E1%3B+SV1%3B+YPC+3%2E2%2E0%3B+%2ENET+CLR+1%2E1%2E4322%3B+yplus+4%2E1%2E00b%29&refv=1&referrer=http%3A%2F%2Fsearch%2Eyahoo%2Ecom%2Fbin%2Fsearch%3Fp%3DSearch%25201x8%2520carpet%252010x10&entry=bonus&SERVER=www%2Ebonus%2Ecom]Click here![/url]

Yeah I know that's a long address. :P
[color=green]
[size=75]mod edit: ever heard of writing "click here" etc. instead of the whole address? :roll: [/size][/color]

I'm wasn't sure what he meant by a single continous cut. I thought it meant a straight line. That would've been impossible  :lol:

[quote="Gyan"]FWIW:  with normal definition of "One continues cut" there can be other solutions too. A simple one is

Assuming the square is at (0,0)(10,0),(10,10),(0,10) Cut along one continous  "cut":
 start from (0,1) go to (10,1) ..
 go to (diagonally)  (9,0)  and then to (9,10) ....
You would now have  a 9x9 square, two (1x9) rectangles and two rightangle triangles( which can be joined to make
 1x1 square.)  .. and that to 1x8 piece and you are done..

(No one said that "one continues cut"  can not intersect itself ..[/quote]

I think that a cut is from a point on the perimeter of the square and terminates at another point on the perimeter.

[quote] think that a cut is from a point on the perimeter of the square and terminates at another point on the perimeter.[/quote]
FWIW  :  Mathenatically speaking, unless you explicitly   define  this  that way, a  "continous cut"  need NOT start from a perimeter (  for example a "hole"  can be cut inside with a  "continous cut ")   or end in a perimeter .. it just means that the cut is continous ....That's why people would use  the term "non itersecting" or  "a cut which starts from perimeter"   etc if they would like to make the problem clear..

Anyway this is matter of difinition so should not be given too much importance..or  taken  as too seriouslly..besides the original question said something like "three pieces" and the solution fails in that way as  there are now "more than four pieces"..:)

[quote="Gyan"]FWIW:  with normal definition of "One continues cut" there can be other solutions too. A simple one is

Assuming the square is at (0,0)(10,0),(10,10),(0,10) Cut along one continous  "cut":
 start from (0,1) go to (10,1) ..
 go to (diagonally)  (9,0)  and then to (9,10) ....
You would now have  a 9x9 square, two (1x9) rectangles and two rightangle triangles( which can be joined to make
 1x1 square.)  .. and that to 1x8 piece and you are done..

(No one said that "one continues cut"  can not intersect itself ..[/quote]

that results in more than three pieces. re-read the original question.

[quote="xxreddevilzxx"]

that results in more than three pieces. re-read the original question.[/quote]
Of course, but if you  [i]too[/i] perhaps re-read the above post, you would know that that point was not missed! :) [quote]Anyway this is matter of difinition so should not be given too much importance..
or taken as too seriouslly..
besides the original question said something like ...
"three pieces" and the solution fails in that way as there are now "more than four pieces"..Smile[/quote]

:)


Does anybody know how to prove "de la Loubere's Algorithm" (also called siamese or staircase methods)?
It states that for empty magic squares of odd side length, a complete magic square can be obtained by the following process:

1. Place a 1 in the top-center box.
2. Move up one square and to the right one square and place the next number in that box. If the box is already filled, then place the next number in the box directly below the previous square.
3. Repeat step 2 until you have a complete magic square.

wait, I don't understand how the algorithm works. I tried it but then I had to guess about what to do when I reached edges.

OMG lol this algorithm is awesome XD
i wish i knew about this algorithm back in elementary school when we were givin those crappy magic square puzzles for like candy and stuff..

yea i don't know how to prove it but I'm very interested to find out. :roll:

It's not an easy proof. One can read about it (and many more nice problems) in "Mathematical recreations and essays" from Ball and Coxeter


what does int mean in the question

limit of int(2x-1) as x approaches 7/2 from the right side?

Most likely the floor function: the greatest integer less than or equal to the argument.


trigonometry
limit
integration
function
calculus
logarithms
real analysis

Evaluate the following limit.

\[\lim_{n\to\infty}\left\{\int^1_0 \left(1+\sin \frac{\pi t}{2005}\right)^n dt\right\}^{\frac{1}{n}}\]

Oops! I mistook where the 2005 was - numerator or denominator. Pretend this post never happened.

But in this particular problem, isn't the answer $1+\sin(\frac{\pi}{2005})=f(1)$? The functon is defined on [$0,1$], isn't it?

Grobber is right. That's the $\infty$ norm of the function in question.

And what is

$\lim_{n\to\infty}(\int_{0}^{1} f^n(t) dt)^{\frac{1}{n}}$

where $f:[0,1]\rightarrow R+$ continous

$\max_{x\in[0,1]}f(x)$?

[I assumed $f$ takes non-negativ values, so we don't have any problems with those $n$'th roots]

I had the proof in the post I erased, so here it is again. Suppose that $f$ is continuous on an interval $I.$

$\max_{x\in I}|f(x)|=M.$

Note that for any $\delta>0,\,\exists$ an interval $J$ of length $|J|>0$ such that for $x\in J,\,|f(x)|>M-\delta.$

$\left(\int_I|f(x)|^pdx\right)^\frac1p\le|I|^{1/p}M.$

Taking limits, $\limsup_{p\to\infty}\left(\int_I|f(x)|^pdx\right)^\frac1p\le M.$

But $\left(\int_I|f(x)|^pdx\right)^\frac1p\ge|J|^{1/p}(M-\delta).$

So $\liminf_{p\to\infty}\left(\int_I|f(x)|^pdx\right)^\frac1p\ge M-\delta.$

Since this is true for arbitrary $\delta>0$, we have that

$\lim_{p\to\infty}\left(\int_I|f(x)|^pdx\right)^\frac1p=M.$

Note that we made very little use of continuity. All we really needed was that $f(x)\le M$ for (almost) all $x$ and that $f(x)>M-\delta$ for $x$ on a set of finite measure. Hence it applies to all $L^{\infty}$ functions, and is the inspiration for the notation of "infinity norm." This is all a standard texbook proof.

Note the absolute values. If you leave them off and allow $f$ to take on negative values, things could get messy. In the original problem, kunny's function was positive, so that's not a worry.

another classical pb:
$f$ continuous. Evaluate $lim (\int_a^b f(t)^{1/n}dt)^n$

We'd better assume that $f>0$ to avoid certain kinds of trouble.

If the previous answer was the infinity norm, this should be the "zero norm", whatever that is. In the continuum of $p$-means, the place of "zero mean" is taken by the geometric mean, so that's our clue about this.

The answer should be $\exp\left(\int_a^b\ln(f(x))\,dx\right)$, provided the integral converges.

I'll post a proof later today or tomorrow.

Better be careful about the measure of the underlying set. Correct that to

$\lim\left(\frac1{b-a}\int_a^b(f(t))^{1/n}dt\right)^n = \exp\left(\frac1{b-a}\int_a^b\ln (f(t))\,dt\right).$

If $b-a\ne1$ then alekk's original limit will in general diverge to either 0 or infinity.


A polynomial $p(x)$ is divided by $(x-1)^4$ and yields the residue

 \[ax^3+bx^2+cx+d.\]

Which residue do we obtain when we divide $p(x)$ by $(x-1)^2$ ?

$P(x)=k\cdot(x-1)^4+ax^3+bx^2+cx+d$
residue of P(x)/(x-1) is equal to residue of ax+bx+cx+d / (x-1)

Which is $(c-2b+3a)x+d-b+2a$ by a simple euclidan division

Hi Peter,

I too thought the same, but then presumed that there must be a 'simpler' way than direct division. Anyone knows any other technique?

Thanks,
Sambit

There is none. The only thing you can do is show that $ax^3+bx^2+cx+d = (ax+b-2a)(x-1)^2+ (c-2b+3a)x+d-b+2a$. If you think that's simpler than an euclidiean division, go ahead :D

really, the division is simple. ;)

Hello, the right technique have to give the right result   : $ (3a + 2b + c)X + (d - b - 2a)$ :D

$\displaystyle R(x) = Q(x)*(x-u)^n + \sum_{i=0}^{n-1}\frac{R^{(i)}_{(u)}}{i!}(x-u)^i$

 :cool: Do you know the derivative ? :cool:

[quote="Diogene"]$\displaystyle R(x) = Q(x)*(x-u)^n + \sum_{i=0}^{n-1}\frac{R^{(i)}_{(u)}}{i!}(x-u)^i$

 :cool: Do you know the derivative ? :cool:[/quote]

Not funny :(

You make me feel dumb. ;)

Ok and now serious, what has this to do with the thing? I honestly don't see the relevance? :?

:cool: I see P_VDD, You dont see that your residue is false  :cool:

Ah that you mean :D

Yeah, I noticed the minus sign. I didn't feel the need to correct it though, the smarter reader will have figured that out himself/herself. ;)

Nevertheless thanks for pointing it out :)


Case:
IA High School mock trial case
Website: http://tinyurl.com/29x6s5
Case Here:
http://tinyurl.com/ytknho

Lawyers/Witnesses:
Depends on how many people sign up

Ok, basically the idea would be for the trial to go the same way a normal one would, with each side giving opening & closing statements and direct & cross examination, all posted on this topic. The speeches and questions wouldn't have to be so formal and polished as in a real competition.

Anyone interested?

[color=green][i]URL's converted to TinyURL's by a moderator to avoid screen-stretching.[/i][/color]

I'm interested. But this goes into the GFF if you ask me.

^^ Agreed. I'm interested. How's this gonna work?

I'm interested. Would we have a judge for the competition, or a jury, though?

Are we using Federal Rules of Evidence, or are there condensed rules?
(I strongly suggest using condensed rules... Federal Rules are easy to misuse in competition)

Wouldn't it be a bench trial, with 3 judges?

I suppose 3 is better than one; I remember having a divorce lawyer judge a competition case for our team once, and it didn't go well (the man seemed to like yelling and brutal accusations much more than intelligent arguments). Not that I expect to see any divorce lawyers on AoPS, but I still hold that 3 is better than 1. If we can find 3 ppl willing to judge.

I call Court Crier, too.

[hide="  :rotfl: "]

All rise and remain standing!

Hear Ye! Hear Ye! The Mock Trial Court of the Art of Problem Solving online forum is now in session. The Honorable Justice [Judge’s screenname] presiding. All persons having business herein can now be heard. You may be seated.

[/hide]


geometry
geometric transformation
reflection
trigonometry
circumcircle
angle bisector

If AD is a bisector of the triangle ABC, which is equal to AC and perpendicular to ON, where O is the center of circumscribed circle and N is the intersection of altitudes then find the angles of the triangle ABC.

It is pretty easy...Since rays $ AH$ and $ AO$ are reflections over the angle bisector, and $ OH$ is perpendicular to $ AD$, $ AH\equal{}AO$....which means $ 2R\cos A\equal{}R$...so $ A\equal{}60$...then $ CAD\equal{}30$, so $ AC\equal{}AD$, $ C\equal{}75$, then $ B\equal{}180\minus{}60\minus{}75\equal{}45$.

A is angle A in the triangle and R is the circumradius.

Why did this have to be solved in fifteen seconds.. Isn't that a bit odd?

No, I used to occasionally make my math team do 15s problems...

Among other things, it helps with:
- Speed Reading
- Accurately reading
- Quickly figuring out the relevant concepts


we divide the set {1,2,...3n} into 3 subsets of n elements . show that there are three numbers x, y, z each from one subset such that x+y=z

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=26762&postorder=asc&highlight=schweitzer&start=15

And you will also find in this link another link to a generalization of that problem.

Pierre.


I don't like the solution on the wiki because it seems a little tedious, does anyone have a better solution?


geometry
perimeter
algebra
polynomial
congruent triangles
geometry proposed

If two triangles have equal area, the perimeter, and the length of median on one side, then the two triangles are Congruent.

Let $ S, s, m$ be the area, semiperimeter, and the C-median of a $ \triangle ABC.$

$ 4m^2 \equal{} 2a^2 \plus{} 2b^2 \minus{} c^2$

$ S^2 \equal{} s(s \minus{} a)(s \minus{} b)(s \minus{} c)$

$ \frac {4S^2}{s} \equal{} (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(s \minus{} c) \equal{} \left(c^2 \minus{} (a \minus{} b)^2\right)(s \minus{} c)$

Substituting $ c^2 \equal{} 2a^2 \plus{} 2b^2 \minus{} 4m^2,$

$ \frac {4S^2}{s} \equal{} \left((a \plus{} b)^2 \minus{} 4m^2\right)(s \minus{} c) \equal{} \left((2s \minus{} c)^2 \minus{} 4m^2\right)(s \minus{} c)$

Therefore, $ c$ is a root of the cubic polynomial

$ P_3(x) \equal{} x^3 \minus{} 5sx^2 \plus{} (8s^2 \minus{} 4m^2)x \plus{} \frac {4S^2}{s} \plus{} 4m^2s \minus{} 4s^3$

$ P_3(x)$ has either one, or three real roots. Some or all of the three real roots can be equal multiple roots. If it has three, not all equal, it has a local maximum and minimum. i.e., $ P'_3(x) \equal{} 3x^2 \minus{} 10sx \plus{} 8s^2 \minus{} 4m^2$ has two different real roots. Assume this is the case, $ x_1 \le x_2 \le x_3$ are the three real roots of $ P_3(x),$ not all equal, and $ x_{ \minus{} } < x_{ \plus{} }$ are the two different real roots of $ P'_3(x).$ Then $ x_1 \le x_{ \minus{} } \le x_2 \le x_{ \plus{} } \le x_3$ and

$ x_{\pm} \equal{} \frac {5s \pm \sqrt {s^2 \minus{} 12m^2}}{3}$

Assume $ c \ge x_{ \minus{} }$:

$ c \ge \frac {5s \minus{} \sqrt {s^2 \minus{} 12m^2}}{3} > \frac {5s \minus{} s}{3} \equal{} \frac {2(a \plus{} b \plus{} c)}{3} \Longrightarrow c > 2(a \plus{} b),$

which is a contradiction. Consequently, $ c$ can be only the smallest root of $ P_3(x),$ i.e., it is unique, given $ S, s, m.$ Then the C-altitude $ h \equal{} \frac {2S}{c}$ is also unique. Given segment $ c \equal{} AB$ with midpoint $ M,$ the remaining triangle vertex $ C$ is one of the intersections of a circle $ (M, m)$ and two parallels to $ AB$ at the distance $ h,$ producing only congruent triangles $ \triangle ABC.$


For any three positive reals $a,b,c$, we have

$\frac{ab}{a^2+2b^2}+\frac{bc}{b^2+2c^2}+\frac{ca}{c^2+2a^2}\le 1$

ok, i 'll try
put $ x = \frac ba , y = \frac cb , z = \frac ac $ and $ xyz = 1$
the inequality is equivalent to
$ \frac {1}{\frac 1x + 2x} + \frac {1}{\frac 1y + 2y} + \frac {1}{\frac 1z + 2z} \leq 1 $

scince $ x + \frac1x \geq 2$
we have
$ \frac {1}{\frac 1x + 2x} + \frac {1}{\frac 1y + 2y} + \frac {1}{\frac 1z + 2z}  \leq  \frac {1}{2 + x} + \frac {1}{2+ y} + \frac {1}{2+ z} $
we have
$\frac {1}{2 + x} + \frac {1}{2+ y} + \frac {1}{2+ z} = \frac { 6 + 4( x+y+z) + ( \frac 1x + \frac 1y + \frac 1z)}{ 8 + 4( x+y+z) + 2( \frac 1x + \frac 1y + \frac 1z)} $
the last expression is obviously $\leq 1 $

is there any mistake  :? ?

$xyz = 1 \Longrightarrow \frac{1}{2+x} + \frac{1}{2+y} + \frac{1}{2+z} \le 1$ is a special case of the following:

[b]Lemma:[/b] Let $n$ be a positive integer and let $x_1, \dots, x_n$ be positive reals with $x_1x_2\dots x_n = 1$. Then

$\frac{1}{n-1+x_1} + \cdots + \frac{1}{n-1+x_n} \le 1$

(Proved in post #8 at http://www.artofproblemsolving.com/Forum/post-286253.html#p286253 )

[quote="Christian Hirsch"][quote="Al-Khawarizmi"]
we have
$\frac {1}{2 + x} + \frac {1}{2+ y} + \frac {1}{2+ z} = \frac { 6 + 4( x+y+z) + ( \frac 1x + \frac 1y + \frac 1z)}{ 8 + 4( x+y+z) + 2( \frac 1x + \frac 1y + \frac 1z)} $
[/quote]
It should be [tex]\frac {1}{2 + x} + \frac {1}{2+ y} + \frac {1}{2+ z} = \frac { 12 + 4( x+y+z) + ( \frac 1x + \frac 1y + \frac 1z)}{ 9 + 4( x+y+z) + 2( \frac 1x + \frac 1y + \frac 1z)} [/tex][/quote]
i m sorry, i m really bad at calculs :oops: thank you christian ;)
it remains then to prove that $ \frac 1x + \frac 1y  + \frac 1z \geq 3$
in fact it's simple to see using AM-GM
we have
 $ \frac 1x + \frac 1y  + \frac 1z \geq 3 \sqrt [3] {(\frac {1}{xyz})} = 3 $
and it's done ;)
thank you mildorf for the generalization


inequalities
floor function
algebra unsolved
algebra

solve the Inequality:$ [x]x^2\leq\sqrt{2007\minus{}x^2}$ :maybe:

[hide="An X-mas quickie :)"]

Let $ \lfloor x \rfloor x =k$, and $ r = x - k \in [0,1)$. Then if $ k\leq 0$ obviously $ kx^2 \leq 0 \leq \sqrt{2007-x^2}$. We only need $ 2007 \geq x^2$, so all $ x\in [-\sqrt{2007}, 1)$ are solutions.

Let $ k\geq 1$. Then $ k^3\leq kx^2 \leq \sqrt{2007-x^2}$, which implies $ k^6<2007$ or ${ k\in\{1,2,3}$. It's easy to check that for $ k=1$ and $ k=2$ values of $ k$ (and $ x$ respectively) the inequality is satisfied. So $ [1,3)$ is part of the solution as well.

For $ k=3$, we have $ x\in [3,4)$ and $ y=x^2 \in [9,16)$, satisfying the inequality $ 3y\leq \sqrt{2007-y}$, or \[ 9y^2+y - 2007 \leq 0,\] therefore $ y \leq \dfrac { -1 + \sqrt { 2007 \cdot 4 \cdot 9 + 1} } {18 }$, which turns out to be between 9 and 16.

So the full range of solutions is
\[ I = \left[ -\sqrt{2007} , \sqrt { \dfrac { -1 + \sqrt{72253}}{18}} \right].\]
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