Given triangle $ABC$ with an arbitrary point $P$ prove that
$\frac{PA^{2}}{b}+\frac{PB^{2}}{c}+\frac{PC^{2}}{a}\ge\min\{a,b,c\}$

[color=purple] It's easy gemath :D.
    We always have that with all M in plane and $x,y,z >0$ then
   $xMA^{2}+yMB^{2}+zMC^{2}\ge \frac{yza^{2}+zxb^{2}+xyc^{2}}{x+y+z}$
(We get the above result by $(x\vec{MA}+y\vec{MB}+z\vec{MC})^{2}\ge 0$ )
 Now replace x=ca,y=ab,z=bc we get
  $caMA^{2}+abMB^{2}+bcMC^{2}\ge \frac{ca^{2}b.a^{2}+ab^{2}c.b^{2}+c^{2}ba.b^{2}}{ab+bc+ca}(*)$
   Your inequality is equipvalent to
  $\frac{MA^{2}ca+MB^{2}ab+MC^{2}bc}{abc}\ge\min{a,b,c}$ Without loss of generality we assume that$c=\min{a,b,c}$ then we need to prove
     $\frac{MA^{2}ca+MB^{2}ab+MC^{2}bc}{abc}\ge c$
  By $(*)$ to prove the following inequality we have to prove
$\frac{ca^{2}b.a^{2}+ab^{2}c.b^{2}+c^{2}ba.b^{2}}{ab+bc+ca}\ge c$
$\longleftrightarrow abc(a^{2}b+b^{2}c+c^{2}a)\ge c(ab+bc+ca)$
$\longleftrightarrow a^{2}b+b^{2}c+c^{2}a \ge abc+c^{2}b+c^{2}a$
$\longleftrightarrow a^{2}b+b^{2}c\ge abc+c^{2}b$
$\longleftrightarrow ab(a-c)+bc(b-c)\ge 0$ which is clearly true because $c=\min{a,b,c}$
   The proff is finihed and by similar method and applying Trebysev inequality we get the similar problem
  $\frac{MA^{2}}{a}+\frac{MB^{2}}{b}+\frac{MC^{2}}{c}\ge \frac{a+b+c}{3}$
:D [/color]


A square has the equation |x+6|+|y-7|=1. What is the product of the lengths of the  diagonals?

[hide="i think this is the answer"]
I changed the equation into $|y-7|=-|x+6|+1$. That gives two equations. $y=-|x+6|+1$ and $y=|x+6|+6$. Solve system and get the intersections as $(-5,7),(-7,7)$  Those two points are opposite each other on the square so they are a diagonal. Use the distance formula and see it is $2$. Since the diagonals in a square are equal, the other is $2$. $2*2=\boxed{4}$[/hide]

What else besides the lengths could it be? widths? then it would be 0

[quote="mathnerd314"]The product of the [b]lengths[/b] of its diagonals?

Just clarifying.[/quote]

I thought I put that?  :?  Well, it's self explanatory :P

Well, you edited it.

[quote="biffanddoc"]$|y-7|=-|x+6|+1$. That gives two equations. $y=-|x+6|+1$[/quote]

Assuming that you took the case where $y-7=-|y-7|$ (because the other case is in the other equation), this gives $-y+7=-|x+6|+1$ and thus $-y=-|x+6|-6$, therefore $y=|x+6|+6$, not $y=-|x+6|+1$. Tell me if I did anything funny.

No, I had $y=|x+1|+6$ in the second case

Oh... duh. :blush:  :oops_sign: I guess I just wasn't reading carefully enough. Or maybe I was reading but I just did something wrong when I assumed that the first case was the negative one.

So the positive case is first? That would explain it.


A clock  here has been experiencing a strange phenomenon
lately. At one time the clock showed the time as 5 minutes to 9, one
minute later it read 4 minutes to 9, two minutes later it showed as 4
minutes to 9, and 1 minute later it read 5 minutes to 9. What exactly
is wrong with the clock?

Please post the reason(s) in your response

[hide]
It is stuck between 5 to 9 and 4 to 9.
Pattern "5 4 5 4 5..."

[/hide]

Does this really have anything to do with math?
[hide]
It's a digital clock, and one of the lights isn't working, so 8s show up as 6s, and 9s show up as 5s.
[/hide][/hide]

[hide][quote] It's a digital clock, and one of the lights isn't working, so 8s show up as 6s, and 9s show up as 5s. [/quote]

If the light is broken so that an 8 appears as a 6  then a 4 wouldn't show up as a 4.[/hide]

The clock is not a digital one.  Sorry for not making it clear.


inequalities
number theory proposed
number theory

if $ a_1,a_2,...,a_n \geq 0$ are such that :
\[ a_1^2\plus{}...\plus{}a_n^2\equal{}1\] then find the Maximum value of product $ (1\minus{}a_1)...(1\minus{}a_n)$

This inequality may be proofed by mixing variable as follow
Let $ f(x) = (1 - x)$
One can prove that $ f(x).f(y) \leq f(0).f(\sqrt { {x^{2} + y^{2}})}$ For $ x,y < 1, \sqrt{x^{2}+y^{2}}<1$ by some simple ways
So we only need to consider the case when $ n-2$ variables are $ 0$
It's a not-hard part
So the Maximum is reached when $ 2$ variable equal to $ \frac {1}{\sqrt {2}}$ and $ n - 2$ others are $ 0$


[hide]
Factor to get $(x-4)(x+2) < 0$, so $-2 < x < 4$. The sum of the values of x is $-1+0+1+2+3$ or $5$.[/hide]

Find the sum of all integral values of x that satisfy:
$x^{2}-2x+6<14$

to explain better

you subtract 14 in both sides to get the right side to become 0.
so it becomes $x^{2}-2x-8$
factor that and follow the above.


Do you have  problems from IMC 2008 longlist?

my team leader said that they stopped publishing longlist at 2007, since some of juries want to use the problems after competition.


Prove that there exists at least one Prime Number between consecutive square numbers.

I think this result is first proven by Erdos.

i am thinking this can be done w/ piegon hole but an not sure how.  :blush:

[quote="B23"]It is a tough [b]unsolved[/b] problem at present.[/quote]

Really? I am pretty sure it's solved :maybe: .

See http://en.wikipedia.org/wiki/Bertrand's_postulate (the link doesn't work :( , but you can copy/paste it in the address bar; there are problems with the ' sign).

[quote]A similar and still unsolved conjecture asks whether for every $n > 1$, there is a prime $p$, such that $n^{2}< p < (n+1)^{2}$. Again we expect from the PNT that there will be not just one but many primes between $n^{2}$ and $(n+1)^{2}$, but in this case the error estimates on the PNT are not (indeed, cannot be) sufficient to prove the existence of even one prime in this interval.[/quote]

See also [url=http://mathworld.wolfram.com/BertrandsPostulate.html]here[/url].

[quote]A related problem is to find the least value of theta so that there exists at least one prime between $n$ and $n+O(n^{\theta})$ for sufficiently large $n$ (Berndt 1994). The smallest known value is $\theta = \frac6{11}+\epsilon$ (Lou and Yao 1992).[/quote]


I know that you can stack the enumerate command, but is there an efficient way to make LaTeX assign roman numerals to the upper level, capital letters to the second level, arabic numerals to the third, and lower-case letters to the fourth? For example,
[code]
I (item)
 A (item)
  1 (item)
  2 (item)
   a (item)
   b (item)
 B (item)
II (item)
...
[/code]

[code]\renewcommand{\theenumi}{[\Roman{enumi}]}
\renewcommand{\theenumii}{\Alph{enumii}}
\renewcommand{\theenumiii}{\arabic{enumiii}}
\renewcommand{\theenumiv}{\alph{enumiv}}[/code]This gives I. (A) 1. a. but if you'd prefer I A 1 a then use
[code]\renewcommand{\labelenumi}{\Roman{enumi}}
\renewcommand{\labelenumii}{\Alph{enumii}}
\renewcommand{\labelenumiii}{\arabic{enumiii}}
\renewcommand{\labelenumiv}{\alph{enumiv}}[/code]


Let a,b,c are positive real numbers.Prove that:
$ \sum \frac{a^3}{b^2\minus{}bc\plus{}c^2} \ge \frac{3(ab\plus{}bc\plus{}ac)}{a\plus{}b\plus{}c}$

[quote="bigbang195"]Let a,b,c are positive real numbers.Prove that:
$ \sum \frac {a^3}{b^2 \minus{} bc \plus{} c^2} \ge \frac {3(ab \plus{} bc \plus{} ac)}{a \plus{} b \plus{} c}$[/quote]

by CS, we have

$ \sum \frac {a^3}{b^2 \minus{} bc \plus{} c^2} \ge \frac {(\sum{a^2})^2}{\sum{a(b^2 \minus{} bc \plus{} c^2)}}$

so, we just need to prove that

$ (\sum{a^2})^2(a \plus{} b \plus{} c) \ge 3(ab \plus{} bc \plus{} ac)\sum{a(b^2 \minus{} bc \plus{} c^2)}, \qquad (1)$

WLOG, assume that $ a \ge b \ge c$, then

$ (1) \iff (2a^2(b \plus{} c) \plus{} a^3 \plus{} bc^2 \plus{} b^2c \minus{} abc)(a \minus{} b)^2 \plus{} ((a \minus{} b)(a^2 \plus{} b^2 \minus{} c^2 \plus{} 3ab \plus{} bc \plus{} ca) \plus{} (2c^2(a \plus{} b) \plus{} c^3 \plus{} ab^2 \plus{} a^2b \minus{} abc))(a \minus{} b)(b \minus{} c) \plus{} (2c^2(a \plus{} b) \plus{} c^3 \plus{} ab^2 \plus{} a^2b \minus{} abc)(b \minus{} c)^2 \ge 0,$

true.

[quote="bigbang195"]Let a,b,c are positive real numbers.Prove that:
$ \sum \frac {a^3}{b^2 \minus{} bc \plus{} c^2} \ge \frac {3(ab \plus{} bc \plus{} ac)}{a \plus{} b \plus{} c}$[/quote]
$ \sum \frac {a^3}{b^2 \minus{} bc \plus{} c^2} \ge a\plus{}b\plus{}c \ge  \frac {3(ab \plus{} bc \plus{} ac)}{a \plus{} b \plus{} c}$

The following nice inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ Prove that
\[ \frac{a^2}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{b^2}{a^2\minus{}ac\plus{}c^2}\plus{}\frac{c^2}{a^2\minus{}ab\plus{}b^2}\geq\frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2}\]

[quote="caubetoanhoc94"]
$ \sum \frac {a^3}{b^2 \minus{} bc \plus{} c^2} \ge a \plus{} b \plus{} c$[/quote]

Do you have a nice proof for that?  :)

[quote="kuing"][quote="bigbang195"]Let a,b,c are positive real numbers.Prove that:
$ \sum \frac {a^3}{b^2 - bc + c^2} \ge \frac {3(ab + bc + ac)}{a + b + c}$[/quote]

by CS, we have

$ \sum \frac {a^3}{b^2 - bc + c^2} \ge \frac {(\sum{a^2})^2}{\sum{a(b^2 - bc + c^2)}}$

so, we just need to prove that

$ (\sum{a^2})^2(a + b + c) \ge 3(ab + bc + ac)\sum{a(b^2 - bc + c^2)}, \qquad (1)$
[/quote]

I think this is a better way to prove this last inequality:

Multiply both sides with $ a + b + c$ and use inequality: $ (a + b + c)^2 \geq 3(ab + ac + bc)$ so we have to prove just this:
${ (\sum_{cyc}{a^2})^2 \geq \sum_{cyc}{a}*\sum_{cyc}a(b^2 - bc + c^2)}$ which is after rewriting equal to:
$ \sum_{cyc}{(a^4 - a^3b - a^3c + a^2bc)} \geq 0$. Clearly true by $ Schur$.


I live in India where the school system is such that just cramming and textbook stuff is stressed. There is absolutely no stress on experiments and most of our labs lack the equipment for the same.

But my [b]curiosity to understand Physics [/b]fully has kindled the interest for experiments in me. I would like to know if anyone can recommend [b]good Physics experiments[/b](not just the trivial one's to find resistance in series / parallel etc.) which I [b]can do myself in my house[/b]. What sort of materials would I require? Please help me!!! :lol:  :help:

[quote="Spoon"]the oil drop experiment is always fun :)[/quote]
Is this the free fall experiment? If so, you will also have to get a very good cemera.

yeah it was a joke :). It's doable with a computer and a CCD camera, but not something that could be done for Rushil.

Most physics experiment require "equipment" other than the very basic ones... so I don't know what to say.

Just beside me... HOW CAN THE MILLIKAN OIL DROP EXPERIMENT BE DONE IN YOUR HOME??????

Also, i'd like more recommendations... Thanks!!!

maybe cavendish's  experiment... measure G?

Wouldn't that require a highly sensitive equipment and condititns... something I doubt is even possible in the best of High school labs.

Does somebody have any unique or difficult experiments or ideas in mind???

well... he just used a mirror to amplify the twisting, no?
though, it might be hard to construct the apparatus, yes... =/ i was thinking about that too

I haven't done any of the "famous experiments" at home or in school (except for some of the really simple ones in school), but I did a lot of interesting experiments that were a problem in some competition - physics olympiad, correspondence seminars, etc. Those were quite simple but some of them very interesting. (Measure the dependence of pressure in a toy balloon from its radius. It's not really as you would expect it to be;-)

And I almost forgot - the Young Physicsist Tournament. [url=http://www.iypt.org/]www.iypt.org/[/url]

There are 17 research-type experimental problems, you have almost the whole year to work on them and prepare your presentation and then you go to the national competition and present & defend your solutions, oppose others solutions, and so on. It is a lot of work behind this, but it is really a great competition. We did some amazing experiments with friends in school (with no really special apparatus available, just the really low standard (which of course helps much compared to home)). We made a magnetic liquid, investigated hydraulic jump, built a large apparatus for the Einstein - DeHaas experiment, played with avalanches, made our own mirage, constructed a windcar (and tested it in out national hydrometeorological labs, where they have a wind tunnel;-), ... if you can get five peole into a team, I really advise you to take part in this competition!!

This is probably exactly what the doctor offered. Thanks, kubus. This competitoin does look inreseting indeed. I'll have to check it's feasibility in INdia.Suggestions are welcome.

Can you post some of your experiments or send them to me so that I could also try them. Thanks!!!

Any more recommendations!!

OH!!! India is not there in the IYPT!!!!! :(  :mad:  :cursing:  :furious:  :noo:  :censored:  :stink:  :read:  :chief:  :help:  :stretcher:

I will have to do a lot to get in but I think it's too late. This years competition has already happened and by next year , I will be going to college. I'm simply not lucky!!!

Anyways, all those experiments look interesting. DId you submit this year , kubus. What was your performace. Can we discuss these experiments on the net. Let me think about them a bit!!

Anyways, any other competitons or experiments (doable on your own in home) in ind?>!?!

I think I wanna bring back this forum.. Please respond Someone. I need you help!!!!

As I've already posted somewhere, check this out:

http://fykos.mff.cuni.cz/en/

If you click "archive" you can read the problem statements and there should be solutions somewhere as well. In each set there is at least one experimental task, although it's probably not exactly what you were looking for... Anyway, you can also try the "current tasks" link and send your solutions by e-mail in English. They'll send you back your solutions corrected and also the proper solutions. (And you should appear in the ranking then. Hopefully :) .) Anyway, I'd give it a try. The experimental problems which you were interested in in this competition are more of a "make up your own experiment" type but I find that much more interesting than following some strict rules on how to do this or that experiment.


geometry
number theory
prime numbers
number theory proposed

What is the quickest method to identify very large prime numbers ?

It is probably a useful method for finding large prime numbers and even more useful for identifying the sequence of prime numbers in areas where thesequence of primes isn't known (since the numbers are so big).

Consider the product 2*3*5*7*11.....p ie the product of all primes from to 2 to some prime number p.

If q is a prime greater than p, than why is it likely that 2*3*5...p + q = a prime number.

For example consider the product 2*3*5*7=210. And let us consider the consecutive prime numbers 11,13,17,19,23,29 and 31. We get

210+11=221
210+13=223
210+17=227
210+19=229
210+23=233
210+29=239
210+31=241

Notice that all of these numbers are prime except for 221. We can do a similar experiment with say 2*3*5*7*11=2310, except this time use the primes 13,17,19,23,29,31,37

2310+13=2323
2310+17=2327
2310+19=2329
2310+23=2333
2310+29=2339
2310+31=2341
2310+37=2347

In this instance were not so lucky in that we only have 2333, 2339, 2341 and 2347 being prime. Obviously this method gets worse the higher you go up. However what makes this procedure useful, is that it enables you to determine consecutive prime numbers (this is why you may have noticed that there are no prime numbers between the four primes written above, or in the six primes in the first example).

Anyone here would like take a stab at why we get these results (it might be easier than you think)?


topology
calculus
derivative
analytic geometry
group theory
abstract algebra
induction

Prove that there are no non-trivial homomorphisms from a compact Lie group to $(\mathbb Z,+)$.

(here, no topology is considered on $\mathbb Z$; "homomorphism" simply means "group homomorphism", i.e. no continuity assumptions are made)

Let $G$ be a Lie group, and let $f$ be a homomorphism from $G$ to $\mathbb{Z}$.
First, suppose that $G$ has dimension at least 1; we will deal with the zero-dimensional case later.
The doubling map $x\mapsto x\cdot x$ on $G$ has derivative $2\cdot I$ at the identity $e$ ($I$ is the identity map from the tangent space to itself). This derivative is invertible, so the doubling map has an inverse $h$ on some neighborhood of the identity. Impose coordinates on that neighborhood $U$, with $e$ mapped to 0. The derivative of $h$ in these coordinates at 0 is $\frac12I$, so it is a contraction on some ball $V$ centered at zero. For any $x\in V$ and all positive integers $n$, $h^{n}(x)$ exists.
The only integer that can be halved arbitrarily many times is zero, so $f(x)=0$ for all $x\in V$.

Now, we need a lemma to continue:
Lemma: If $G$ is any Lie group, the connected component $H$ of $G$ containing the identity is a closed, open, normal subgroup of $G$. The quotient $G/H$ is discrete.
[hide="Proof"]Let $U$ be a connected neighborhood of the identity. Let $V=\{x: x\in U\text{ or }x^{-1}\in U\}$. $V$ is a union of two intersecting open connected sets, so it is open. Let $V_{n}$ be the set of products of $n$ elements of $V$. $V_{n}$ is the image of the connected set $V_{n-1}\times V$ under multiplication, so it is connected (by induction on $n$). The union $W$ of the $V_{n}$ is a union of connected sets which all intersect at $e$, so $W$ is connected. $W$ is a subgroup by construction, and $W$ is open sice $W=\bigcup_{x\in W}Vx$. $W$ is also closed, since its complement is a union of cosets. An open, closed, connected set is a connected component, and $W=H$. Finally, $H$ is normal since $x^{-1}Hx$ is also an open, closed, and connected subgroup; by the above argument, $x^{-1}Hx=H$.
The quotient $G/H$ is discrete because $H$ is open; this makes $\{e\}$ open in the quotient, and therefore all sets are open.[/hide]

Now that $f$ is zero on a neighborhood of $e$ (which generates $H$), it is zero on all of $H$, and $f$ descends to a homomorphism from the discrete group $G/H$ to $\mathbb{Z}$. A quotient of a compact group is compact, so $G/H$ is compact and we have reduced to the zero-dimensional case.
All zero-dimensional Lie groups are discrete. A compact discrete group is finite. Every homomorphism from a finite group to $\mathbb{Z}$ is zero since all elements of finite order must be sent to zero, and we are done.

Thanks. My proof was quite different:

Just as you observed, if the group is finite dimensional, then it must be finite and we're done. Otherwise, the connected component of the group has some positive dimensional maximal torus $T$. Since a torus is divisible, the restriction of the homomorphism to $T$ is trivial. The homomorphism is constant on every conjugacy class, and since, by a well-known result of Cartan (I think), every element of the connected component is conjugate to some element in $T$, the restriction of the homomorphism to the connected component is trivial. Finally, we conclude basically as you did: the homomorphism is constant on each connected component, and there are only finitely many connected components, so it must be trivial.

The interesting thing is that the result is still true for arbitrary compact Hausdorff groups. A few days ago, when I posted the problem, I knew about the general version but couldn't prove it, so I posted this weaker version.  So let's state the follow-up:

Prove that there are no non-trivial homomorphisms from a compact Hausdorff group to $\mathbb Z$. (AMM, problem 6246, proposed by L. Washington and W. Parry)

(my proof of this has nothing to do with the problem posted above, concerning Lie groups :); they can be regarded as being completely different problems)


I want to post some problems which are the last post of inequalities of my in this forum. All of them have a very nice solution,

[b]Problem 1.[/b] Suppose that $a,b,c$ are three positive real numbers satisfying that \[(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=13.\] Find the maximum and minimum value of \[P=\frac{a^{3}+b^{3}+c^{3}}{abc}.\] [b]Problem 2.[/b] Prove that for all non-negative real numbers $a,b,c$ then \[\frac{ab}{a+4b+4c}+\frac{bc}{b+4c+4a}+\frac{ca}{c+4a+4b}\le \frac{a+b+c}{9}.\] [b]Problem 3.[/b] Suppose that $a_{1},a_{2},...,a_{n}$ are positive real numbers and \[a_{1}+a_{2}+...+a_{n}=a_{1}^{-1}+a_{2}^{-1}+...+a_{n}^{-1}=n+2.\] Find the maximum and minimum value of \[P=a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}+2a_{1}a_{2}...a_{n}.\] (another as follow is also nice, but you can try or not)

[b]Additional problem.[/b] If $a,b,c,d\ge 0$ and $a+b+c+d=4$ then \[(1+a^{2})(1+b^{2})(1+c^{2})(1+d^{2}) \ge (1+a)(1+b)(1+c)(1+d).\] Also, there're quite alot my problems which is unsolved in this forum now, some of them are very nice. Try to prove it!

[quote][b]Additional problem.[/b] If $a,b,c,d\ge 0$ and $a+b+c+d=4$ then \[(1+a^{2})(1+b^{2})(1+c^{2})(1+d^{2}) \ge (1+a)(1+b)(1+c)(1+d).\] [/quote]

To Thuan. This problem was used to introduce the book in Vietnamese (actually, I use exponent 4 and 3 replacing for 2 and 1). Actually, I create this problem first and by Holder, I replace to \[(1+a^{4})(1+b^{4})(1+c^{4})(1+d^{4}) \ge (1+a^{3})(1+b^{3})(1+c^{3})(1+d^{3}).\] because I hope some one can solve it by a simple way than me (I use SMV). Hi, not because I dont think it, (oh, exponet 4  is so hard to check by SMV, I only want to hide the simple problem, Thuan).

There're six problems used, Problem 2 is one of them.

For the minimum, we have the following inequality: \[(a+b+c)\left(\frac{1}{a}+\frac1b+\frac1c\right)-\frac{a^{3}+b^{3}+c^{3}}{abc}\leq6.\] The additional problem (hung cited) was proposed by Phan Thanh Viet.

but we can not find the minimum using your inequality pvthuan because your equality case is a=b=c which doesn't satisfy (a+b+c)(1/a+1/b+1/c)=13

I think problems are also in English Version of your book  :D
For problem 1:
$11-2\sqrt{3}\leq P \leq 11+2\sqrt{3}$
And my solution use substation and daravion method.  :D


so I took the AMC 12A and barely qualified for AIME with a score of 105. I heard the USAMO index is usually around 205+. So, I would need an 11 out of 15 on the AIME I to guarantee qualification into USAMO. However, I have not done any prep at all for AIME. What do you guys suggest to do? Is it worth it to spend the time needed to practice problems and study for the AIME, which is on March 17th or so, in hopes of qualifying for USAMO? (which I really want badly) the only problem is...how should I go about studying in 1 month to be proficient enough to get around 11 right out of 15 on the test? thanks!

u could try again with amc 12 b

hey u never know

[quote="the future"]one of my friends got a 100.5 on the AMC 12 last year and made MOP...so its possible[/quote]

really? I got 115.5 in AMC 12 and I'm desperate now. :o

[quote="the future"]one of my friends got a 100.5 on the AMC 12 last year and made MOP...so its possible[/quote]

who was this friend?

His initials are JS and he was Red Mop (yes he also took the 10-so that helped his index, but still with that 100.5, he ended up getting 9-10 on the AIME).

Obviosuly cognos doesnt have to check on the website to find out who he is  :P

Wuyonglin2004, I jsut read your other post and you are <10th grade?  Oh then you should defiently not be worried.  A 115.5 is a very good score for your grade, and you should definetly take the 10B (dont even think abotu the 12B) to try and improve on the index.  Just to give you some experience.  Last year I got a 109.5 on the 12A, but then took the 10B, and got a 144 (with 8 on AIME), so you obviously have a chance to improve your index to give you the best shot at making the math olympiad.  Just forget about your 12A score.

Hi "The future"
Can I ask you why you take AMC 12A last year instead of 10A?

[quote="Brut3Forc3"][quote="MathleteWannabe"]I have similar questions:

Is it too late to sign up for the B?

If our school doesn't offer the B and we take the AMC B at a local college location and qualify for AIME, are we allowed to take the AIME at our regular school when our school offers it?[/quote]
It really depends on the place where you have to sign up. Look around in your local community (mainly colleges/universities), ask your teachers... but do hurry since it's coming up soon.

For the second question: yes, you can take it at your school, even if you qualify for AIME at a different location.[/quote]

Thanks. They offer it nearby at a local college. It's actually an hour away, but I consider that lucky since it's the only one in the state. I'll just pray they are still accepting people.


[quote="Andreas"]The sequence of integers $u_{0}, u_{1}, u_{2}, u_{3}, ...$ satisfies $u_{0}= 1$ and \[u_{n+1}u_{n-1}= ku_{n}\] for each $n \geq 1$, where $k$ is some fixed positive integer. If $u_{2000}= 2000$, determine all possible values of $k$.[/quote]
[hide] Let $a=u_{2}$. Then we can solve for $u_{n}$ in terms of $k$ and $a$.
$u_{0}=1$, $u_{1}=\frac{a}{k}$, $u_{2}=a$, $u_{3}=k^{2}$, $u_{4}=\frac{k^{3}}{a}$, $u_{5}=\frac{k^{2}}{a}$, $u_{6}=1$
etc. It repeats every 6 terms.
$u_{2000}\equiv u_{2}=a$, which is not dependent on $k$, therefore, $k$ can be any positive integer as long as $u_{2}=2000$. [/hide]

[quote="lotrgreengrapes7926"][hide] Let $a=u_{2}$. Then we can solve for $u_{n}$ in terms of $k$ and $a$.
$u_{0}=1$
$u_{1}=\frac{a}{k}$
$u_{2}=a$
$u_{3}=k^{2}$
$u_{4}=\frac{k^{3}}{a}$
$u_{5}=\frac{k^{2}}{a}$
$u_{6}=1$
etc. It repeats every 6 terms.
$u_{2000}\equiv u_{2}=a$, which is not dependent on $k$, therefore, $k$ can be any positive integer as long as $u_{2}=2000$. [/hide][/quote]

[quote="Andreas"]The sequence of [b]integers[/b] ...[/quote]

[hide] The terms with fractions, $\frac{a}{k}, \frac{k^{3}}{a}, \frac{k^{2}}{a}$, must all be integers. If $\frac{k^{2}}{a}$ is an integer, then $\frac{k^{3}}{a}$ is also.

$\frac{a}{k}=\frac{2000}{k}$, so $k$ must be in the form of $2^{x}\cdot 5^{y}$, with $x=0,1,2,3,4$ and $y=0,1,2,3$.
$\frac{k^{2}}{a}=\frac{k^{2}}{2000}=\frac{2^{2x}\cdot5^{2y}}{2000}$, so $2x\ge4\Longrightarrow x\ge2$ and $2y\ge3\Longrightarrow y\ge2$.

Therefore, $k$ can be any number in the form $2^{x}\cdot5^{y}$ with $x=2,3,4$ and $y=2,3$. [/hide]


MATHCOUNTS induction quadratics algebra polynomial symmetry percent

[quote="jhredsox"]I'm just wondering...how much could the use of a good TI calculator (like a graphing one over the simple TI 31) w/ programs and features can help your Mathcounts score, mostly at state and National levels?
-jorian[/quote]
Oh, my, gods.
I don't get why everyone puts such a big groking emphasis on calculators. It matters for grok what kind of calculator you have as long as it can do the grokking arithmetic. All those groking programs are groking pointless if you know how to do the problem. Even the problem that was posted earlier that had 47 as a factor, 47 is the 15th prime, it isn't that hard to find that it's a factor.

[quote="7h3.D3m0n.117"]Well i think calculators make a difference. Since I'm in 8th grade I don't know all of the stuff the MATHCOUNTS Competition gives you so my friend (who is in 10th grade) programs formalaes we didn't memorize or we don't know yet.[/quote]
Oh, my, gods.
Formulas are useless if you can't derive them (note that I didn't say prove).
It's all well and all if you can prove a formula with induction, and it is a common proof method for contests, but the thing is, if you have to come up with the formula, you're in trouble. That is why being able to derive the formula is important.

hey, i was just asking how much they COULD help

jorian

p.s. i have a ti 31 or so:)

Actually, I do think calculators helpful to prime factorize and find roots of polynomials. As for formulas, it's generally a good idea to know their derivation, but it would be tedious to have to re-derive say, the sin addition formula at every contest.

[quote="sonny"][quote="7h3.D3m0n.117"]Well i think calculators make a difference. Since I'm in 8th grade I don't know all of the stuff the MATHCOUNTS Competition gives you so my friend (who is in 10th grade) programs formalaes we didn't memorize or we don't know yet.[/quote]

Yeah, don't memorize formulas that help solve a problem. Try to remember or derive the methods needed to solve one.[/quote]

I was just commenting on the memorization of formulas. Don't get me wrong, calculators help but memorizing formulas isn't too good.

Actually, in my opinion, there's really nothing wrong at all with memorizing formulas, just as long as you [b]understand[/b], and better yet, can [i]derive[/i] them. In fact, memorizing formulas can actually be helpful, as they can often save you a great deal of time in many competitions.
However, you really need to be sure you know where the formulas are coming from because that can be much more useful.
For example, many problems don't use a formula/theorem in their solution but rather the derivation (or possibly even the proof) of it.

[quote="jhredsox"]thats a great point neal:)

lemme change my question
Can a good calculator help you more than just a regular one? (it can, right?)

-jorian

@nebula, did u go on ur site?[/quote]
Maybe like a half point. Maybe. If you know how to solve the problems it won't help you.

[quote="jhredsox"]ok. i was just wondering if there are any features on ti models above 31 that help u in target:)

-jorian[/quote]
Well, there is this one button that looks like a ^ that is pretty useful.

Not really. As long as it is scientific, it doesn't really make a difference. (I solved a problem last year using trig)...

Ok, first of all, everyone memorizes formulas. I don't think you want to be trying to derive quadratic in the middle of a sprint test. The important thing is to know why they work and how to apply them.
Second, the only thing a calculator will help with is making you go faster. Instead of doing out quadratic, just use a program. Instead of using shoelace, use a program. There won't be anything that will automatically solve a problem.
You really don't need to worry about getting a fancy calculator. Just make sure you know how to do the problems. Instead of worrying about how a calculator will help you on mathcounts, see how well you yourself can do the problems.

I feel like a hypocrite right now, but I don't like the use of all these high-tech calculators.

Yes, arithmetic sucks. Use a calculator to find 7^8. You don't learn anything multiplying out. However, things like finding roots of really busy polynomials, if they appear on a contest, there has to be a quick way of doing it (symmetry, perfect squares/cubes, etc.).

Fine, get a ti-89. I'd just like to say I got first in MA with a TI-30 last year, as I forgot my TI-84, with a perfect target. In national practices, there was a grand total of two problems in which I utilized the TI-84's special properties.

Plus, for nationals they give you a graphing calculator (before the test....right?), so that certain people won't have advantages

im glad to say i use a

ti30
-jorian

Umm.. a graphing calculator is a HUGE improvement over anything else, that's a basic requirement.

[quote="jhredsox"]@ignite: a graph calc is a necessity?
-jorian[/quote]
No it isn't. Anyways, certain people with 89s still have advantages at nationals.

havn't they given calcs better than ti89's before?

-jorian

They usually give out the Calcs after the test, although I feel like they gave them out before hand last year????

In the last 9 years they have given out TI89s 4 times, TI84+silvers 2 times, TI83+silvers 2 times and TI92s once.  The 92 is not allowed on any test since it has a regular QWERTY keypad.

Pretty sure those are right since I've had a kid make Nats each of the last 9 years.

You are all making good points.  Calcs can make things faster, but if you can't do a problem without the calc, you probably aren't going to realize you could even use the calc to do it.

Mathcounts at least has come right out and said they are trying to write their Target and Team round questions so that the only need for the calculator would be computation.  While you might find that you need them for older tests you use to practice, newer tests use them less.  I rarely use a calc at all on Targets and Team questions except to Calulate percents or something.

Think of it this way though.  A carpenter is going to buy the best power drill he can afford.  He(or she) will want the most options and power.

If the question is "which Calc?"..get the best one you can afford.  If the question is "do I need a calc?" or "will it help?"  then that depends on the individual.

yes, i'm prob getting a ti-89(i made a remark in another topic)
-jorian

[quote="egghead91"]oh. I was told graphing calculators were required in MTHCOUNTS, or was I thinking of Math A?[/quote]

Calculators aren't required for MathCount. You should have one but you dont HAVE to have one or a graphing calculator.

note: i took the 2001 target w/o a calc and got a 1...mostly because i didn't know what i was doing
-jorian


[quote="bodan"]Suppose $a,b,c,d$ are positive real numbers with product 1. Prove the inequality \[\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+d)}+\frac{1}{d(1+a)}\geq2.\] More generally, if $a_{1},a_{2},\ldots,a_{n}$ are positive real numbers with product 1, prove the inequality \[\frac{1}{a_{1}(1+a_{2})}+\cdots+\frac{1}{a_{n}(1+a_{1})}\geq\frac{n}{2}.\] [/quote]
If I am not mistaken, this is shapiro's inequality.
If $a_{1}=\frac{x_{1}}{x_{n}},$ $a_{2}=\frac{x_{2}}{x_{1}},...,a_{n}=\frac{x_{n}}{x_{n-1}}$ then
$\frac{1}{a_{1}(1+a_{2})}+\cdots+\frac{1}{a_{n}(1+a_{1})}\geq\frac{n}{2}\Leftrightarrow$
$\Leftrightarrow\frac{1}{\frac{x_{1}}{x_{n}}\cdot(1+\frac{x_{2}}{x_{1}})}+...+\frac{1}{\frac{x_{n}}{x_{n-1}}\cdot(1+\frac{x_{1}}{x_{n}})}\geq\frac{n}{2}\Leftrightarrow$
$\Leftrightarrow\frac{x_{n}}{x_{1}+x_{2}}+...+\frac{x_{n-1}}{x_{n}+x_{1}}\geq\frac{n}{2}.$
If so, the second inequality is not true. :wink:


[quote="Yurmun Gur"]let rearrangement be a method to solve this problem: we know that a_i is bigger than 1. lets say its bigger than 2!
so we know that the sum of all a_i is bigger than 1.
let all this be sum of 1+2+3... and so on.
dealing with this mustery proves that the solution of the problem is quite obvious. its trivial. :roll:  :bored:
i have another problem for you:
let JARID be a regular pentagon, let ALPHA be an angle. prove that BETA is bigger than 104 and smaller than 106 degrees.
good luck!!
youll need it.[/quote]

I have a better idea, please never post again.


I saw a mexican movie (a good one) a few days ago with a friend, who is not native spanish speaker, but he learned not long ago, and he was not sure about the translation of the title of the movie to English, and he missed. Nevertheless, he missed by a blank space.

Let me explain: The movie is called [b]Amar te duele[/b] that translated means [b]To love hurts you[/b], and he asked if the title was [b]It huts to love you[/b], that in spanish would be [b]Amarte duele[/b] (you can expressed different, but this one is correct). If you read carefully, the difference between both sentences is just a space, but the sence changes a lot.

I was wondering if you can give examples in your own language of sentences or phrases that change their meaning if you change an accent, introduce a space or introduce a punctuation symbol, but not doing any other modification.

BTW, English is one totally acceptable language, if you find some phrase like that in English, but I can't think in one of them now.

[quote="djimenez"]BTW, English is one totally acceptable language, if you find some phrase like that in English, but I can't think in one of them now.[/quote]English is weird, for example in order to [b]fill out[/b] a form you have to [b]fill it in[/b]. :D

[quote="stupidkid"]In Chinese accents are so important.  For every syllable there are four accents (in Mandarin at least), so if the accent is different, most times the meaning is different.[/quote]

Yes, I have heard that for example the word for mother, if it is mispronounced it may mean some animal (I don't remember what animal, but that was told to me by a chinease), then, when foreing people is learning Mandarin they some time say stupid things, because of the difference in the accents, when they want to say something really different!

Mispronouncing the word "Mother" in Chinese might result in saying the word "Horse". (Both are pronounced 'Ma', but with different accents)

There are actually quite a few jokes about Chinese accents. I remember this one part of this joke: someone ordered some buns, in return he got newspaper. (Bao Ji?)

Derek's joke is a good example, if you didn't get it.  One "accent" way means a type of food (like a bun, with stuff inside), and the other means "newspaper".

I remember reading something about politician's trying to woo the "Hispanic vote" on their side by campaigning in Spanish. I remember reading something about some politician saying "embarazada", which means "pregnant" in Spanish, when he was trying to say the English word "embarrassed".

Well, actually, there is a nice comercial about that in Brazil. There is a Brazilian guy in an Argentinan beach and the guy goes to a place where they have a telephone say that he wants to do a "ligacion",  in Portuguese, "ligaçao" means call, in Spanish, "ligar" means in the popular language, flirt, and of curse, the girl gets unconfortable, and then the guy say "perdon, eu no queria dejarla embarazada" mixing portuguese and spanish, and wanting to say "I am sorry, I didn't want to embarass you", but saying "Sorry, I didn't want to get you pregnate".

Also, something that sometimes is quite funny is to see the translations that people do of some messages, specially safety messages, on some places. A few months ago in Kroger, there was a message saying: "Danger, hazardous floor" and the translation to Spanish: "Cuidado, piso azarozo" ("Danger, random floor")! The worst part is that someone got a lot of money for doing such messages.

One president - I don't remember which one - once tried to say "I desire to be with you" to a group of Poles in Polish but it translated to "I desire you."

In most newspapers, editors shorten the headlines in order to save money. In some cases, the headline can have several different meanings!

[url=http://www.funny2.com/headlines.htm]Funny Headlines[/url]
[/list]

会中文的朋友：
“老爷们”怎么念？
呵呵can mean both "noble man/male" depends on where do you stress(first/second)
you can feel it pretty interestingif you know Chinese!
haha
I'm really proud that I know this language though I feel bad cuz my Chinese is becoming worse...
I'd like to talk to Chinese people!!!
People speak Chinese, come to our school!!!
[u]www.lakefieldcs.on.ca[/u]

And I think English is pretty amazing that you can actually find a lot of "opposite" stuff
just like mentioned above-- "fill out"=fill in something.

[quote="bleumoose"]ah! another spanish one, im surprised you didn't mention, david:
the famous ad "Got Milk?" was translated into spanish, "?Tienes leche?" Which means, very literally, "do you have milk?" Sounds right, right? WRONG! That phrase, in general Spanish, means "Are you lactating?" BIG mistake.[/quote] :rotfl: :rotfl: :rotfl: :rotfl:   That's all I have to say.


vector
rotation
linear algebra
matrix
geometry
geometric transformation
reflection

I know this is really elementary but its also elegant. ;)

Consider a Vector $ \vec{v} \equal{} x\hat{i} \plus{} y\hat{j}$
In the same $ xy$ plane let both the co-ordinate axes be rotated by an angle of $ \theta$ anticlockwise. Let the new axes be $ \hat{i'}$ and $ \hat{j'}$.

$ \hat{i'} \equal{} Cos \theta \hat{i} \plus{} Sin \theta \hat{j}$
$ \hat{j'} \equal{} \minus{} Sin \theta \hat{i} \plus{} Cos \theta \hat{j}$

Let us find the component of $ \vec{v}$ along both the new axes and since these 2 are perpendicular we will have no overlap.

Component along $ \hat{i'} \equal{} \vec{v} . \hat{i'} \equal{} xCos \theta \plus{} ySin \theta$
Component along $ \hat{j'} \equal{} \vec{v} . \hat{j'} \equal{} \minus{} xSin \theta \plus{} yCos \theta$

Hence the Vector in terms of the new system is
$ \vec{v} \equal{} (xCos \theta \plus{} ySin \theta) \hat{i'} \plus{} ( \minus{} xSin \theta \plus{} yCos \theta) \hat{j'}$

Any comments/criticism/modifications/additions are welcome.

Comment: since there's a $ 2\times 2$ matrix (of a very special structure) sitting before us, why not use matrix notation?

---

Among $ n\times n$ real matrices, the set of [i]orthogonal matrices[/i] (conventionally denoted $ \text{O}(n)$) consists of all of the rotations and reflections of $ \mathbb{R}^n.$ A matrix $ A\in\text{O}(n)$ iff $ A^TA\equal{}I.$ If we want to leave the reflections out and just have the rotations, we also impose the condition $ \det(A)\equal{}1,$ cutting down to the [i]special orthogonal matrices[/i], or $ \text{SO}(n).$

It is well known that $ A\in\text{SO}(2)$ iff $ A$ can be written as $ \begin{pmatrix}\cos\theta&\minus{}\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}$ for some angle $ \theta.$


Prove or disprove.

For 0 \leq a<b<c \leq 2p+1   then a^p+b^p \leq c^p

All numbers are integers.

It seems to be true. All that we have to do is showing that

(2p+1)^p>=(2p)^p + (2p-1)^p
or (1+1/2p)^p > 1 + (1 - 1/2p)^p

I think it's not difficult.

Namdung


geometry
3D geometry
prism
circumcircle
inradius
geometric transformation
reflection

Let $ M$ be an arbitrary point on the circumcircle of $ ABC$ and let the tangents from this point to the incircle of the triangle meet the side $ BC$ at $ X_{1}$, and $ X_{2}$. Prove that the second intersection of the circumcircle of triangle $ MX_{1}X_{2}$ with the circumcircle of $ ABC$ (different from $ M$) is fixed and it coincides with the tangency point of the mixtilinear incircle in angle $ A$ with the circumcircle.

Note: The [b]mixtilinear incircle[/b] in angle $ A$ is the circle tangent to the sides $ AB$, $ AC$ and to the circumcircle of $ ABC$ internally.

Cosmin, I could  not resist. :D

Let $ (I)$ be the common incircle of the $ \triangle ABC, \triangle MX_1X_2$ with inradius $ r.$ Let $ D, E, F, Y_1, Y_2$ be the tangency points of $ (I)$ with $ BC \equiv X_1X_2, CA, AB, MX_1, MX_2.$ Inversion with center $ I$ and power $ r^2$ takes the vertices $ A, B, C, M, X_1, X_2$ to the midpoints $ A', B', C', M', X_1', X_2'$ of the sides $ EF, FD, DE, Y_2Y_1, Y_1D, DY_2$ of the contact $ \triangle DEF, \triangle DY_2Y_1.$ The circumcircles $ (O), (P)$ of the $ \triangle ABC, \triangle MX_1X_2$ go into the circumcircles $ (O'), (P'),$ of the $ \triangle A'B'C', \triangle M'X_1'X_2',$ identical with the 9-point circles of the two contact triangles with the common circumcircle $ (I),$ both with radius $ \varrho \equal{} \frac {_r}{^2},$ congruent. The incircle tangents $ BC \equiv X_1X_2, CA, AB, MX_1, MX_2$ go to the circles $ \Gamma_a, \Gamma_b, \Gamma_c, \Omega_1, \Omega_2$ with diameters $ ID \equal{} IE \equal{} IF \equal{} IX_1 \equal{} IX_2 \equal{} r,$ all congruent to $ (O'), (P').$ Since $ (A'I \equiv AI) \perp (EF \parallel B'C'),$ $ B'I \perp C'A',$ $ C'I \perp A'B',$ the inversion center $ I$ is the orthocenter of the $ \triangle A'B'C'$ and similarly, the orthocenter of the $ \triangle M'X_1'X_2'.$ The congruent circles $ (O'), \Gamma_b, \Gamma_c$ with radii $ \varrho \equal{} \frac {_r}{^2}$ meet at $ A'.$ A circle $ (A', r)$ with center $ A'$ and radius $ r$ is tangent to all three at points diametrically opposite to $ A'.$

The internal bisector $ AI$ of the $ \angle A$ passing through the inversion center $ I$ is carried into itself. The mixtilinear incircle $ (K)$ and the mixtilinear excircle $ (L)$ of the $ \triangle ABC$ in the $ \angle A$ are the only two circles centered on $ AI$ and tangent to $ CA, AB, (O).$ Since the inversion center $ I$ is the similarity center of a circle and its inversion image, only the images $ (K'), (L')$ of $ (K), (L)$ are centered on $ AI$ and tangent to $ \Gamma_b, \Gamma_c, (O').$ The mixtilinear excircle $ (L),$ lying outside of the circumcircle $ (O)$ and outside of the inversion circle $ (I),$ has both intersections with $ AI$ on the ray $ (IL.$ Since the inversion in $ (I)$ has positive power $ r^2,$ its image $ (L')$ also has both intersections with $ AI$ on the ray $ (IL$ and it is centered on the ray $ (IL.$ It cannot be identical with the circle $ (A', r)$ centered on the opposite ray $ (IA.$ Therefore, the circle $ (A', r)$ is the inversion image of the mixtilinear incircle $ (K)$ in the $ \angle A.$ The inversion image of the tangency point $ Z$ of $ (K), (O)$ is the tangency point $ Z'$ of $ (A', r), (O'),$ the diametrically opposite point of $ (O')$ with respect to $ A'.$

Let $ A_0, B_0, C_0, O_1, O_2$ be centers of the congruent circles $ \Gamma_a, \Gamma_b, \Gamma_c, \Omega_1, \Omega_2.$ Since $ \Gamma_a$ is a reflection of $ (O')$ is $ B'C'$ and $ A'O', A'I$ are isogonals of $ A'B', A'C',$ the quadrilateral $ B'Z'C'I$ is a parallelogram, therefore, its diagonals $ B'C', IZ'$ cut each other at half at the midpoint of $ B'C'.$ Then the quadrilateral $ IA_0Z'O'$ is also a parallelogram, because its diagonals $ IZ', O'A_0$ cut each other in half at the midpoint of $ B'C',$ and $ IA_0 \parallel O'Z',$ $ IA_0 \equal{} O'Z' \equal{} \varrho.$ Since $ A_0I \equal{} A_0X_1' \equal{} O_1I \equal{} O_1X_1' \equal{} \varrho,$ the quadrilateral $ A_0IO_1X_1'$ is a rhombus and $ O_1X_1' \parallel IA_0 \parallel O'Z'.$ In addition, $ O_1X_1' \equal{} O'Z' \equal{} \varrho,$ hence, the quadrilateral $ O_1X_1'Z'O'$ is a parallelogram and $ O'O_1 \parallel Z'X_1',$ $ O'O_1 \equal{} Z'X_1'.$ Obviously, $ P'M' \equal{} P'X_1' \equal{} O_1M' \equal{} O_1X_1' \equal{} \varrho,$ the quadrilateral $ P'X_1'O_1M'$ is a rhombus and $ P'X_1' \parallel M'O_1.$

As a result, the $ \triangle P'X_1'Z' \cong \triangle M'O_1O'$ with $ O'O_1 \parallel Z'X_1',$ $ O'O_1 \equal{} Z'X_1'$ and $ P'X_1' \parallel M'O_1,$ $ P'X_1' \equal{} M'O_1 \equal{} \varrho$ are congruent by SAS and $ P'Z' \equal{} M'O' \equal{} \varrho.$ But this means that $ Z' \in (P')$ and consequently, $ Z \in (P),$ which is what we were supposed to prove.

Thank you for the nice solution, Vladimir! I was pretty sure you were going to [i]invert[/i] things around here.  :)

It appears that I had the same solution as Vladimir. It would be interesting to see a proof without inversion.

Dear Cosmin and Mathlinkers,
a syntetical proof without inversion of this result can be found on on my website :
http://perso.orange.fr/jl.ayme   vol. 4   A new mixtilinear incircle adventure III, p. 21
Sincerely
Jean-Louis

Mr Zhonghao Ye found an interesting result about the problem:
The circumcircle of triangle $ MX_{1}X_{2}$ is always tangent to a fixed circle internally.
The circle is constructed as follows:
Let the mixtilinear incircle in angle $ A$ touch the circumcircle at $ A'$. Let the tangent of the incircle which is parallel to $ BC$ cut the circumcircle at $ B',C'.$ then the fixed circle is the mixtilinear incircle in angle $ A'$ of triangle $ A'B'C'.$

Another similar problem (Zhonghao Ye's discovery):
When moving point $ M$ lies on the another circle (call it $ \omega$) which is different from the circumcircle of triangle $ ABC$. We have
(I): There exist two fixed circles which are tangent to the circumcircle of triangle $ MX_1X_2.$ ( The two fixed circles are also tangent to $ \omega$)
(II):The locus of the circumcenter of triangle $ MX_1X_2$ is an ellipse.

This is not correct. The $ \triangle ABC$ and its circumcircle $ \odot(ABC)$ are irrelevant here, just a fixed tangent $ l$ of $ (I)$ is required. For $ M \in \omega,$ the circumcenter of $ \triangle MX_1X_2$ lies on a conic; this can be either ellipse or hyperbola, even if $ (I)$ is outside of $ \omega.$ It can be only hyperbola, when $ (I)$ is inside of $ \omega.$ That's because $ \odot(MX_1X_2)$ is tangent to 2 fixed circles $ (F_1), (F_2),$ their centers being foci of the conic.

[color=blue]The point is to show that the foci $ \color{blue} F_1, F_2$ are collinear with the center $ \color{blue}I$ of the circle $ \color{blue}(I).$ Moreover, the circles $ \color{blue}(I), (F_1), (F_2)$ have a common similarity center, and when the arbitrary circle $ \color{blue}\omega$ degenerates to an arbitrary line, this similarity center falls on the tangent $ \color{blue}l$ of $ \color{blue}(I)$.[/color]

The general problem has 2 important special cases:

[b](1)[/b] $ (I)$ and $ \omega$ are incircle and circumcircle of some triangle $ \triangle ABC$ with $ l \equiv BC.$ Then one of the circles $ (F_1), (F_2),$ degenerates to a point on $ \omega,$ the tangency point of the A-mixtilinear incircle of $ \triangle ABC.$

[b](2)[/b] The arbitrary circle $ \omega$ degenerates to a line. This leads to Thebault theorem and more:
Let $ \omega$ be a line through the vertex $ A$ of a $ \triangle ABC,$ $ (O), (I)$ its circumcircle and incircle, $ (F_1), (F_2)$ circles tangent to $ \omega, BC$ and internally tangent to $ (O).$ Then $ I, F_1, F_2$ are collinar (Thebault theorem). In addition: If $ A' \in \omega$ is arbitrary, tangents to $ (I)$ from $ A'$ cut $ BC$ at $ B', C',$ then circumcircle $ (O')$ of $ \triangle A'B'C'$ is also tangent to $ (F_1), (F_2).$

See [url]http://www.mathlinks.ro/viewtopic.php?t=275799[/url] for an approach. The only bit not proved there follows from a simple lemma: Let $ (I, r)$ be an arbitrary circle and let $ \mathcal S_1, \mathcal S_2$ be 2 concentric circles centered at an arbitrary point $ S,$ such that the difference of their radii is equal to $ r,$ but otherwise arbitrary. Let $ (F_1), (F_2)$ be inversion images of $ \mathcal S_1, \mathcal S_2$ in $ (I).$ Then the circles $ (I), (F_1), (F_2)$ have a common similarity center.

[b]Another synthetical proof:[/b]
First I will change the names of points (see in the figure)
Let $ ME\cap (O)\equal{}\{P\}, MF\cap (O)\equal{}\{N\}$. It's easy to see that $ PN$ is tangent to $ (I)$, therefore $ (I)$ is the incircle of triangle $ MNP$.
Denote $ K\equal{}(MEF)\cap (O), L$ the midpoint of arc $ BC$ which contains $ A$.
So we will show that $ K,I,L$ are collinear which follows that $ K$ is the tangency of A-Mixtilinear incircle and $ (O)$.
$ \Leftrightarrow KI$ is the bisector of $ \angle BKC (1)$
We have $ \angle KBE\equal{}\angle KNC, \angle BEK\equal{}\angle KMF\equal{}\angle KCN$ therefore $ \angle BKE\equal{}\angle CKN$
Thus $ (1)\Leftrightarrow KI$ is the bisector of $ \angle EKN. (2)$
Denote $ MI\cap (MEF)\equal{}\{Q\}, MI\cap (O)\equal{}\{R\}$
We get $ \angle EKQ\equal{}\angle EMQ\equal{}\angle RKN$
Thus $ (2)\Leftrightarrow KI$ is the bisector of $ \angle RKQ$.
$ \Leftrightarrow \frac{KR}{KQ}\equal{}\frac{IR}{IQ} (3)$
Since $ \Delta PKN\sim \Delta EKF$ we get $ \frac{KR}{KQ}\equal{}\frac{RP}{QE}\equal{}\frac{RI}{QI}$
Therefore $ (3)$ is true. We are done.

My solution:

Let $ I_a $ be the $ A $ -excenter of $ \triangle ABC $ and $ T=AI \cap \odot (ABC) $ .
Let $ I' $ be the incenter of $ \triangle MX_1X_2 $ and $ K=\odot (BCII_a) \cap \odot (X_1X_2II') $.
Let $ Y=AI \cap BC, Y'=MI' \cap BC, P=\odot (MX_1X_2) \cap (ABC) $ .

Since $ \angle IKI'=\angle IKI_a=90 $ ,
so $ K, I', I_a $ are collinear .
Since $ (I,I';Y',M)=(I,I_a;Y,A)=-1 $ ,
so $ AM, BC, I_aK $ are concurrent at $ Z $ ,
hence $ AZ \cdot ZM=BZ \cdot ZC=I_aZ \cdot ZK $ . i.e. $ A, K, M, I_a $ are concyclic
Consider $ \{\odot (ABC), \odot (MX_1X_2), \odot (X_1X_2II') \} $ and $ \{ \odot (ABC), \odot (MX_1X_2), \odot (BCII_a) \} $
we get $ S=BC \cap PM \cap IK $ is the radical center of $ \{ \odot (ABC), \odot (MX_1X_2), \odot (X_1X_2II'), \odot (BCII_a) \} $ .

Since $ SI \cdot SK=SP\cdot SM $ ,
so $ I, K, P, M $ are concyclic ,
hence $ \angle TPI=\angle TPM+\angle MPI=\angle I_aKM+180-\angle IKM=90 $ .
i.e. $ P $ lie on $ \odot (IT) $ or $ P $ is the tangent point of $ A- $ Mixtilinear circle with $ \odot (ABC) $

Q.E.D


number theory
greatest common divisor
trigonometry
geometry
3D geometry
sphere
ratio

Hi,

I'm working through some problems in Ahlfors' Complex Analysis, and some feedback would be appreciated!

1.  Let the roots of z^n=1 be denoted as the nth roots of unity, call them "w"

Now, I'm given the following equation:

1-w^h + w^(2h)-w^3h+.....+(-1)^(n-1)w^(n-1)h

where gcd (h,n)=1

I'm supposed to find the value of the above expression.  I'm stuck what to do. I did the "obvious" thing and simply plugged in w, in terms of cosines and sines, and tried to evaluate (applying DeMoirver's Formula along the way), but I couldn't obtain any sort of cancellation with the values.  Is there something I'm missing or doing wrong?

Riemann Sphere:

1. Show z and z' correspond todiametrically opposite points on the Riemann sphwere if and only if z  times (conjugate of z')=-1

I don't even know where to begin...

2.

Find the radius of the spherical image of the circle in the plane with center a and radius R.

I don't understand the question, I think.  Am I supposed to write out the equation of a circle in the complex plane, apply the stereographic projection and ..?  Well, anyways, any thoughts would be appreciated...

But it's alternating..I don't see how one can apply facts from a geometric series to this...

Alternation is just a multiple of $ \minus{}1$ in the ratio.

I don't see why that should be true,

so for instance, if we have,

1+z+z^2+......z^(n) we can apply the geometric series identity.  So the sum should be

(1-z^(n+1))/(1-z)

But if the terms are alternating, I don't see why the identity, (with a plus/minus 1) should still old.  Sorry to belabor a point, but I don't know what that should hold, if I'm understanding you correctly.

For #1 regarding the Riemann sphere, first just do a cross section of it along the real axis.  In the figure below I let $ z\equal{}2$.  Now, can you interpret the diagram below and the equation $ z\overline{w}\equal{}\minus{}1$ implies that the projections of $ z$ and $ w$ onto the Riemann sphere are opposite one another along an equatorial circle on the sphere at least in the case when $ z$ is on the Real axis?

Repeating what I said- it's still a geometric series, but the ratio is $ \minus{}z$ rather than $ z$.

#2 on Riemann Sphere:

News to me circles make circles but most of what I read in here is news.  The plot below maps the brown circle $ (x\minus{}2)^2\plus{}y^2\equal{}1$ onto the Riemann sphere as the red circle via the transformations:

$ u\equal{}\frac{2x}{x^2\plus{}y^2\plus{}1}$

$ v\equal{}\frac{2y}{x^2\plus{}y^2\plus{}1}$

I projected the extremities of the brown circle as blue and green lines intersecting the red circle.  Using this geometry and perhaps Law of Sine or Cosine, I suspect the diameter of the red circle could be determined.  Not sure though.

The riemann sphere questions can be done by raw calculation. For #1 just map $ (x_1, x_2, x_3)$ and $ (\minus{}x_1, \minus{}x_2, \minus{}x_3)$ with stereographic projection and see that the relation holds.

For #2 you can calculate the projected circle as the intersection of a plane and the sphere. Follow the normal vector from the origin to the plane to find the midpoint of the projection. Then calculate the radius. Of course, I admit this is quite messy ...

I'm sure that's true without a doubt.  Oh yea, I guess $ (\minus{}x,\minus{}y,\minus{}z)$ really is on the other side . . . thanks.  I'd like to propose another way which I haven't verified yet but I think would work:

Express the circle in the complex in the form: $ C: (x\minus{}h)^2\plus{}(y\minus{}k)^2\equal{}r^2$.  Now draw a line from the origin through the center of this circle.  The equation of that line is $ L: y\equal{}k/hx$.  Now solve these two equations simultaneously to obtain the points of intersection on the circle: $ P_1: (a,b)$ and $ P_2: (c,d)$.  Project then these two points onto the Riemann surface to obtain the points: $ R: (x,y,z)$ and $ T: (u,v,s)$.  Then the diameter of the circle on the Riemann surface is just the distance between these two points: $ d\equal{}|R\minus{}T|$.  I'm not sure about this though.

Why would $ R$ and $ T$ be diametrically opposed wrt the circle on the riemann sphere?

I'm not sure.  Kinda' new for me.  Would they not?  Would it be so only when the circle in the Complex Plane is outside the unit circle?  Maybe I need to think about it more.   Hopefully too, lil' Fermat is catching all this and gaining something from it as it's his thread. :)


Let be $ ABC$ a triangle with $ AB\equal{}AC$, $ \angle A\equal{} 20^ \circ$ and $ M \in (AB)$ so that $ AM\equal{}BC$. Fiind the measure of $ \angle AMC$.

[hide="Hint=Construction"]Erect equilateral $ \Delta ACX$ with $ X,B$ on opposite sides of $ AC$[/hide][hide="Proof"]$ \angle MAX \equal{} 60^0 \plus{} 20^0 \equal{} \frac {180^0 \minus{} 20^0}{2} \equal{} \angle ABC$.
$ \Delta MAX\equiv \Delta ABC$

$ XC \equal{} XM$
$ \angle AMC \equal{} 80^0 \plus{} \frac {180^0 \minus{} (60^0 \minus{} 20^0)}{2} \equal{} \boxed{150^0}$[/hide]

I guess you`re wrong. The answer of this problem is $ \angle AMC \equal{} 150^ \circ$

[quote="I"][hide="Hint=Construction"]Erect equilateral $ \Delta ACX$ with $ X,B$ on opposite sides of $ AC$[/hide][hide="Proof"]$ \angle MAX \equal{} 60^0 \plus{} 20^0 \equal{} \frac {180^0 \minus{} 20^0}{2} \equal{} \angle ABC$.
$ \Delta MAX\equiv \Delta ABC$

$ XC \equal{} XM$
$ \angle AMC \equal{} 80^0 \plus{} \frac {180^0 \minus{} (60^0 \minus{} 20^0)}{2} \equal{} \boxed{70^0}$[/hide][/quote]
I was too stupid.
$ 80^0 \plus{} \frac {180^0 \minus{} (60^0 \minus{} 20^0)}{2} \equal{} 150^0\not \equal{} 70^0$
 :P
I've edited

Draw the equilateral triangle BCD, with A and D on the same side of BC, see that the triangles AMC and CDA are equal, hence <ACM = 10 deg and <AMC = 150 deg.

[i]Note: [b]D is the excenter of the triangle BCM, hence <BMC = 30 deg and, subsequently, <AMC = 150 deg.[/b][/i].

Best regards,
sunken rock


The coefficients of the equation $ ax^2\plus{}bx\plus{}c\equal{}0$, where $ a\ne 0$, satisfy the inequality $ (a\plus{}b\plus{}c)(4a\minus{}2b\plus{}c)<0$. Prove that this equation has $ 2$ real distinct solutions.

Let $ f(x)\equal{}ax^2\plus{}bx\plus{}c$, $ (a\plus{}b\plus{}c)(4a\plus{}2b\plus{}c)<0\Longleftrightarrow f(1)f(2)<0$.


geometry
quadrilateral
midpoints
IMO Shortlist

$ ABCD$ is a quadrilateral with $ BC$ parallel to $ AD$. $ M$ is the midpoint of $ CD$, $ P$ is the midpoint of $ MA$ and $ Q$ is the midpoint of $ MB$. The lines $ DP$ and $ CQ$ meet at $ N$. Prove that $ N$ is inside the quadrilateral $ ABCD$.

Let's denote $S_{F}$ the area of figure $F$.
It is obvious one of the angles $BCD$ or $CDA$ is acute.
Therefore we easily obtain that point $N$ is between lines $CD$, $BC$, $AD$.
Suppose point $N$ is outside or the quadrilateral $ABCD$ $\Rightarrow$ $S_{TBC}+S_{TCD}+S_{TDA}>S_{ABCD}$.
$BQ=QM\Rightarrow S_{BNQ}=S_{NQM}$, $S_{BQC}=S_{QCM}$ $\Rightarrow S_{NBC}=S_{NCM}$
Similarly $S_{NDM}=S_{NDA}$.
$CM=MD \Rightarrow S_{NBC}=S_{NCM}=S_{NMC}=S_{NDA}=X$
$S_{TBC}+S_{TCD}+S_{TDA}=4X$
Let the distance between $N$ and $AD$ be $y$ and between lines $BC$ and $AD$ be $h$.
$S_{AND}=X=\frac{1}{2}yAD=\frac{1}{2}(h-y)BC \Rightarrow y=\frac{h BC}{BC+AD}$
$S_{ABCD}<S_{TBC}+S_{TCD}+S_{TDA}=4X=4\frac{1}{2}\frac{hBC}{BC+AD}AD=\frac{2h AD BC}{AD+BC}<h\frac{AD+BC}{2}=S_{ABCD}$ we used here AM-GM. Contradiction.

[quote=polya78]I'm not sure why no one has posted a solution here yet. The problem is not very hard.  Let the lines through $P,Q$ parallel to $\overline {BC}$ and $\overline{AD}$ intersect $\overline {CD}$ in $P',Q'$ and let the parallel line through $N$ intersect $\overline{AB},\overline{CD}$ in $Q,L$.   Assume WLOG that $CD=4$, and let $r=BC$,$s=AD$, $y=CL$.

We need to show that $LN \le LQ$.  Because of the similarity of triangles $CQQ'$ and $CNM$, we have that $LN=ry/2$.  Similarly, $LN=(4-y)s/2$.  Solving yields $y=4s/(r+s)$ and $LN=2rs/(r+s)$.  Now $LQ=(sy+r(4-y))/4 =(r^2+s^2)/(r+s)$, so the desired result follows from the AM-GM inequality.[/quote]
you said that $\triangle PQQ'\sim \triangle PMN$ means $QQ'||MN\implies MN||AD$ how do you proved that?

we will use the coordinates $(D;\vec{DA},\vec{DC})$
we put
D$(0,0)$
A$(1,0)$
C$(0,c)$
B$(b,c)$
and thus
M$(0,c/2)$
P$(1/2,c/4)$
Q$(\frac{b}{2},3c/4)$
solve for N
N$(\frac{2b}{b+1},\frac{bc}{b+1})$
now it's clear the the coordinates of $N$ are positive
so it suffices to show that $N$ is to the left to $AB$ and down of $CB$
the equation of $AB$ is $y=\frac{cx-c}{b-1}$
so the if a point is to the left of that line it must satisfy the inequality:
$y<\frac{cx-c}{b-1}$ if $b<1$
or
$y>\frac{cx-c}{b-1}$ if $b>1$
[hide]because of the sloop of $AB$[/hide]
which is clear if we put $N$ into the inequality
and because $c>\frac{bc}{b+1}$
$N$ is down of $CB$
and we win :D
[hide]we can suppose WLOG that $c=1$ but what I did is safer[/hide]

It's obviously sufficient to show that $N$ lies on the same side of $AB$ as $D$. We use barycentric coordinates with respect to $\triangle ABD$. Let $C=(-c,1,c)$. Take $c>0$, then we have to show that the $z$-coordinate of $N$ is non-negative. Indeed, $M=(-c:1:1+c),P=(2-c:1:1+c),Q=(-c:3:1+c)$, so $\overline{DP}: x=(2-c)y$ and $\overline{CQ}: 0=(1-2c)x+cy-2cz$. Intersecting these two lines yields $N=\left(\frac{c(2-c)}{c+1},\frac{c}{c+1},\frac{(c-1)^2}{c+1}\right)$ and the last coordinate is $\geq 0$, as desired.

A nice problem for coordinates!
anyway D=(0,0) A=(1,0) and C=(0,m) and B=(n,m). now M=(0, m/2) and P=(1/2, m/4) and Q=(n/2, 3m/4) so DP is y=mx/2, and CQ slope is m/2n, now if we solve for N, we get N=(2n/(n+1), mn/(n+1)), now we need to show that N is below BC, so we need to prove m>(mn)/(n+1) which follows because 1>n/(n+1)

yay


Solve in positive integers: $7^x+x^4+47=y^2$

Why isn't it possible, in a bit more detail?

[quote="Singular"]Consider modulo 4; then x must be even.  Since the LS is even, then y must be even, so the RS is 0 mod 4.  But this isn't possible.  Hence, no solution.[/quote] :|

$7^4+4^4+47=52^2$

$(x,y)=(4,52)$ is a solution...

Well, clearly $x > 0$. Note that $(-1)^x + x^4 - 1 \equiv y^2 \equiv 0,1 \pmod{8}$ so $x$ must be even ($x$ odd yields $-1+1-1 \equiv -1$ for the LHS)

For even $x$, we have

$(7^{x/2})^2 < 7^x+x^4+47 < (7^{x/2}+1)^2 = 7^x + 2\cdot 7^{x/2} + 1$ for $x > 4$. So $y$ can't be an integer ($7^x+x^4+47$ lies between two consecutive squares)

We can check up to 4 to get the solution (4, 52).


Prove the following property of determinants \[ \det (A \cdot B) = \det A \cdot \det B .  \]

In a way, my teacher gave it "as is", with no proof for $n > 2$.

I guess the proof should appear in most textbooks, but I don't have one which includes it... I have just an older one

because i tried to prove it, but things got very messy. hence i want to know the proof because i don't know how to prove it

and I guess it shouldn't be too hard

Do row elimination on $A.$ Each row operation can be considered left multiplication by an elementary matrix $E_k$ There are three types of row operations. Each of the three types has a predictable and provable effect on a determinant, and each corresponding type of elementary matrix has a readily computable determinant. We verify that the effect of the row operation on $A$ is exactly the same as multiplying by the determinant of the elementary matrix. (This has to be done three times, once for each case.)

Two things can happen at the end of the row elimination for $A.$ Either we can complete factor this as $A=E_kE_{k-1}\cdots E_2E_1,$ or we wind up with $A=E_kE_{k-1}\cdots E_2E_1 P$ where $P$ has a row of zeros across the bottom.

Let's concentrate on the first case, which happens iff $A$ is invertible. In that case, our work referenced above shows us that $\det A=\det E_k\cdots \det E_2\cdot\det E_1$ Suppose that $B$ is also invertible. Then $B=F_j\cdots F_2F_1$ and $\det B=\det F_j\cdots\det F_2\cdot\det F_1.$ Then we have an elementary factorization of $AB$ and can say that $\det AB=\det E_k\cdots \det E_2\cdot\det E_1\cdot\det F_j\cdots\det F_2\cdot\det F_1=\det A\cdot\det B.$

As for the other case: we have (from this very work) that $\det A=0$ iff $A$ is not invertible. We can also show that $AB$ is not invertible iff either $A$ is not invertible or $B$ is not invertible. Thus $\det AB=0$ iff $\det A=0$ or $\det B=0.$

Thanks for the proof, Kent. It's great. I knew about that "elimination" algorithm, but I didn't think it had so nice applications.

Also, is it a standard textbook proof of this property? I'm just curious, because all calculation-based attempts seem doomed to me

Yes, this is a standard textbook argument.


On the sides of the triangle ABC,in the exterior,build squares.Let $ A_1$ be the center of the square oposite to A.The same way is $ C_1$ and $ B_1$.Prove that $ AA_1,BB_1,CC_1$ are passing thru the same point V.

I will prove that $ B_1C_1\perp AA_1$ (and similarly for the other pairs). That will mean that $ AA_1, BB_1, CC_1$ will be concurrent at the orthocenter of $ A_1B_1C_1$.

Suppose that we have built the square $ ABDE$. Then it is easy to prove that $ AB_1C_1$ will be transformed   to $ EAC$ by a spiral similarity with center A, angle 45 and ratio $ \frac{\sqrt{2}}{2}$. Then we can prove that $ BEC$ can be transformed into $ BA_1C$ by a spiral similarity centered at $ B$, with angle 45 and ratio $ \sqrt{2}$. Thus by the composition of these two spiral similarities it is clear that $ B_1C_1$ can be transformed into $ AA_1$ by a spiral similarity with angle $ 45\plus{}45\equal{}90$ (and ratio $ \frac{\sqrt{2}}{2}*\sqrt{2}\equal{}1$).

Therefore $ AA_1\perp B_1C_1$ (and equal) which is what we wanted :)

q.e.d.


Putnam
vector
linear algebra
matrix
college contests

How would you approach and motivate the solution to this problem?

$ M$ and $ N$ are $ n \times n$ matrices. If $ (MN)^2 \equal{} 0$, must $ (NM)^2 \equal{} 0$?

First thought: we can probably find such matrices. (Because matrices are super uncommutative)

Let's just look at matrices that only take unit vectors to 0 or to other unit vectors, because that's the easiest thing to deal with. A little experimentation makes me think that we're not gonna get anywhere with two by two matrices - although, I did notice that if $ N$ takes $ e_1 \to e_2 \to 0$, $ M$ takes $ e_1 \to e_1$, $ e_2 \to 0$, then we get $ NM \ne 0$, $ MN \equal{} 0$.

So it makes sense to try something similar with three by three matrices: $ N$ takes $ e_1 \to e_2 \to e_3 \to 0$, $ M$ takes everything to itself except for $ e_3$, which it takes to $ 0$. Then, since $ (MN)^2$ takes $ e_1$ to $ 0$, it definitely takes everything else to $ 0$ (because having a higher subscript only brings you to $ 0$ faster), while $ (NM)^2$ brings $ e_1$ to $ e_3$, and thus is nonzero. Hooray!

Well, my approach:

At first we note that this is true for exactly a single value of $ n$ and false otherwise. So we study the problem in two different cases.

[b]Lemma 1:[/b] If $ A$ and $ B$ are $ 2 \times 2$ matrices and $ (AB)^{2} \equal{} O_{2}$ then $ (BA)^{2} \equal{} O_{2}$.

[i]Proof:[/i] $ (BA)^{3} \equal{} BABABA \equal{} B(ABAB)A \equal{} B(AB)^{2}A \equal{} BO_{2}A \equal{} O_{2}$. But it is well-known that if a $ 2 \times 2$ matrix $ M$ satisfy $ M^{3} \equal{} O_{2}$ then also $ M^{2} \equal{} O_{2}$. If it is not well known a simple computation with the entries of the matrix show us that result. Then $ (BA)^{2} \equal{} O_{2}$ as desired.

From the [b]Lemma 1[/b] the partial answer to the problem is, "Yes it must, if $ n \equal{} 2$".

But this is not the case when $ n > 2$. In fact, you can construct easily a counterexample.


function
complex analysis
complex analysis unsolved

Can anyone help in solving this question
Show that an isolated singularity of f can not be a pole of e^f.
\

thx in adavance

$e^x$ has an essential singularity at $\infty$. The range of an analytic function in a neighborhood of an essential singularity is dense in the plane.


limit
parameterization
function
calculus
calculus computations

I am new to small oh or big Oh. I found the following statement somewhere

[quote]
...we have
$A = \frac{b}{8c} + \sqrt{\frac{3(1-p)}{4cp} + \left(\frac{c-10}{3c}\right)^2}$.
Observe that
$A = \sqrt{\frac{3}{4cp}} + o(1/\sqrt{p})$,
for sufficiently small p.
[/quote]

The $b, c$ are constant scalars and $0 \leq p \leq 1$. I see that

$\lim_{p \to 0} \frac{A - \sqrt{\frac{3}{4cp}}}{\frac{1}{\sqrt{p}}} = \lim_{p \to 0} \frac{\sqrt{p}b}{8c} + \sqrt{\frac{3p(1-p)}{4cp} + p\left(\frac{(c-10)}{3c}\right)^2} - \sqrt{\frac{3p}{4cp}} = 0$.

But how the author figured out $1/\sqrt{p}$? I find $1/p^n$ for $n \geq 1$ is also valid.

[quote="Hazel"] But how the author figured out $1/\sqrt{p}$? I find $1/p^n$ for $n \geq 1$ is also valid.[/quote]
He just wanted to show that the error is less than the main term in the right hand side expression. As to $1/p^n$ with $n\geq 1$, don't forget that $p$ tends to $0$, not to $\infty$, so to be $o(1/\sqrt p)$ is a stronger property than to be $o(1/p^n)$ with $n\ge 1$.

Could you show me how to arrive at $1/\sqrt{p}$? (Is there a standard way of computing $o(\cdot)$?)

You mentioned about [b]stronger property[/b]; how do you compare one is better the other?

[quote="Hazel"]Could you show me how to arrive at $1/\sqrt{p}$? (Is there a standard way of computing $o(\cdot)$?)[/quote]
This is just the computation of the limit you did, neither less, nor more. The notation $f(x)=o(g(x))$ as $x\to a$ means, by definition, that $\lim_{x\to a}\frac{f(x)}{g(x)}=0$.
[quote="Hazel"]
You mentioned about [b]stronger property[/b]; how do you compare one is better the other?[/quote]
If $f(x)=o(g(x))$ as $x\to a$ and $h(x)>g(x)$ for all $x$ close to $a$, then $f(x)=o(h(x))$ as well.

By the way, have you read Kent's stickie about big Oh, little oh? If you are still confused with what these symbols mean, it'll, probably, answer most of your questions.

[quote="fedja"][quote="Hazel"]Could you show me how to arrive at $1/\sqrt{p}$? (Is there a standard way of computing $o(\cdot)$?)[/quote]
This is just the computation of the limit you did, neither less, nor more. The notation $f(x)=o(g(x))$ as $x\to a$ means, by definition, that $\lim_{x\to a}\frac{f(x)}{g(x)}=0$.[/quote]
What my question was, for example $\sqrt{n^2 + 1} = n + O(?)$, as $n \to \infty$,  how do one determine the $``?\"$ ?

[quote="fedja"]
[quote="Hazel"]
You mentioned about [b]stronger property[/b]; how do you compare one is better the other?[/quote]
If $f(x)=o(g(x))$ as $x\to a$ and $h(x)>g(x)$ for all $x$ close to $a$, then $f(x)=o(h(x))$ as well.
[/quote]
So $g(x)$ is stronger. I got you now. Thanks.

One more question: Does $O(1)$ mean not larger than a constant regardless of any parameter?

[quote="Hazel"]
What my question was, for example $\sqrt{n^2 + 1} = n + O(?)$, as $n \to \infty$,  how do one determine the $``?\"$ ?
[/quote]
You mean how to determine the best possible $``?\"$? Sometimes it is a tricky business; usually people think of some parametrized scale of functions (like $n^\alpha$, $\alpha\in\mathbb R$) and try to figure out what the best possible value of the parameter is. Sometimes this is impossible and you just write the best $O$ you can prove. Sometimes you just need some fixed order of growth/decay of the error in your formula and just prove it without caring whether you can do any better (this is definitely what happened with the formula in your first post: one just wanted to know that the error was much less then the main term and did not even try to get the best possible estimate for it). In your last example, however, everything is fairly straightforward: you can use the Taylor expansion of the square root and get $\sqrt{n^2+1}=n\sqrt{1+\frac 1{n^2}}=n(1+O(n^{-2}))=n+O(n^{-1})$.  Here $O(n^{-1})$ can actually be replaced be $\frac 1{2n}+O(n^{-3})$, so you know that the "error" is, in fact, comparable to $n^{-1}$, not just dominated by it. Taylor expansions are where the $O$ notation appears most often.
[quote="Hazel"]
One more question: Does $O(1)$ mean not larger than a constant regardless of any parameter?[/quote]
Some people use it this way, though it is more traditional to use $O(1)$ in the same way as any other $O$, i.e., in expressions like $f(x)=O(1)$ as $x\to a$ (say, $1/x=O(1)$ as $x\to\infty$, but not as $x\to 0$).


The sequence $ (x_n)$ is given by

$ x_1\equal{}\dfrac{1}{2}$ and $ x_n\equal{}\dfrac{2n\minus{}3}{2n}\cdot x_{n\minus{}1}$, for all $ n\ge 2$, $ n\in \mathbb N$.

Prove that the following inequality holds for all $ n\in \mathbb N$:

$ x_1\plus{}x_2\plus{}...\plus{}x_n<1$

[quote="Inequalities Master"]The sequence $ (x_n)$ is given by

$ x_1 \equal{} \dfrac{1}{2}$ and $ x_n \equal{} \dfrac{2n \minus{} 3}{2n}\cdot x_{n \minus{} 1}$, for all $ n\ge 2$, $ n\in \mathbb N$.

Prove that the following inequality holds for all $ n\in \mathbb N$:

$ x_1 \plus{} x_2 \plus{} ... \plus{} x_n < 1$[/quote]
We can prove that $ x_n \leq \dfrac{1}{n(n \plus{} 1)}$ .After that ,we have easy problem  :) .

[quote="babyandprettyboy"]
We can prove that $ x_n \leq \dfrac{1}{n(n \plus{} 1)}$ .After that ,we have easy problem  :) .[/quote]
It is not true .

 $ x_n \equal{} \frac{(2n\minus{}3)!!}{(2n)!!}$   ;   $ \frac{1}{4n\sqrt{n\minus{}1}} \le x_n \le \frac{1}{2n\sqrt{2n\minus{}1}}$ , $ (n \ge 2)$

$ \sum_{i\equal{}1}^n \ x_i \equal{} 1\minus{} \frac{(2n\minus{}1)!!}{(2n)!!}$ :)


Here is an interesting one I found, can't get the answer, but feel free to try it out.

The question is:
Given secA = -17/15 and cotB = -12/5 where neither A nor B is in quadrant II, find the exact value of csc(A-B) preferrably without the use of a calculator.

Enjoy!

lol...the same problem, that was a weird coincidence

[url]http://mb.sparknotes.com/mb.epl?b=72&m=1149505&t=326827[/url]


Diophantine equation
number theory unsolved
number theory

$ x,y\in\mathbb{N}$
$ y(y\plus{}x)\equal{}x^{2}\plus{}1$ find all solutions to  this equation

analyze the discriminant wrt $ x$ then use the following problem to find all solutions:
http://www.mathlinks.ro/viewtopic.php?t=150786

[quote="cnyd"]$ x,y\in\mathbb{N}$
$ y(y \plus{} x) \equal{} x^{2} \plus{} 1$ find all solutions to  this equation[/quote]

This is equivalent to $ (2y\plus{}x)^2\minus{}5x^2\equal{}4$

The Pell equation $ u^2\minus{}5v^2\equal{}4$ is easy to solve starting from $ (3,1)$ and using the transformation $ (u,v)\to(9u\plus{}20v,4u\plus{}9v)$

And so the original equation is easy to solve starting from $ (1,1)$ and using the transformation $ (x,y)\to(13x\plus{}8y,8x\plus{}5y)$

And so the solutions $ (1,1),(21,13),(377,233),...$

[quote="cnyd"]$ x,y\in\mathbb{N}$
$ y(y \plus{} x) \equal{} x^{2} \plus{} 1$ find all solutions to  this equation[/quote]

the equation is equivalent to $ x^2\minus{}xy\minus{}y^2\plus{}1\equal{}0$
$ \Delta  \equal{} 5y^2\minus{}4$
we need $ 5y^2\minus{}4\equal{}k^2$ ,with $ k \in N^*$ which is a Pell equation


Given $ABC$ and a point $P$ inside it. Denote by $d_{a}, d_{b}, d_{c}$ the distances from $P$ to $BC, CA, AB$. Show that

$2(d_{a}+d_{b}+d_{c})+a+b+c < 2\sqrt{2}(PA+PB+PC)$

Hi treegoner, hope you had an easy flight back at home :)

The problem is really nice.
Suppose $T$ is projection of $P$ on the side $AB$. Note $AT=u$, $BT=v$, then $AP^{2}=u^{2}+d_{c}^{2}$ and $BP^{2}=v^{2}+d_{c}^{2}$.
Using Cauchy-Buneakovsky: $AP\cdot BP\geq uv+d_{c}^{2}$

Summing we get $(AP+BP)^{2}\geq (u+v)^{2}+4d_{c}^{2}=c^{2}+4d_{c}^{2}\geq \frac{1}{2}(c+2d_{c})^{2}$

Thus $\sqrt{2}(AP+BP)\geq c+2d_{c}$ and summing over all we get the stated inequality.

To prowler,

Thank you very much. Your solution is very similar to mine.  :)

Thank prowler for the good solution, with this menthod we can see this inequality is also true with any $P$ in plane!  :)


I am looking for your idea... Urgly? Very urgly? ...
Thank you very much. :)

You did this in just an hour? I can even wright that fast.

[quote="manlio"]Dear can_hang2007,

I checked your solution (two hours ) and it is OK, but I cannot imagine how did you create it.

You really shocked me. :o[/quote]
 :blush:  :blush:


hi all

[b]are there at IMO some contestants using left-hand to write? :D [/b]
left-hander are more intelligent  :D

you agree with me? :blush:

Well I am left handed guy, but I wouldn't say that the hand with which you write influences what you write  ;)

and you ever have been at IMO? :)

your intelligent was been to create this site :rotfl:

I´m left handed also.
Left hand Rules!!
Also, i have notice that the percent of left handed in math olympiad is more or less 16%. Greater than 10%

I ask the expert  [img]http://artofproblemsolving.com/Forum/images/avatars/45961890041e4a6671adbd.jpg[/img]
[quote]I use right paw for mouse, left paw for keyboarding and mouth for catching frisbees  -  proof that  I  can  solve complicated diff. eqs to calculate trajectry in real tiime. ..[/quote]

I'm left-handed too. I got bronze in IMO, which is exactly what I expected. But I still think I'm smart :D

[quote="Valentin Vornicu"]I am curious Dave, how did you actually compute this percentile? :)[/quote]
Is just what i have noticed. Where i have asked there is usually one left handed for each 6 people (usually one per team). We could make a poll to see how many mathlinkers are left handed.

One guy from our team: Tigran Sloyan(Tiks in mathlinks) is also left handed. :)


Suppose x1, x2, ..., xn > 0 such that x1x2...xn = 1. Show that

\sumk = 1, 2, ..., n(n - 1 + xk)-1 \leq 1.

(Thank you Harazi for the correction)

That was indeed a typo. Nice solution, Harazi! Is there a more natural solution?

In my opinion, this is the most natural. I've seen a splendid solution using AM-GM but that cannot be called natural at all.

That splendid solution I was speaking about is the following:
  x_1/(n-1+x_1)>=x_1^z/((x_1)^z+(x_2)^z+...+(x_n)^z)) and so on, where z is 1-1/n. Sum up these inequalities and you will find that sum x_1/(n-1+x_1)>=1, from where sum 1/(n-1+x_1)<=1.

[quote="harazi"]Something is strange with this inequality:
  If it is sum 1/(n+1-x_k) then this one is at least n^2/(n^2+n-sum x_k) and this one is at least 1. So, in this case the problem is trivial and incorrect stated.
  If it is sum 1/(n-1+x_k), then a simple solution is the following:
  Consider p(x)=(x+x_1)...(x+x_n). We have to prove that P(n-1)>=P'(n-1), which is true since sum x_i1*x_i2*..x_ik>=kCn for all k.[/quote]

I'm sorry. Could you please explain this in detail? I see that it is necessary and sufficient to prove P(n-1) >= P'(n-1), but I do not see how sum x_i1*x_i2*..x_ik>=kCn for all k solves the inequality.

Just write P(X)=X^n+S_1X^n-1+...+S_n-1X+S_n, where S_k=sum x_i1*...*x_ik. Then proving that P(n-1)>=P'(n-1) comes down to
  S_2*(n-1)^n-3+...+S_n-2*(n-1)(n-3)+S_n-1*(n-2)>=(n-1)^n-1, which is true since S_k>=kCn. Use the binomial theorem.


calculus
derivative
analytic geometry
graphing lines
slope
trigonometry
logarithms

If $ f(x) \equal{} x^3 \minus{} 3x \plus{} 5$, find $ (f^{ \minus{} 1})'(x)$ at $ a \equal{} f(2)$.

I'm kinda confused but here's what I tried:

$ a \equal{} f(2) \equal{} 2^3 \minus{} 3(2) \plus{} 5 \equal{} 7$

Assuming $ f^{ \minus{} 1}(x) \equal{} f(y)$, so

$ (f^{ \minus{} 1})'(x) \equal{} \cfrac{1}{f'(y)}$

$ (f^{ \minus{} 1})'(2) \equal{} \cfrac{1}{f'(7)} \equal{} \boxed{\cfrac{1}{144}}$

The answer key says otherwise.. what did I do incorrectly?

Ok, the simplest way to solve this is to set $ f\equal{}f(x) \Rightarrow x\equal{}x(f)$. Now:

$ f\equal{}x^3(f)\minus{}3x(f)\plus{}5$

$ 1\equal{}3x^2(f)x'(f)\minus{}3x'(f)$

$ x'(f)\equal{}\frac{1}{3x^2(f)\minus{}3}\equal{}\frac{1}{3\cdot2^2\minus{}3}\equal{}\frac{1}{9}$

Another way to think about it: you want the slope of the line tangent to the graph of $ y \equal{} f^{\minus{}1} (x) \implies f(y) \equal{} x$ at the point $ (f(2),2)$.

This is the inverse of the slope of the line tangent to $ y\equal{}f(x)$ at the point $ (2,f(2))$.

$ (x^3\minus{}3x\plus{}5)'  \equal{} 3x^2 \minus{} 3$, so at $ (2,f(2))$, the slope of the tangent to $ y\equal{}f(x)$ is $ 3 \cdot 2^2 \minus{} 3 \equal{} 9$

We want the reciprocal of that.

haha it took me forever to figure that what $ x(f)$ meant (idk why.. this is easy :| ), but I got it.. Thanks for the help! Both methods are great.

I have one more question - finding the derivative of $ f(x) \equal{} x^2\cos^{ \minus{} 1}{(\ln\sqrt {x})}$, (not an inverse though - I wasted 20 minutes thinking it asked for the derivative of the inverse) I tried:

$ f'(x) \equal{} x^2\cdot\left( \minus{} \cfrac{1}{\sqrt {1 \minus{} (\ln\sqrt {x})^2}}\right)\cdot\cfrac{1}{\sqrt {x}}\cdot\cfrac{1}{2\sqrt {x}} \plus{} 2x\cos^{ \minus{} 1}{(\ln\sqrt {x})}$

$ \equal{} \boxed{ \minus{} \cfrac{x}{2\sqrt {1 \minus{} (\ln\sqrt {x})^2}} \plus{} 2x\cos^{ \minus{} 1}{(\ln\sqrt {x})}}$.

Did I do something wrong again, or is it really just an ugly problem?


analytic geometry
graphing lines
slope
AMC

[quote="vinle"]Hello. I'm new here.
The problem can be found on this link.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=750073#p750073
I saw the solutions to the problem and did not get them, even though I got the right answer also.
What I did was I set an equality after drawing the square and corresponding them with their respective logs.
The respective graphs all have (1,0) and no vertical or phase shift, so I compared them. Since B is farther from the y axis than A, I saw that the slope of 2loga(x) is closer to the x axis than loga(x).
Therefore, I concluded that a must be less than 1. That leaves me with only A as an answer.
Does my way work? I still think it was a lucky guess.
Thank you.[/quote]

It was a lucky guess, because $ \sqrt[6]{3}$ is not less than 1.

As was already pointed out, your reasoning is not correct.  In particular, one should observe that your assumption that $ B$ is to the right of $ A$ is incorrect, because in fact that is not possible.

Thank you, I think I get the problem now.
I "knew" it was a lucky guess, too.  :rotfl:
At least I got over 100 for the AMC 12 (though not by much).
The first 15 were simple, and I solved 3 more  (including that one, which I was so rushed of time by then) and checked my answers.
I got 16 right. :lol:


prove that in triangle ABC:

                                      A(O,H,I,Ib)=-1

hmmm...ure right.i didnt explain it. :(

i mean O,H,I,Ib, are harmonic. and if u draw lines AO,AH,AI,AIb,u can see that for each line that is intersect by these

four lines,we have the intersection points will make four harmonic points.

i note some usefull theorems in: http://mathlinks.ro/forum/viewtopic.php?t=51865


Find all a,b (positive intergers) for which:
$a^{3}+b^{3}+3a^{2}b$ is a square of interger

We have $a^{3}+b^{3}+3a^{2}b=x^{2}$
$(a+b)^{3}=x^{2}+3ab^{2}$ I believe that it is a very hard and interesting problem..... :ninja:


I am reviewing for a test thats tomorrow and am stuck on a basic part of the problem... I just need to know how
1/cos2x=1/2cos2x-1     ... I know I am missing something simple, but I just can't make the connection. If anyone could help me, I'd greatly appreciate it.

Thanks!!!
tina

i think it's
$\frac{1}{cos2x}=\frac{1}{2cos^{2}x -1}$
it


Δίνω μια ανισότητα προς αποδειξη.Αν μπορεί κάποιος ας την ξαναγράψει σε κανονικη μορφή.
ΝΔΟ ΑΝ c>o και n=1,2,3... TOTE c^(2n)+c^(2n-1)+...+c^2+c+1>=(2n+1)c^n

[quote="DIMITRIS T"]
ΝΔΟ ΑΝ $ c > o$ και $ n \equal{} 1,2,3...$ τότε $ c^{2n} \plus{} c^{2n \minus{} 1} \plus{} ... \plus{} c^2 \plus{} c \plus{} 1\geq(2n \plus{} 1)c^n$[/quote]

Δίνω και ένα hint:
[hide]Η [i][u]επαγωγή[/u][/i] σε τέτοιες περιπτώσεις δουλέυει σχεδόν πάντα :wink: [/hide]

Παίρνουμε $ c^{2n}\plus{}1\geq 2c^n$  , $ c^{2n\minus{}1}\plus{}c\geq 2c^n$ κτλ και θα μείνει και ένα στην μέση το $ c^n$ Με προσθεση το ζητούμενο

ΘΑ ΧΡΗΣΗΜΟΠΟΙΗΣΟΥΜΕ ΤΗΝ ΑΝΙΣΟΤΗΤΑ α+1/α>=2 ΓΙΑ ΚΑΘΕ α>0, ΘΕΤΟΝΤΑΣ ΤΗΝ ΠΡΩΤΗ ΦΟΡΑ α=C,ΥΣΤΕΡΑ α=c^2,...,α=c^n ΕΤΣΙ ΕΧΟΥΜΕ c+1/c>=2,c^2+1/c^2>=2,...,c^n+1/c^n>=2
ΠΡΟΣΘΕΤΟΥΜΕ ΚΑΤΑ ΜΕΛΗ ΑΥΤΕΣ ΤΙΣ ΑΝΙΣΟΤΗΤΕΣ ΜΕΤΑ ΠΡΟΣΘΕΤΟΥΜΕ ΤΟ 1 ΣΤΑ ΔΥΟ ΜΕΛΗ ΤΗΣ ΑΝΙΣΟΤΗΤΑΣ ΠΟΥ ΠΡΠΟΚΥΠΤΕΙ ΚΑΙ ΥΣΤΕΡΑ ΠΟΛΣΙΑΖΟΥΜΕ ΜΕ ΤΟ c^n ΣΤΑ ΔΥΟ ΜΕΛΗ ΤΗΣ ΚΑΙΝΟΥΡΙΑΣ ΑΝΙΣΟΤΗΤΑΣ ΚΑΙ ΠΡΟΚΥΠΤΕΙ Η ΖΗΤΟΥΜΕΝΗ.
ΑΠΛΗ ΑΥΤΗ Η ΛΥΣΗ ΑΛΛΑ ΗΘΕΛΕ ΦΑΝΤΑΣΙΑ!!!

Έγραψα ακριβώς την ίδαι λύση ένα ποστ παραπάνω... :wink:


Hi everybody
I am looking for a software accessory for pdf to help me introduce changes on pdf documents
Thanks.
SENOUY


MATHCOUNTS
articles
algorithm
modular arithmetic
function
SFFT
special factorizations

I'm talking about problems of the form "What is the remainder when x^y is divided by z?". Since I never really learned how to solve them, I ended up creating my own method with my logic.

My logic: The remainder when x^y is divided by y is equal to the remainder when x is divided by y.

Using my logic, I solve problems like this:

Example Problem: "What is the remainder when 6^25431 is divided by 7?"

25431 is divisible by 7, so the problem "What is the remainder when 6^3633 is divided by 7?" has the same answer as the problem "What is the remainder when 6^25431 is divided by 7?" since 25431/7 = 3633.

3633 is divisible by 7, so the problem "What is the remainder when 6^519 is divided by 7?" has the same answer as the problem "What is the remainder when 6^3633 is divided by 7?" since 3633/7 = 519.

518 is the largest multiple of 7 less than 519. Thus, the problem "What is the remainder when 6^75 is divided by 7?" has the same answer as the problem "What is the remainder when 6^519 is divided by 7?" since (519 -518) + 518/7 = 75.

70 is the largest multiple of 7 less than 75. Thus, the problem "What is the remainder when 6^15 is divided by 7?" has the same answer as the problem "What is the remainder when 6^75 is divided by 7?" since (75 - 70) + 70/7 = 15.

14 is the largest multiple of 7 less than 15. Thus, the problem "What is the remainder when 6^3 is divided by 7?" has the same answer as the problem "What is the remainder when 6^15 is divided by 7?" since (15 - 14) + 14/7 = 3.

6^3 = 216. The remainder when 216 is divided by 7 is 6. Thus, the answer to the problem "What is the remainder when 6^3 is divided by 7?" as well as the original problem "What is the remainder when 6^25431 is divided by 7?" is 6.

So that's how I solve those types of problems?. What do you think of my method? Is there a faster/easier way I should know about? Who knows, my method might become like Simon's Favorite Factoring Trick and be used quite often  :D .

EDIT: I have found a simpler way to state my method.

For any problem of the from "What is the remainder when x^y is divided by z?", find the greatest multiple of (z - 1) that is less than or equal to y. Subtract that multiple from y. Now solve the much-simplified problem, which will yield the same answer as the original problem. Keep in mind that this method only works when z is not a factor of x. If z is a factor of x and y > 0, the remainder is 0. If z is a factor of x and y = 0, the remainder is 1.

I should call my method Mark's Marvelous Modular MathCounts Method  :D . Maybe it will become an article in the AoPS Wiki.

The method is great, the underlying may be a case of [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's Little Theorem[/url].

But FLT only works for prime values of y that aren't factors of x; For example, $ 2^4$ does not leave remainder $ 2$ when divided by $ 4$.

EDIT: Yey you figured out a simple algorithm that [b]always works[/b] for mods!
This is new  :D .

EDIT2: [quote=12markkram34]Keep in mind that this method only works when z is not a factor of x. If z is a factor of x and y > 0, the remainder is 0. If z is a factor of x and y = 0, the remainder is 1. [/quote]
Ok, never mind. Still cool though.

Hmm...it appears that I have come up with Fermat's Little Theorem. Darn...Fermat beat me to it  :( .

What is the remainder when 6^25431 is divided by 7?

Faster way. 6=-1(mod7)
so 6^25431(mod 7)=(-1)^25431(mod 7)
since the exponent is odd we have
6^25431=x(mod 7)
-1=x(mod 7)
so x equals 6.

Yes, before doing MMMMM (:D), it is often useful to check if the base is $ 1$ or $ \minus{}1 \pmod n$.

What if the problem asked to find the remainder when $ 2^{89}\plus{}3$ is divided by $ 7$

[hide="Click here after answering, Please confirm if I did it right"]
$ 2^{89}\plus{}3 \equiv n \mod 7$

$ 2^{89} \equiv n\minus{}3 \mod 7$

$ 2^{87}\cdot4 \equiv n\minus{}3 \mod 7$

$ 8^{29}\cdot4 \equiv n\minus{}3 \mod 7$

$ 8^{29} \equiv \frac{n\minus{}3}{4} \mod 7$

$ 1 \equiv \frac{n\minus{}3}{4} \mod 7$

$ 4 \equiv n\minus{}3 \mod 7$

$ 0 \equiv n \mod 7$

Thus, $ 2^{89}\plus{}3$ divides $ 7$.
[/hide]

Was that how you would've done it, Izzy? (because if that's the method, than it will always work, but is limited to numbers that has powers that are either 1 greater or less than the modulus).

eh easier way

$ 2^3\equiv 1\mod 7$
$ 2^{87}\equiv 1\mod 7$
$ 2^{89}\equiv 4\mod 7$

$ 4 \plus{} 3 \equal{} 7 \equiv 0 \mod 7$

Wait... I'm not sure I understand your method well.

So, the basis for your first steps is that $ (6^k)^7 \equiv (6^k) \pmod{7}$, or, in generally, $ n^p \equiv n \pmod{p}$, for some prime $ p$. That's FELT (Or, I suppose, just FLT).

Now you have $ 6^{519}$. You're splitting this into $ 6 \times 6^{518} \equal{} 6 \times 6^{7 \times 74} \equiv 6 \times 6^{74} \equal{} 6^{75}$. That's applying the previous.

You repeat to get $ 6^3 \pmod{7}$, which you solve to get 6.

Aha! Now that I have wasted a post writing it out, I think I understand it. You came up with this by yourself, without knowing FELT? Bravo.

[quote="RisingFlame"]Generally, the fastest way for bigger numbers is to use Euler's theorem: for relatively prime numbers $ a$ and $ b$,  $ a^{\varphi(b)} \equiv 1 \pmod b$.
$ \varphi(b)$ is the totient function which tells how many integers less than or equal to $ b$ are relatively prime to it.[/quote]

Indeed, but if you don't happen to remember the algorithm for finding $ \phi(n)$ on the spot in a contest, this progressive method is a bit easier to remember than both FELT AND the algorithm to find $ \phi(n)$.

[quote="isabella2296"]The way I solve problems like these is sort of similar to yours, but I'm not sure if mine always works. For example:

Find the remainder when $ 2^{87} \plus{} 3$ is divided by 7.

First we note that 87 is divisible by 3, so we write this as $ 2^3 \plus{} 3$ or $ 11$, which when divided by 7, yields $ \boxed{4}$.

Is this just sort of a coincidence that it works so far all the time or is does it always work?[/quote]

For that I would find the remainder when 2^87 is divided by 7 using my method, then add 3 to it and check if the sum is 7 or greater.


geometry
circumcircle
quadratics
3D geometry
algebra
quadratic formula

Mandelbrot Round 3

4. Compute the number $(\frac{r}{s})^3$ if $r$ and $s$ are numbers satisfying $\frac{1}{r + s} = \frac{1}{r} + \frac{1}{s}$.

5. The measures of the angles of $\triangle ABC$ are $\angle A = 60^{\circ}$, $\angle B = 40^{\circ}$, and $\angle C = 80^{\circ}$. Let $O$ be the center of the circle circumscribed about $\triangle ABC$. If it has radius $1$, then find the radius of the circumcircle of $\triangle BOC$.

7. Through the interiors (not just vertices) of how many lattice squares does the segment connecting $(0,0)$ and $(119,153)$ pass? (A lattice square has sides of length $1$ and vertices with integer coordinates.)

8. Determine the value of the infinite products $(2^\frac{1}{3})(4^\frac{1}{9})(8^\frac{1}{27})(16^\frac{1}{81})$

edit: fixed typo

Number 5:

Have to go right now, so can someone finish this up?

First we want to find the number of degrees in everything if we draw in the radius, so we can set up an algabraic expression:

$x+y+z=90$
$x+z=40$
$z+y=80$
$x+y=60$

So we have that $z=50$, $y=30$ and $x=10$ degrees. So we have that triangle $BOC$ has angles of $30,30,$ and $120$, so by making two $30,60,90$ triangles, we see that the sides are $1,1,\sqrt{3}$,

So now that we know the sides, we use the equation: $R=\frac{abc}{4A}$ and ${A=\frac{\sqrt{3}}{4}}$, so we can solve for $R$.

#4 has a typo. I believe it should be, compute the number $\Big(\frac{r}{s}\Big)^3$

For #5, there's an easier way to find the angles of $\triangle BOC$.

[hide]Note that $\triangle BOC$ is an isosceles triangle ($BO=CO$).
Also, $\angle BOC=2\angle A=120^\circ$.
So $\angle BCO=\angle CBO=30^\circ$ ;) [/hide]

[hide="7"]it will intersect 119 squares going horizontally, and 153 squares going vertically.  However, $(119, 153)$ of these will be intersected twice (i.e. at a vertex).  So the answer is $119+153-17= \boxed{255}$[/hide]

[quote="kunny"]4. Compute the number $(\frac{r}{s})^3$ if $r$ and $s$ are numbers satisfying $\frac{1}{r + s} = \frac{1}{r} + \frac{1}{s}$.

[hide]$\frac{1}{r+s}=\frac{r+s}{rs}\ (rs\neq 0)\Longleftrightarrow (r+s)^2=rs.$

$\Longleftrightarrow \left(\frac{r}{s}+1\right)^2=\frac{r}{s}\Longleftrightarrow \left(\frac{r}{s}\right)^2+\frac{r}{s}+1=0\Longleftrightarrow \frac{1-\left(\frac{r}{s}\right)^3}{1-\frac{r}{s}}=0,$ yielding $\left(\frac{r}{s}\right)^3=\boxed{1}.$[/hide][/quote]

Sorry I'm not that strong at math! But may someone explain how you go from (r/s)^2 + r/s + 1 = 0 to the next step!

Thanks!

[quote="kunny"]You can use the formula: $a+ar+ar^2+\cdots+ar^{n-1}=\frac{a(1-r^n)}{1-r}\ (r\neq 0).$ :)[/quote]
oh wow now i see it omg thanks so much :)

It's difference of cubes.

$x^3 - y^3 = (x-y)(x^2 + xy + y^2)$

In this case, $x=1$ and $y=\frac{r}{s}$.  Recognize the form and you can rewrite

$x^2 + xy + y^2 = \frac{x^3-y^3}{x-y}$

[quote="kunny"]

$\frac{1-\left(\frac{r}{s}\right)^3}{1-\frac{r}{s}}=0,$ yielding $\left(\frac{r}{s}\right)^3=\boxed{1}.$[/quote]

there is one thing that bothers me, namely, that expression becomes undefined

i think you have to use the quadratic formula when you solve for $\frac{r}{s}$

If you use the quadratic formula, you'll end up getting the extraneous solution of $-1$.  $\left(\frac{r}{s}\right)^3 = 1$ is right.

[quote="Altheman"][quote="kunny"]

$\frac{1-\left(\frac{r}{s}\right)^3}{1-\frac{r}{s}}=0,$ yielding $\left(\frac{r}{s}\right)^3=\boxed{1}.$[/quote]

there is one thing that bothers me, namely, that expression becomes undefined[/quote]The expression is defined.  The cube of $\frac{r}{s}$ is 1, but $\frac{r}{s}$ itself is not 1.
.


modular arithmetic
number theory proposed
number theory

Are there infinitely many positive solutions to
\[
2^n \equiv -3 \pmod{n}?
\]

Internet Relay Chat, more specifically the channel #math at EFNet.


linear algebra
matrix
linear algebra unsolved

Consider the parametric equations
$ (x,y,z) \equal{} (0,0,1).t$
$ (x,y,z) \equal{} (1,2,0) \plus{} (1,0,0).t$
$ (x,y,z) \equal{} (1,1,1)\plus{}(0,1,0).t$
$ (x,y,z) \equal{} (1,0,0) \plus{} (1,1,1).t$
How many lines intersect the 4 lines above?
Well, I have the solution of the problem here, but it's hard to understand.
The author considers 4 points which lie in each line above respectively. $ A \equal{}(0,0,a)$, $ B\equal{}(1\plus{}b,2,0)$, $ C\equal{}(1,1\plus{}c,1)$, $ D\equal{}(1\plus{}d,d,d)$. And then he makes an assumption that these points must be collinear if and only if the matrix:
\[ \left(
\begin{array}{cccc}
0 & 0 & a & 1 \\
1\plus{}b & 2 & 0 & 1 \\
1 & 1\plus{}c & 1 & 1 \\
1\plus{}d & d & d &1 \end{array}
\right)\]
has rank 2. But why does this matrix need to have rank 2 ? I thought to have the collinear points, the determinant should be zero.

$ A, B, C$ and $ D$ are collinear $ \Longleftrightarrow$ the points $ (0, 0, a, 1), (1 \plus{} b, 2, 0, 1), (1, 1 \plus{} c, 1, 1)$ and $ (1 \plus{} d, d, d, 1)$ are coplanar. Then each of them is a linear combination of any other two (it is easy to see these points are pairwise distint, therefore LI), so the rank of the matrix is $ 2$. On the other hand, if the rank of the matrix is $ 2$, there are two points that generate the other two. And if $ X \in span\{Y, Z\}$, then $ X, Y$ and $ Z$ are coplanar.

All right?

And how does the solution end?

[quote="feliz"]$ A, B, C$ and $ D$ are collinear $ \Longleftrightarrow$ the points $ (0, 0, a, 1), (1 \plus{} b, 2, 0, 1), (1, 1 \plus{} c, 1, 1)$ and $ (1 \plus{} d, d, d, 1)$ are coplanar.[/quote]
Why of this?

I'm sorry... Replace "coplanar" with "coplanar with the origin."

$ A, B, C$ and $ D$ are collinear $ \Longleftrightarrow$ the points $ (0, 0, a, 1), (1 \plus{} b, 2, 0, 1), (1, 1 \plus{} c, 1, 1)$ and $ (1 \plus{} d, d, d, 1)$ are collinear $ \Longleftrightarrow$ the points $ (0, 0, a, 1), (1 \plus{} b, 2, 0, 1), (1, 1 \plus{} c, 1, 1)$ and $ (1 \plus{} d, d, d, 1)$ are in a plane passing through the origin. Then each of them is a linear combination of any other two (it is easy to see these points are pairwise distint, therefore LI), so the rank of the matrix is $ 2$. On the other hand, if the rank of the matrix is $ 2$, there are two points that generate the other two. And if $ X \in span\{Y, Z\}$, then $ X, Y, Z$ and the origin are coplanar.