Let $n\geq 3$ be natural numbers, and let $a_1,\ a_2,\ \cdots,\ a_n,\ \ b_1,\ b_2,\ \cdots,\ b_n$ be positive numbers such that $a_1+a_2+\cdots +a_n=1,\ b_1^2+b_2^2+\cdots +b_n^2=1.$ Prove that $a_1(b_1+a_2)+a_2(b_2+a_3)+\cdots +a_n(b_n+a_1)<1.$

see http://www.mathlinks.ro/Forum/viewtopic.php?t=39364 .

  dg

solution
$S=\sum_{i=1}^n a_ib_i+\sum_{i=1}^{n-1} a_ia_{i+1}+a_na_1$
objective is to prove S<1

Consider,$(\sum_{i=1}^n a_i)^2=1 $
$\implies \sum_{i=1}^n a_i^2+2\sum_{i\neq j}^n a_ia_j=1$
$\implies \sum_{i=1}^n a_ia_{i+1} +a_na_1<1/2-\sum_{i=1}^n a_i^2/2$
Therefore, $S<\sum_{i=1}^n a_ib_i+1/2-\sum_{i=1}^n a_i^2/2$
Now $\sum_{i=1}^n (b_i-a_i)^2>0$
$\implies \sum_{i=1}^n a_i^2+\sum_{i=1}^n b_i^2-2\sum_{i=1}^n a_ib_i>0$
$\sum_{i=1}^n a_ib_i<1/2+\sum_{i=1}^n a_i^2/2  (because \sum_{i=1}^n b_i^2=1)$
plugging the inequality in the inequality derived before
$S<1/2+\sum_{i=1}^n a_i^2/2+1/2-\sum_{i=1}^n a_i^2/2$
$\implies S<1$

By Cauchy we get

$$\sum a_i b_i \le \sqrt{\sum a_i^2}.$$

Hence it suffices to prove that $\sqrt{\sum a_i^2} + \sum a_ia_{i+1} \le 1.$

We have that $\sum a_i a_{i+1} \le \frac12 ((\sum a_i)^2 - (\sum a_i^2)) \le \frac{(\sum a_i)^2 - (\sum a_i^2)}{\sqrt{(\sum a_i)^2} + \sqrt{\sum a_i^2)}}.$

This is just $\sum a_i - \sqrt{\sum a_i^2},$ and so we're done.

$\square$


What percent of one hour is five seconds? Round your answer to the nearest tenth.

[hide] In $ 1$ hour, there are $ 3600$ seconds. Therefore, $ \frac{5}{3600}\equal{}\frac{1}{720}\approx\boxed{.1\%}$.[/hide]


linear algebra
matrix
algebra
polynomial

If $x$ is $n \times 1$-matrix on a field, prove that \[\det (I_{n}+xx^{t}) = 1+x^{t}x\]

I would say calculate the eigenvalues, that would give you the characteristic polynomial and hence the result :
$1+x^{t}x$

[quote="Omid Hatami"]Because of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87254]this topic[/url] you see that $\det (I_{n}+xx^{t})=ch_{xx^{t}}(1)=ch_{x^{t}x}(1)=\det(I_{n}+x^{t}x)=1+x^{t}x$

P.S.
Of Course the solution you wrote is the hint I gave at that topic[/quote]

I'm sorry. I do not have any information about your topic.

[quote="mofidy"]I'm sorry. I do not have any information about your topic.[/quote]
Just click on the link I have given ([url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87254]this topic[/url])


Firstly, how do you think the difficulty of 2002 and 2003 quarterfinals are compared to 2009 f=ma?

Secondly, do you think it would be a good idea to go through the quarterfinals to help prepare for the 2010 F=ma? Or do they stress different stuff?

well it definitely wouldn't hurt if you have nothing else to do to prepare...I think it would probably be more helpful to just review all the concepts from previous f=ma tests and ingrain in your head the harder to derive and useful equations...since there are only a couple of days left to take the 2010 prelim exam.


Chords $AA^{\prime}$, $BB^{\prime}$, $CC^{\prime}$ of a sphere meet at an interior point $P$ but are not contained in a plane.  The sphere through $A$, $B$, $C$, $P$ is tangent to the sphere through $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $P$. Prove that $\, AA' = BB' = CC'$.

Ok!
Construct plane $ABPA'B'$, then intersection of this plane and big sphere  is a circle, so we conclude $AP\cdot PA'=BP\cdot PB'$ [b](*)[/b]. Moreover, intersection with small spheres is a two tangent circles. On this figure it is easy to show that $AB||A'B'$. It follows that $AP:BP=A'P:B'P$ [b](**)[/b].
From [b](*)[/b] and [b](**)[/b] we conclude $AA'=BB'$ :)

[quote="grobber"]Or, you could invert wrt $P$ with power $PA\cdot PA'$. The two tangent spheres turn into two parallel planes, and it's clear from here.[/quote]
Hmm... In my opinion you use the same arguments...

I see that now, but I hadn't read your message when I posted mine. :)

A detailed solution:

Consider the inversion with pole $P$ fixing the sphere. It sends $X$ to $X'$ and vice-versa for all $X\in\{A,B,C\}$. Since the spheres $(PABC),(PA'B'C')$ are tangent, the planes $(A'B'C'),(ABC)$, which are the images of those spheres respectively, must be parallel, and we're done: on the one hand, we have $PA\cdot PA'=PB\cdot PB'$, and on the other hand we have $\frac{PA}{PA'}=\frac{PB}{PB'}$, so $\{PA,PA'\}=\{PB,PB'\}\Rightarrow AA'=PA+PA'=PB+PB'=BB'$, and we do the same to prove $AA'=CC'$.

Invert at $P$ with power $\sqrt{PA \cdot PA'}$ followed by reflection in $P$. The spheres $\mathcal{S}$ through $A, B, C, P$ and $\mathcal{S}'$ through $A', B', C', P$ are mapped to planes parallel to each other and tangent to the original sphere. It follows that $PA \cdot PA'=PB \cdot PB'$ and $$\frac{PA}{PA'}=\frac{PB}{PB'},$$ so $AA'=BB'$. Similarly, $AA'=CC'$ and we may conclude.

Solved with the help of 17 people (mostly with their cameras off and no participation)

[asy]
draw(circle((-2,0),1));
draw(circle((1,0),2));
pair A,B,C,D,P;
A=(-2,1);
B=(-2.455111059789108,-0.8904346821608);
C=(1,-2);
D=(1.910222119578216,1.780869364392161);
P=(-1,0);
dot(A^^B^^C^^D^^P);
label("$A$",A,N);
label("$B$",B,S);
label("$A'$",C,S);
label("$B'$",(1.910222119578216,1.921869364392161),NW);
label("$P$",(-0.7,0),E);
draw(A--B--C--D--B--C--A--B--C--D--A--B--D);
[/asy]

Consider the cross-section going through the plane defined by the two intersecting lines $AA'$ and $BB'.$ By Power of a Point, we have $$PA~ \cdot ~PA' = PB~ \cdot ~PB' ~~(\clubsuit)~~ \implies \frac{PA}{PB}=\frac{PB'}{PA'} ~~(\alpha)~~\implies \triangle PAB \sim \triangle PB'A'.$$ From here, we present two solutions:
First, note that $$\triangle PAB \sim \triangle PB'A' \implies \angle PAB = \angle PA'B' \qquad (\heartsuit).$$ Since $A,B,A',B'$ were on the circumference of a sphere and the cross-section of a sphere is a circle, we have that $ABA'B'$ is a cyclic quadrilateral and that $$\angle PAB = \angle PB'A' \qquad (\spadesuit).$$ Together $\heartsuit$ and $\spadesuit$ imply that $$\angle PA'B' = \angle PB'A' \implies PB'=PA' \stackrel{\alpha}\implies PA=PB.$$

Or, we can proceed without angles:
$$\triangle PAB \sim \triangle PB'A' \implies AP ~ \cdot ~ B'P=BP ~\cdot ~A'P ~~(\diamondsuit).$$ Adding $\clubsuit$ and $\diamondsuit$ and rearranging yields $$PA~ \cdot ~PA' ~+~AP ~ \cdot ~ B'P= PB~ \cdot ~PB' ~+~BP ~\cdot ~A'P \implies PA(PA'~+~PB') = BP(PA'~+~PB') \implies PA = BP \stackrel{\alpha}\implies PB'=PA'.$$

Either way, we obtain $PA'=PB'$ and $PA=PB,$ therefore $$PA+PA'=PB+PB' \implies AA'=BB'.$$ We can derive $BB'=CC'$ in the exact same manner, concluding our proof.

$\mathbb{Q.E.D.}$

[quote=JAnatolGT_00]By PoP $|PA|\cdot |PA'|=|PB|\cdot |PB'|$ and by homothety $\frac{|PA|}{|PA'|}=\frac{|PB|}{|PB'|}.$ Hence $|PA|=|PB|\implies |AA'|=|BB'|.$
Analogously $|BB'|=|CC'|.$[/quote]

Seems to be what I did except you just made me realize I could've multiplied the first two equations you wrote down and square-rooted. :roll:

[hide=Another solution]Note that inversion $A\leftrightarrow A'$ wrt $P$ maps sphere $ABCP$ onto plane $A'B'C',$ and sphere $A'B'C'P$ onto $ABC.$
Hence these two planes are parallel, and conclusion follows easily.[/hide]

Do we really need to use inversion?! I guess not. Here another simple approach.

[asy]
draw(circle((-2,0),1));
draw(circle((1,0),2));
pair A,B,C,D,P;
A=(-2,1);
B=(-2.455111059789108,-0.8904346821608);
C=(1,-2);
D=(1.910222119578216,1.780869364392161);
P=(-1,0);
dot(A^^B^^C^^D^^P);
label("$A$",A,N);
label("$B$",B,S);
label("$A'$",C,S);
label("$B'$",(1.910222119578216,1.921869364392161),NW);
label("$P$",(-0.7,0),E);
draw(A--B--C--D--B--C--A--B--C--D--A--B--D);
[/asy]

Draw a common tangent passing from point $P$ and suppose that the tangent line cuts segments $BA'$ and $AB'$, respectively, at point $M$ and $N$. Then, $\angle{PB'A'}=\angle{BAP}=\angle{BPM}=\angle{NPB'}=\angle{PA'B'}=\angle{ABP}$ from which we get that triangles $ABP$ and $A'B'P$ are isosceles, and have the same base angles. From here we get that $AA'=AP+A'P=BP+B'P=BB'$.


[quote="caubetoanhoc94"][quote="bigbang195"]$ a,b,c \ge 0$. Prove that :
$ \sum_{cyclic} \sqrt []{\frac {a(b \plus{} c)}{a^2 \plus{} bc}} \ge 2$[/quote]

It's old problem of Vasc  :)[/quote]

why !!. I don't understand . (em ko hiểu, anh làm rõ đi ạ  :wink: )

[quote="bigbang195"]$ a,b,c \ge 0$. Prove that :
$ \sum_{cyclic} \sqrt []{\frac {a(b \plus{} c)}{a^2 \plus{} bc}} \ge 2$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87751


I'm not sure if there might be restrictions or something, since it's so soon after the competition. I wanted to see how many people can solve certain problems, and how much time it takes them.

Probably. Although I think ARML will post them up, like they did with Last year's problem, on their site at arml.com.

I was thinking of having more discussion here, though, unless arml.com has a forum.

I dunno. But if it is allowed, then you should post it on Other US Contests and Programs Forum because i think that is where it is mostly suitable.

2005 ARML problems are posted on the ARML website, so you can post any questions you have.  Attached are the 2005 Questions as well as the Answers.  For a complete index of all the rounds of the 2005 ARML, click [url=http://www.arml.com/2005contest.htm]here.[/url]


function
real analysis
calculus
calculus computations

I'm new to delta-epsilon proofs, and I'm struggling to prove that this statement is true:
$ \exists \epsilon > 0, \forall \delta > 0$ such that $ \exists x \in (0,1), \exists y \in (0,1), |x\minus{}y| < \delta$  	  implies $ |\frac{1}{x} \minus{} \frac{1}{y}| < \epsilon$.

Thanks for the help.

For good reason. That's not true.

In writing it down formally, you mixed up the quantifiers; whatever your $ \epsilon$, I can choose $ \delta\equal{}1$, make $ y$ close to zero and have $ \left|\frac1x\minus{}\frac1y\right|$ arbitrarily large.

OK, we fix that obvious mistake and match the definition of a limit: for any $ \epsilon>0$, there exists a $ \delta>0$ such that...
Then this is the definition of uniform continuity, and it's still not true; $ \frac1x$ is not uniformly continuous on $ (0,1)$. For any $ \delta$ and any $ \epsilon$, we can choose $ x$ and $ y$ close to zero so that $ \left|\frac1x\minus{}\frac1y\right|>\epsilon$; $ x\equal{}\frac1N$ and $ y\equal{}\frac2N$ work for large enough $ N$.

[quote="perlypython"]I'm new to delta-epsilon proofs, and I'm struggling to prove that this statement is true:
$ \exists \epsilon > 0, \forall \delta > 0$ such that $ \exists x \in (0,1), \exists y \in (0,1), |x \minus{} y| < \delta$  	  implies $ |\frac {1}{x} \minus{} \frac {1}{y}| \geq \epsilon$.[/quote]

I think it should be >= rather than <.

Then it's the negation of uniform continuity, and I just gave you a construction.

[Added in edit]
The part after "such that" needs cleaning up too; it's ungrammatical with both "exists" and "implies" there. For what you're asking, the full statement should be:
There exists an $ \epsilon>0$ such that for any $ \delta>0$, there exist $ x,y\in (0,1)$ with $ |x\minus{}y|<\delta$ and $ \left|\frac1x\minus{}\frac1y\right|\ge\epsilon$.

Uniform continuity of a function $ f$ on an interval $ I$ would be:
For any $ \epsilon>0$, there exists a $ \delta>0$ such that for all $ x,y\in I$ with $ |x\minus{}y|<\delta$, $ |f(x)\minus{}f(y)|<\epsilon$.

I strongly recommend using words rather than symbolic quantifiers if you're not clear. It's much easier to see what statements make sense.

If we didn't have to prove uniform continuity and just had to show continuity, could we do this if we set $ \delta \equal{} xy\epsilon$:

$ |\frac{1}{x} \minus{}\frac{1}{y}| \equal{} |\frac{y\minus{}x}{xy}| \le |\frac{x\minus{}y}{xy}| < \frac{\delta}{xy} \equal{} \epsilon$ ?

You don't get to choose both variables before choosing $ \delta$- that just doesn't make sense.

How am I suppose to make it make sense then?


Harry Potter has n mixtures in front of him, arranged in a row. Each mixture has one of 100 different colors (colors have numbers from 0 to 99).

He wants to mix all these mixtures together. At each step, he is going to take two mixtures that stand next to each other and mix them together, and put the resulting mixture in their place.

When mixing two mixtures of colors a and b, the resulting mixture will have the color (a+b) mod 100.

Also, there will be some smoke in the process. The amount of smoke generated when mixing two mixtures of colors a and b is a*b.

Find out what is the minimum amount of smoke that Harry can get when mixing all the mixtures together.
Input

There will be a number of test cases in the input.

The first line of each test case will contain n, the number of mixtures, 1 <= n <= 100.

The second line will contain n integers between 0 and 99 - the initial colors of the mixtures.
Output

For each test case, output the minimum amount of smoke.
Example

Input:
2
18 19
3
40 60 20

Output:
342
2400

In the second test case, there are two possibilities:

    * first mix 40 and 60 (smoke: 2400), getting 0, then mix 0 and 20 (smoke: 0); total amount of smoke is 2400
    * first mix 60 and 20 (smoke: 1200), getting 80, then mix 40 and 80 (smoke: 3200); total amount of smoke is 4400

The first scenario is a much better way to proceed.

Does any one have a idea?

the source is spoj.. I posted it earlier and the idea is Dynami c Programming..check tripleM solution

For the DP, consider dp[a][b], which holds the minimum smoke for mixing the mixtures between index a and index b. Then consider states in order of length, i.e. (b-a).


calculus
integration
trigonometry
calculus computations

Hi, could you please help me on this problem.
Integrate $ \int F(x)$
where $ F(x) \equal{} \sqrt {4 \minus{} x^2}$


What is the sum of the first 80 positive odd integers subtracted from the sum of the first 80 positive even integers?

[hide] We know that the sum of the first $ n$ positive odd integers is $ n^2$, and that for even integers it is $ n(n\plus{}1)$. Solving this general case, we have $ n(n\plus{}1)\minus{}n^2\equal{}n^2\plus{}n\minus{}n^2\equal{}n$. Applying this to our specific case, we see that $ n\equal{}\boxed{80}$.[/hide]


integration
calculus
calculus computations

If $ f(x)$ is continuous and $ |f(x)| \le 1/(|x|^3\plus{}1)$ for all $ x$ show that,
$ \int \int \int \bigtriangledown \cdot f \,dx \equal{} 0$

There is some truth to this statement, but strictly speaking, it's false. First, $ \nabla \cdot \vec f$ does not necessarily exist. Second, the equality is true only if $ \iiint_{\mathbb R^3}$ is interpreted in a particular way.


Found this neat game, currently stuck on lvl. 22.

http://www.newgrounds.com/portal/view/408447

OMGOMGOMGOMGOMG I luv this game!!!!!!!

Minor note: stuck on lvl 13.

yayyy thx i rly love this game :D i didnt know there a sequel.

[img]http://img66.imageshack.us/img66/1216/logicdc1.jpg[/img]

p.s. lvl 22 and 23 seem to be a bit dificult but once u pass them u storm away in minutes :D


geometry
inradius
area of a triangle
Heron\u0027s formula
geometry unsolved

[quote="yrhc@eht"]in a triangle ABC, $\mid AB \mid$ =2cm, $\mid AC \mid$ =4cm and the inscribe circle in ABC has a radius 1cm, find $\mid BC \mid$.[/quote]

National olympiad???
Apply Heron's formula for the area of a triangle and the fact that the area equals inradius multiplied by semiperimeter. The rest is computation...

  darij


AOPS should make a book just full of problems..( and a solutions book) with nothing else.....

Well, that's not the real point of a book. What kind of problems should there be, what level, etc, are all questions you can ask.
There are a lot of questions available on the internet, so you can always find questions.

And I'm sure there are already tons of books that are full of just problems and solutions. The great thing about the AoPS books is that they [b]teach[/b] the concept very efficiently and then have problems at the end of each chapter that test if you truly grasped/understood the chapter's concepts.


how do you draw lines in latex? like, as in line AB, the little line above AB... thanks. :D

EDIT never mind.

Use
[code]\bar{x}[/code]
or
[code]\overline{AB}[/code]
to get $\bar{x}$ and $\overline{AB}$.


induction
inequalities open
inequalities

For any natural numbers $ n\ge 2$,

prove that $ (n\plus{}1)^{n\minus{}1}(n\plus{}2)^n > 3^n (n!)^2$


geometry
AMC
USA(J)MO
USAMO
AMC 10
AMC 8

I'd like to hear some recommendation on how appropriate the AMC 10 is for middle schoolers based on their AMC 8 scores.  For example, should student who scored say 15 on the AMC 8 be encouraged to take the AMC 10?  Or only those with scores of ~20?

My reasoning:  The AMC 10 and 12 exams have not been offered ever at any of the high schools in our town (according to the helpful AMC 10 registration coordinator).  I managed to get my 8th grade son's middle school to offer the AMC 8 by paying the registration fee, and a number of students took the test.   Two scored highly enough (22) that I know they should take the AMC 10, but I'm not sure about the rest.  I'm trying to decide whether to lobby the middle school to offer the AMC 10 or to approach one of the high schools instead.  The middle school will be more inclined to agree if they have several students who should take it.  And though a high school offering might be more broadly appropriate, I have much better contacts at the middle school.

Thanks!

I would encourage you to take the AMC 10 as a middle schooler just for the experience. It's different from AMC 8, but it isn't too difficult. I wouldn't know the exact conversions from AMC 8 to AMC 10. Take the AMC 10 because there are also more opportunities from doing well on it.

I think letting them take the AMC 10 is an excellent idea. As long as the middle schoolers have a sound knowledge of basic Algebra 1 or even just Pre-algebra, they wouldn't do too badly on the AMC 10. At my town's middle school, we had everybody who received greater than a 20 on the AMC 8 take the AMC 10, but that restriction was only due to the fact there were over 1000 kids at the school and many got 10-20. Theoretically, I think a person who got 15 on the AMC 8 can figure out at least 7 or 8 problems on the AMC 10. Those with more than 20 can range from 6 right to 15+ right.

i would have to say it would be a great thing to them get a feel of taking the AMC-10 for the future. Just make sure that they have a solid base on math topics like geometry, algebra. Also just to mention, AMC-8 scores definately don't correlate to AMC-10 scores. For example, i knew 8th grades  who got a 24 on AMC-8 and an eight-something on AMC-10, and an 8th grader who got a 22 on AMC-8 and 144 on AMC-10  (actually they got to the USAMO too, but that's a completely different case). Also make sure that the middle schoolers just can't get down on themselves if they don't get it some problems, especially towards the end with the geometry question as I have experienced before as a 7th grader. I wish them luck on the AMC-10!


[quote="onlylove_math"][quote="M.A"]Let $ a_{1},a_{2},...,a_{n} > 0$ such that $ a_{1}a_{2}...a_{n} \equal{} \frac {1}{2}$ then find the minimum value of
\[ \sum_{cyc}\sqrt {a_{1} \plus{} \frac {1}{a_{1}}}
\]
[/quote]
It 'll be solve by Minkopski[/quote]
Detail it my friend,I only solve it by AMGM or C_S.Help me

[quote="Allnames"][quote="onlylove_math"][quote="M.A"]Let $ a_{1},a_{2},...,a_{n} > 0$ such that $ a_{1}a_{2}...a_{n} \equal{} \frac {1}{2}$ then find the minimum value of
\[ P \equal{} \sum_{cyc}\sqrt {a_{1} \plus{} \frac {1}{a_{1}}}
\]
[/quote]
It 'll be solve by Minkopski[/quote]
Detail it my friend,I only solve it by AMGM or C_S.Help me[/quote]
$ P \equal{}\sum_{cyc}\sqrt {a_{1} \plus{} \frac {1}{a_{1}}} \ge\ \sqrt {(\sqrt {a_{1}}\plus{}\sqrt {a_{2}}\plus{}...\plus{}\sqrt {a_{n}})^2 \plus{} (\frac {1}{sqrt {a_{1}}}\plus{}\frac {1}{\sqrt {a_{2}}}\plus{}...\plus{}\frac {1}{\sqrt {a_{n}}})^2}$
$ P \ge\ \sqrt {n^2\sqrt[n]{a_{1}a_{2}...a_{n}} \plus{} \frac {n^2}{\sqrt[n]{a_{1}a_{2}...a_{n}}}}$
$ P \ge\ n\sqrt {\frac {1}{\sqrt [n]{2}}\plus{}\sqrt [n]{2}}$


calculus
rotation
geometry
geometric transformation
reflection

Consider the region bounded by f(x)=x^2 and g(x)=5x. Find the volume of the solid formed by rotating the region about the line x=2.

You know how to do general rotational volumes, right?  The problem for this one is that it's self-intersecting . . . when you rotate it, you end up filling in volume you've already filled in.  Thus, you have to break it into two seperate pieces and rotate them seperately.  I recommend that you reflect the two graphs about the line x = 2 in order to find a nice way of splitting.

Actually, on further thought, you might want to take the volume through normal means, and then compute the volume of the overlap seperately.

Make sure that you consider all the area on one side of the line of rotation -- that is, reflect part (either the bottom or the top).  Otherwise its volume will count for negative, which will be bad.

By the way, I get a very ugly answer.

I understand that it is self-intersecting. I did it, with some decimal approximations (my calc teacher assigned this, but I doubt that she meant it), and I wanted to know if I got it right.

Decimal approximations :(

well, what did you get?

129.949 or so. I lobbed decimals at 4 places, and lobbed the answer at 3


I can't get the \boxed{whatever i want boxed} to work. The error window pops up and says "undefined control sequence." Is there a package I need to include?

\boxed is an amsmath command so have you got \usepackage{amsmath} in your document's preamble?


linear algebra
matrix
vector
linear algebra unsolved

Let $A$ an hyperplan of $M_n(K),($where $K$ is $\Re$ or $C)$,which is stable by product.Show that $I_n \in A$.

It is known that the hyperplane will contain an invertible matrix, thus all the subalgebra generated by this matrix, thus also its inverse (which is in the subalgebra), thus their product, $I_n$.

In fact, I think I've managed to prove that for $n>2$ there aren't such hyper- planes. ;)

What I said in the first message is total non-sense. But what I said in the second remains true.

Could someone please delete this?  :blush: :)

I have another solution.
Suppose that $ R $ is a hyperplane in $ M_{n}(K) $, which is
stable by product. Then the linear relation which defines $ R $
can be written as $ tr(XA) = 0 $ , $ X \in M_{n}(K) $, for some
matrix $ A $. Then if we take some $ Y \in R $ then for any $ X
\in M_{n}(K) $ we have $ XY \in M_{n}(K) $, so  $ tr(XYA) = 0 $.
So every $  X \in M_{n}(K) $ satisfies $ tr(X(YA)) = 0 $,
therefore because $ R $ is a hyperplane, we have that $ XA =
\lambda A $ for some $ \lambda \in K $, for every $ X \in M_{n}(K)
$. Now if we denote by $ T \subseteq M_{n}(K) $ the set of all
matrices $ X $, such that $ XA $ is proportional to $ A $, then it
is a linear subspace of $ M_{n}(K) $. Also it is proper, because $
A \neq 0 $, so there exists some vector $ v $, such that $ w = Av
\neq 0 $, therefore if we take some linear operator $ X \in
M_{n}(K) $, such that $ Xw $ is not proportional to $ w $, then $
XA $ will not be proportional to $ A $. Therefore $ T $ is a
proper linear subspace of $ M_{n}(K) $, which contains a
hyperplane $ R $, therefore it is equal to $ R $. Now, if we take
$ X = I_{n} \in M_{n}(K) $, then of course $ XA = A $ is
proportional to $ A $, so $ I_{n} \in T = R $.

I hope it is correct.

Killer PM-ed me to tell me what was wrong with my solution, and I felt like an idiot :). Anyway, I wanted to post the solution along the same lines as my first post, which will hopefully be deleted by a moderator :).

Assume our hyperplane $\mathcal H$ does not contain $I_n$. Then every $n\times n$ matrix $A$ can be written uniquely as $M_A+\lambda_AI_n$ for a matrix $M_A\in\mathcal H$ and $\lambda_A\in K$. This means that for all $i\ne j$, our hyperplane will contain a matrix $A_{ij}$ with some $\lambda_{ij}$ on the main diagonal and a $1$ in position $(i,j)$. Since $A_{ij}^k\in\mathcal H,\ \forall k\in\mathbb N^*$, we immediately find that for all $i\ne j$ our hyperplane contains $M_{ij}$, the matrix with $0$ everywhere except for the position $(i,j)$, where it has a $1$, and from here it's easy: we add such matrices to get two different permutation matrices which are inverse to one another, and their product, which is $I_n$, will also belong to $\mathcal H$, contradiction.

Hopefully, it's Ok this time.


geometry
trigonometry
geometric transformation
homothety
ratio
reflection
geometry unsolved

[color=darkblue]Let $ ABC$ is a triangle. Let $ (O;R)$ is a change circle but $ (O;R)$ always pass point $ A$ and center $ O$ of circle lie on $ BC$. Let $ (O;R)$ meet $ AB, AC$ at $ M$ and $ N$ respectively. Let $ H$ is orthocenter of $ AMN$ triangle. Find locus of point $ H$.[/color]

Hi is it necessary circle $ (O)$ ? Let $ \angle BAC\equal{}\alpha,d$ be bisector of $ \angle A$ and $ f_1\equal{}H_{A}^{2|\cos\alpha|}$ is homothecy center $ A$ ratio $ 2|\cos\alpha|$ and $ f_2\equal{}R_d$ is reflection axis $ d$ then $ H\equal{}f_1\circ f_2(O)$ so $ \{H\}\equal{}f_1\circ f_2(BC)$.


geometry
circumcircle
cyclic quadrilateral
geometry unsolved

[quote="behdad.math.math"]consider ABC is an triangle with acute angles, and o is circumcenter of ABC triangle.
S is the circumcircle of ABO.lines CA and CB hit S in P and Q.prove that CO and PQ are perpendicular.[/quote]
It is true, for every circle with cord the side-segment $ AB$ of $ \bigtriangleup ABC,$ which intersects its sidelines $ AC,\ BC,$ at points $ P,\ Q,$ respectively.

This is the well known property of the [b][size=100]Antiparallel lines[/size][/b]. The line segment $ PQ,$ is always parallel to the tangent line of the circumcircle $ (O)$ of $ \bigtriangleup ABC,$ at vertex $ C$ ( easy to prove from the cyclic quadrilateral $ APQB$ ) and so, we have $ CO\perp PQ.$

The line segment $ PQ$ is called, the antiparallel line of the line segment $ AB,$ with respect to the angle $ \angle C$ ( = with respect to the rays of the angle $ \angle C$ ).

Kostas vittas.


Starting with n = 0

-1, 0, 1/2, 1/6, ...

Find the function in terms of n, and then, prove:

lim f(n) = 0
n-->infinity

Have fun! Easy once you find the pattern!

I can't prove things :(  But I think I have the pattern!!!1
[hide] are you adding 1/2 times the last thing you added to any number n starting with -1 as a(when a is the first term) and 1 being the first add-on(i know there has to be a better way to say this...
so that is is
-1,0,1/2,1/6,1/14 and so on...?(i'd put more but it's soon bed time...)[/hide]

hmm... I think there is a problem with this, I mean all you did was give us the first four terms,  so like I don't really know what kind of pattern your asking for.  For instance, those four points define a cubic, so if I felt so inclined, I could just figure out what that was.  Or it could possibly be some sort of recursion that we could find a closed formula for I mean it kind of looks like [hide] A_n = A_(n-1) - A_(n-2) x 1/n[/hide] but again I don't really know what your asking for.

[quote="neelnal"]f(n) = (n-1)/n!

As n approaches infinity, f(n) approaches 0.[/quote]

The problem with all these "find the pattern" problems is that there are lots of different solutions.

I have a pattern that correctly generates the given terms:

Criterion 1: The first term will be -1.
Criterion 2: The second term will be 0.
Criterion 3: The third term will be 1/2.
Criterion 4: The fourth term will be 1/6.
Criteiron 5: The subsequent terms will all be 42.

I know that there are ways to get around problems like this, but you really just have to go for the most sensible solution.

Is it...
[hide]$(n-1)/n!$[/hide]


The number of points with positive rational coordinates selected from the set of points in the xy-plane such that $x+y\leq 5$, is:

$\text{(A)} \ 9 \qquad \text{(B)} \ 10 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{infinite}$

$E$, as we can use negative coordinates. :P

*Edit* I can't read  :blush:

...Though still $E$

[hide]
(B) 10

Based on the conditions:
$x>0$
$y>0$
$x+y\le15$

This makes a triangle with 10 lattice points in it.

So the answer is 10.
[/hide]

[quote="Xantos C. Guin"](B) 10

Based on the conditions:
$x>0$
$y>0$
$x+y\le15$

This makes a triangle with 10 lattice points in it.

So the answer is 10.[/quote]Careful. It said positive rational coordinates. ;)

Oh duh I was thinking integers.

[hide]
(E) Infinite since there are an infinite number of rational coordinates.
[/hide][/hide]


topology
advanced fields
advanced fields unsolved

Let $X,Y$ be two topological spaces such that $X$ is homeomorphic to a subspace of $Y$, and $Y$ is homeomorphic to a subspace of $X$. Prove that $X$ can be partitioned into two parts $X_1,X_2$, and $Y$ can be partitioned into $Y_1,Y_2$ such that $X_i$ is homeomorphic to $Y_i$ for $i=1,2$.

Could you show me an example of the partition about X=R, Y=[0,1]?  I found it at least not very trivial, if possible.


What is the greatest value of $ x$ such that $ 10x$ is a divisor of the sum of the first twenty natural numbers?

The sum of the first twenty natural numbers is $ \frac{20\cdot21}2\equal{}210$. Since $ 10x|210\implies x|21$, it is easy to see that $ x\leq\boxed{21}$.


videos
probability
expected value
Support

I just finished the 60 diagnostic problems a couple of hours ago and have been watching the videos. I have watched three so far (Venn Diagrams Part II, Counting part II, and Expected Value part I), but my karma is at 990 (while one would expect it to be at 985). Also, I rated all the videos I watched, and commented on Venn Diagrams II. Interestingly, the ratings "disappear" when I refresh the page, my comment doesn't show up, and my karma only goes down  :huh:

$ E(\text{My karma every time I watch a video})\equal{}\minus{}5$  :rotfl:

On a serious note: Y'know, fix it.

A little more background information: Firefox 3.0.4 on Mac OS X 10.5.5. No weird extensions or anything, cookies+javascript enabled, latest flash player, etc, etc.

No, I don't believe I've rated any questions. I'll try that and post back here if I see any positive/negative result.

Thanks.

Edit:
This is getting weird...
My karma has dropped to 945, which is odd, because all I've done so far is a few problems and watched a couple of videos...
I left a comment on a video and rated a couple of videos but, as I mentioned before, the ratings/comments disappear when I refresh the page, and my karma only goes down.

Define "a few".  Because your karma goes down by $ 5$ for getting a question.

after you watch a video, do you click the back button?
if you do, maybe the rating doesn't save if you go back
i usually just click the get problem button and it seems to work fine

Yes, I do click the back button, but I can't imagine why that would affect the rating (once I submit a rating, it should be passed to the server-side script that inserts it into their db, so me going back should not change that).

I think I did 9 problems (my problem count is 69 I think). The thing is, I rated every solution and every video that I watched, so my karma should still be 1000.

To repeat:

[quote="levans"]Thanks for the report. We'll look into the video ratings not storing next week.[/quote]


floor function
limit
real analysis
real analysis unsolved

For $n \in N$ let $f(n)=\sum_{i=1}^{n}\frac{1}{\lfloor 10^{i}\pi \rfloor }$

Clearly $f$ is bounded.
Find \[\lfloor \lim_{n \rightarrow \infty}f(n) \rfloor\]

Er... is there some mistake?

$\lfloor 10^{i}\pi \rfloor > 3\cdot 10^{i}$ and hence $\frac{1}{\lfloor 10^{i}\pi\rfloor}<\frac{1}{3\cdot 10^{i}}$.

so, $\lim f(n) < \sum_{i=1}^{\infty}\frac{1}{3\cdot 10^{i}}<1$. Therefore the answer is $0$ since the sum is obviously positive.


linear algebra
matrix
geometry
geometric transformation
homothety
linear algebra unsolved

Find all $n \in \mathbb{N}$ for which there exist square $n \times n$ matrices $A$ and $B$ such that
$\mathrm{rank}\, A+\mathrm{rank}\, B\leq n$ and every square real matrix $X$ which commutes with $A$ and $B$ is of the form $X=\lambda I,\,\lambda \in \mathbb{R}.$

Each integer n is appropriate:
take A=diag(0,1,...,n-1) and B=[Bij] with Bij=1 for all i,j.
Rank(A)=n-1,rank(B)=1,UA=AU iff U is a diagonal matrix.
If U is a diagonal matrix then BU=UB implies that U is an homothety (cf. the first row of UB-BU).


There are 3 hats with numbers 1,2, and 3 on them, owned by Mr. Ratteh Relwob. He also owns 3 bowling balls numbered 1,2,3. If Mr. Relwob insists on storing his bowling balls in his hats and does not wish to put a ball in a hat with the same number, how many ways can he do it? (He makes large enough hats to put 3 bowling balls in)

Billy

Can one hat go waste in a condition Billy?

Yes, you don't have to use all the hats.

Im getting $2\cdot3 + 2 = 8$ combinations... is that right?

[hide="Answer"]Well, the hats may be large enough to put 3 bowling balls in, but he can't put all three in one hat, or else the number will overlap. ;) Let's make a chart:

[code] 1  2  3
---------
 B  C  A
 B A/C
 C  A  B
   A/C B
    C A/B
 C    A/B
B/C A
B/C    A[/code]

And that covers everything. There are three combinations of two, and each combination of two can only go in one hat; but, while that combination of two is in one hat, there are two places for the other ball to go, so we have $6$. Then, there are two places where each ball can go individually, and two combinations will cover that, we our answer is $\boxed{8}$.[/hide]


trigonometry
modular arithmetic
LaTeX
calculus
function

hey guys i was just wondering how this is true;

(i^(n-k)-(-i)^(n-k))/(2i)
	=	sin[1/2(n-k)pi].

Let me be sure I got this right...

$\frac{i^{n-k}-(-i)^{n-k}}{2i} = \sin{\frac12(n-k)\pi}$

Hmm... I'll need to work on that one

yeh thats what it is

except
everything after sin is in brackets

the second part should be sin(1/2(n-k)pi)

[hide]
Let $p=n-k$ and try mod 4.
This should be helpful because you can simplify $i^p$ based on what $p$ is equivalent in mod 4.
For instance, if you know $p=1\pmod{4}$, $i^p=i$, etc...
[/hide]

[hide="Hint"]Remember that $i^n$ is cyclic - $i^1=i,i^2=-1,i^3=-i,i^4=1$, then $i^5=i$ again.[/hide]

[hide]

It isn't true for all real numbers, but it is true for integers.  It's very easy to find a conterexample using your calculator.  Try $n-k=.2$ and you'll see for yourself.

However I will show it is true for all integers.

For ease of computations, let $x = n - k$, since $n-k$ is never actually manipulated.

$\frac{i^x - (-i)^x}{2i} = \frac{i^x - (i^3)^x}{2i} = \frac{i^{x+1} - i(i^3)^x}{-2} = \frac{i^{3x + 1} - i^{x+1}}2 = \frac{i^{x+1}(i^{2x} - 1)}2 = \frac{i^{x+1}(i^x-1)(i^x+1)}2$

Now we examine this casewise.

If $x \in \{...,-4,0,4,8,...\}$ then $i^x = 1$, and the numerator (which includes $i^x - 1$) becomes zero.  $\sin{\frac12x\pi} = \sin{2k\pi} = 0$ and it is equivilant.

If $x \in \{...,-3,1,5,9,...\}$ then $i^x = i$, and the numerator becomes $-(i-1)(i+1) = -i^2 + 1 = 2$, and since the denominator equals 2, the whole thing equals 1.  $\sin{\frac12x\pi} = 1$ and it is equivilant.

You cans work out the last two yourself; they're very similar.

[/hide]

Edit: I should have hid this the first time.

[quote="boxedexe"]Let $x=n-k$
$\\\frac{i^{n-k}-(-i)^{n-k}}{2i}=\frac{(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})^x-(\cos\frac{\pi}{2}-i\sin\frac{\pi}{2})^x}{2i}\\=\frac{e^{ix\frac{\pi}{2}}-e^{-ix\frac{\pi}{2}}}{2i}=\frac{\sinh(ix\frac{\pi}{2})}{i}=\sin(x\frac{\pi}{2})=\sin(\frac{1}{2}\pi(n-k))$

Q.E.D

EDIT: I had several problems with latex, so calculus lover beated me. :(
Masoud Zargar[/quote]

I'm not sure that hyperbolic functions generally, let alone of a complex argument, are for an intermediate topic.

ok i understand now :lol:
thank you everybody  :)

[quote="chess64"][quote="boxedexe"]beated[/quote]

:D[/quote]

You're right.  beat would be the correct one according to the illogical language of english. (illogical$\rightarrow$ :()

English is illogical*, so English=illogical.

Illogical = impractical by the fundamental theorem of illogicalness and impracticalness, so English = impractical by transivity.

Impractical things should not exist, so English should not exist, by transivity again. $\boxed {}$

*[size=75]English is illogical because ${\small \text {``beated''} \ne \text { past participle of ``beat''.}}$ For other examples see the proof that English is illogical in reference [1].[/size]

[u]References[/u]
1. English is illogical by chess64, published 2007.

[i]Dear Sen[/i], here is a particular case of a  more general question:

$z=\cos x+i\sin x\Longrightarrow z ^p=\cos px+i\sin px,\ \overline z ^p=\cos px-\sin px\Longrightarrow$
$\sin px=\frac{z^p-\overline z^p}{2i}$. For $z: =i\ (\ x=\frac{\pi}{2},\ \overline z =-i\ )$ and $p: =n-k$ we get:

$\boxed {\ \sin \frac{(n-k)\pi }{2}=\frac{i^{n-k}-(-i)^{n-k}}{2i}\ }$


Let a,b,c are positive real satisfied (ab) 2 +(bc) 2 +(ca) 2  \geq (abc) 2 .
Prove that:
               1/(c 3*(a 2 +b 2 )+1/(b 3*(c 2 +a 2)+1/(a 3*(b 2 +c 2 ) \geq  \sqrt 3/2

From what I have seen, this magazine has beautiful problems. This is one of them. Let 1/a^2=x and so on. Then the inequality becomes sum xsqrt(x)/(y+z)>=sqrt(3)/2. But sum xsqrt(x)/(y+z)=sum x^2/[sqrt(x)(y+z)]. But sum sqrt(x)(y+z)=sum sqrt(y+z)*sqrt(x(y+z))<=sqrt(sum (y+z))*sqrt(sum x(y+z))=2sqrt[(sum x)(sum xy)]<=2sqrt[(sum x)*(sum x)^2/3]<=2sqrt[(sum x)^4/3]=2/sqrt(3)*(sumx)^2 and the solution is finished. I have used that sum x>=1.


Can the Ti-89 calculator be used on the PSAT and SAT tests?

rly? i just took my SAT's this morning. u don't need it. i didn't take a calculator at all.

Well, you don't need it. That's not what he was asking though... for the PSAT, SAT Math, and the SAT Math Level 2, you can use a TI-89 Calculator.


geometry
incenter
geometric transformation
homothety
trapezoid
analytic geometry
reflection

Let $\omega$ be incircle of $ABC$. $P$ and $Q$ are on $AB$ and $AC$, such that $PQ$ is parallel to $BC$ and is tangent to $\omega$. $AB,AC$ touch $\omega$ at $F,E$. Prove that if $M$ is midpoint of $PQ$, and $T$ is intersection point of $EF$ and $BC$, then $TM$ is tangent to $\omega$.
[i]By Ali Khezeli[/i]

Let $I$ be the incenter and $L$ is the tangency point of $\omega$ with $BC.$ Now let $AL\cap \omega =S.$ Now it is easily seen that $FLES$ is harmonic quadrangle and so $TS$ is tangent to $\omega.$ (It follows easily from polus and polar) Let $AL\cap PQ=R.$ Now it is easily seen that a homothety $h$ with center $A$ which maps $\omega$ to $\omega_{a}$ also maps $PQ$ to $BC.$ So now we conclude that if $PQ\cap\omega =W$ we have $PR=WQ.$ Because of the fact that $PQ\parallel BC$ it follows that $WL$ is diameter of the incurcle and therefore $\angle WSR=\pi/2.$ But if $TS\cap PQ=M_{1}$ in the right-angle triangle $RSQ$ we have $M_{1}W=M_{1}S$ (because $SM_{1}$ and $M_{1}W$ are tangents to $\omega$) and from well-known problem we conclude that $M_{1}$ is the midpoint of $RW.$ Thus $M_{1}P=M_{1}R+RP=M_{1}W+WQ=M_{1}Q$ and so $M_{1}=M.$ QED

I have a solution with complex numbers!
If the origin of complex plane is the incenter, and the incircle is the unit circle, we will have:

$m = \frac{x(xz+xy-2yz)}{(x-z)(x-y)}$ and $t = \frac{x(xy+xz-2yz)}{x^{2}-yz}$,

where y = E, z = F and x = K (K is the intersection of the incircle with BC)!

To prove that TM is tangent to the incircle is easy!

Another "projective" solution, a little simpler from the one bilarev gave, should be by considering $N$ the midpoint of $(BC)$, and denote $V \in BQ \cap CP$.

Because $BCQP$ is a trapezium, we have that $A,M,V,N$ are collinear and $(A,M,V,N)$ form a harmonical division.

Now because $BCQP$ is a circumscribe quadrilateral from the Newton theorem, we have that $V \in EF$, thus $T(A,M,V,N)$ is a harmonical division, and now by intersecting it with $AD$ we have that $(AX_{1}YD)$ is a h.div, where $X \in TM \cap AD$ and $Y \in TV \cap AD$.

But because $EF$ is the polar of $A$, we have that $(AX_{2}YD)$ is a h.div, where $X_{2}\in AD \cap w$. Thus $X_{1}\equiv X_{2}$, but now $TX_{2}$ is trivially tangent to $w$, because $AD$ is the polar of $T$, thus the point $X_{2}\in AD \cap w$, is the place where the second tangent cuts the circle, thus $TM$ is tangent to $w$.

Actually I improved my solution and now I think this is the shortest one:

Just consider $ X \in AD \cap w$ and $ Y \in AD \cap EF$, because that $ AD$ is the polar of $ T$ we have that $ TX$ is tangent to $ w$, but we have that $ (AXYD) \equal{} \minus{} 1$, so $ B(AXYD) \equal{} \minus{} 1$, therefore by intersecting it with $ PQ$, we have that $ (P,M,Q, \infty) \equal{} \minus{} 1$, thus $ M$ is the midpoint of $ (PQ)$., and thus the problem is solved. (I used that $ BY$ cuts $ PQ$ in $ Q$ because of the Newton theorem).

[size=130][color=darkred][b]PP ([u]Iran TST 2007[/u]).[/b] Let $\omega$ be the incircle of the triangle $ABC$. Let $P\in AB$ and $Q\in AC$ be two points such that

$PQ\parallel BC$ and $PQ$ is tangent to the circle $\omega$. The lines $AB$, $AC$ touch the circle $\omega$ at $F$, $E$ respectively. Denote

the middlepoint $M$ of $[PQ]$ and $T\in EF\cap BC$. Prove that the line $TM$ is tangent to the circle $\omega$ ([u]Ali Khezeli[/u])[/color][/size]

[size=115][color=darkblue][b][u]Proof[/u].[/b] Denote $D\in BC\cap w$ , $N\in IM\cap BC$ and the midpoints $S$ , $V$ of the segments $[AD]$ , $[BC]$ respectively. Observe that

$M\in AV$ , $IM=IN$ . From the well-known properties $\boxed{\ S\in IV\ }\ (1)$ and $\boxed{\ TI\perp AD\ }\ (2)$ obtain immediatelly that $MN\parallel AD$ ,

i.e. $TI\perp MN$ and the line $TM$ is tangent to the circle $w$ because the lines $TM$ , $TN$ are symmetrically w.r.t the line $TI$ .

$(1)\blacktriangleright$ Suppose w.l.o.g. $b\ne c$ . Denote the intersection $L\in AI\cap BC$ . Prove easily that $VL=\frac{a|b-c|}{2(b+c)}$ , $VD=\frac{|b-c|}{2}$ ,

$\frac{IA}{IL}=\frac{b+c}{a}$ . Apply the [u][b]Menelaus[/b]' theorem[/u] to the transversal $\overline{VIS}$ and the triangle $ADL$ , i.e. $\frac{VD}{VL}\cdot\frac{IL}{IA}\cdot\frac{SA}{SD}=1$ $\Longleftrightarrow$ $S\in IV$ .

$(2)\blacktriangleright$ The line $EF$ is the polar $\alpha$ of the point $A$ w.r.t. the circle $w$ and the point $T\in\alpha$ $\implies A$ belongs to the polar $\tau$ of the point $T$

w.r.t. the circle $w$ and $D\in\tau$ $\implies$ the polar of $T$ is $\tau=AD$ $\implies$ $TI\perp AD$ . [u][b]Otherwise[/b][/u] prove easily that $TA^{2}-TD^{2}=IA^{2}-ID^{2}$ .

[u]See [b]PP14[/b][/u] from[/color] [url=http://www.artofproblemsolving.com/blog/101045][color=red][b]here[/b][/color][/url][/size]

let $AD$ intersect incircle at $S$ and $PQ$ at $D'$ and consider incircle is tangent to $PQ$ at $R$ , we know

that  $MD'=MR$ and $\angle RSD=90^{\circ}$  hence $MD'=MR=MS$ also we have $\angle ADB=\angle MD'S=\angle D'SM$

hence $MS$ is tangent to incircle , let $MS\cap BC=T'$ , polar of $T'$ pass through $A$ so polar of $A$ (EF)

pass through $T'$ hence $T'\equiv T$

[quote="Joao Guerreiro"]E.lopes, how do you prove that the line TM is tangent to the unit circle?[/quote]

if $z+t.z' = c$ is the equation of TM (where $z'$ is the conjugate of $z$), the foot of the perpendicular, trought the incenter, to TM, will have cordinates $\frac{c}{2}$. So, we can prove that $|\frac{c}{2}| = 1$!

hoch ghaste tohin be aghaye khezeli ro nadaram valiiiiiiii
in soal akhe chie ke dadan bara TS....
mmidadan marhale 2 laaghal mishod ye karish  kard:d

Another approach:

By direct calculation we get $RM\cdot TD=IR^2$, i.e. $\triangle RIM \sim \triangle DTI$, done.

Best regards,
sunken rock

What is R?
What is D?
What is I?
Your solution, sunken rock, is very unclear! :(

@CDP100:
Sorry, $\{R\} \equiv \omega \cap PQ$, $I$ - the center of $\omega$, $\{D\} \equiv \omega \cap BC$.
Further, after having seen, by compiling $TD$ and $RM$ that $IR^2=RM\cdot TD$, the reflection of $D$ in $IT$ coincides with the reflection of $R$ in $IM$, which solves the problem.

[b]Note[/b]: It has been a [b][u]typo[/u][/b], I edited it. Sorry for any inconveniences caused!

Best regards,
sunken rock

Dear Mathlinkers,
1. let I the center of the incircle, D be the point of contact of the incircle and BC, D' the antipode of D, D" the second point of intersection of AD with the incircle and D+ the point of intersection of AD and PQ.
2. It is known that TD" is tangent to the incircle at D"
3. M is the midpoint of D'D+
4. IM is parallel to AD and perpendicular to D'D"
5. By symmetry wrt MI, MD" is tangent to the incircle at D"…
and we are done
Sincerely
Jean-Louis

[quote="sunken rock"]@CDP100:
Sorry, $\{R\} \equiv \omega \cap PQ$, $I$ - the center of $\omega$, $\{D\} \equiv \omega \cap BC$.
Further, after having seen, by compiling $TD$ and $RM$ that $IR^2=RM\cdot TD$, the reflection of $D$ in $ID$ coincides with the reflection of $R$ in $IM$, which solves the problem.

Best regards,
sunken rock[/quote]

Thank you very much!

Let $BQ \cap CP = X$.
$X \in EF$ (by Brianchon on $PQECBF$).
Since $X \in AM$ we have $TM$ is tangent to $\omega$ (by converse of Brianchon on $TBAEQM$).

Let $AD$ intersect $TM$ at $U$, $D$ be the $A$-intouch point, and $DG$, where $G$ is the antipode of $D$ on $\omega$, intersect the line parallel to $BC$ at $Y$.  Let the concurrency point of $EF, DG$, and $AN$ (where $N$ is the midpoint of $BC$) be $X$, and let $FE$ intersect $AD$ be $V$.

Then observe $$(A, V; U, D) = (A, X; M, N) = (Y, X; G, D) = (P_{\infty}, N; AG \cap BC, D) = -1.$$  As $V$ lies on the polar of $A$, $U \in \omega$, and as $AD$ is the polar of $T$, this is equivalent to $U$ being the other tangency point of $T$.

Harmonic cookie :) (also probably the most complicated solution with a million intersections)

Let $K=\overline{AD} \cap \omega$, $(K,D;F,E) = -1 = (\overline{TT} \cap \omega,D;F,E)$ so $TK$ is tangent to $\omega$. Let $D'$ be the andipode of $D$ in $\omega$ which is the touchpoint of $\overline{PQ}$ with $\omega$, also let $N$ be the midpoint of $\overline{BC}$, $D'' = \overline{AD'} \cap \overline{BC}$, $R=\overline{DD'} \cap \overline{EF}$ and $S=\overline{KD} \cap \overline{EF}$. By incircle concurrency lemma $R$ lies on $\overline{AN}$.
[b]Key Claim: [/b] $(A,R;M,N) = -1$.
[i]Proof: [/i]
$$(A,R;M,N) \stackrel{D'}{=} (D'' , D;\infty , N) = -1$$
where $N$ is midpoint of $\overline{D' D''}$ because of a well known config. Then note that
$$(A, R ; \overline{TK} \cap \overline{AN}, N) \stackrel{T}{=} (A,S; K,D) = -1 = (A,R;M,N)$$
So $M \equiv \overline{TK} \cap \overline{AN} \implies T-K-M$ and thus $TM$ is tangent to $\omega$ at $K$.

This is such a bruh solution

Let $N,N'$ be touch point of $\omega$ with $AB,PQ.$ Let $TR$ tangent to $\omega$ at $R\neq N,$ and redefine $M$ as $RT\cap PQ.$
Observe that $NFRE$ is harmonic, so by duality wrt $\omega$ $$(PQM\infty )=N'(EFRN)=-1,$$ i.e. $RT$ bisects $PQ$ as desired.

[asy]
/* TAKEN FROM V_ENHANCE AND EDITED */
size(10cm);
defaultpen(fontsize(9pt));
pair A = dir(130);
pair B = dir(210);
pair C = dir(330);
draw(A--B--C--cycle);

draw(incircle(A, B, C), blue);
pair I = incenter(A, B, C);

pair F = foot(I, A, B);
pair E = foot(I, A, C);
pair T = extension(E, F, B, C);

draw(E--T--B);

pair P = extension(A, B, I, I+rotate(-90)*(I-B));
pair Q = extension(A, C, I, I+rotate(90)*(I-C));
draw(P--Q);
pair M = midpoint(P--Q);
draw(T--M, red);

pair D = foot(I, B, C);
draw(A--D, dotted);

pair X = extension(A, D, M, T);
pair Y = 2*I-D;
pair Sp = extension(A,D,P,Q);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$F$", F, dir(150));
dot("$E$", E, dir(45));
dot("$T$", T, dir(T));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(20));
dot("$M$", M, dir(90));
dot("$D$", D, dir(D));
dot("$S$", X, dir(170));
dot("$R$", Y, dir(90));
dot("$S'$", Sp, dir(110));

/* Source generated by TSQ */
[/asy]
Define $D$ the tangency of $\omega$ with $BC.$ Let line $AD$ intersect $\omega$ at $S$ and $PQ$ at $S'.$ Let $R$ be the tangency of $PQ$ with $\omega.$

Since $AF$ and $AE$ are tangents of $\omega$ at $E$ and $F,$ it follows $FDES$ is a harmonic quadrilateral. By well known properties, this implies $TS$ is tangent to $\omega$ at $S.$ It suffices to show $MS$ is tangent to $\omega.$ First, consider the homothety $\mathcal{H}$ centered at $A$ bringing $\omega$ to the $A$-excircle. It follows $\mathcal{H}(S')=D$ and $\mathcal{H}(R)=R',$ where $R'$ is the tangency point of $BC$ and the $A$-excircle.

By excircle properties, it follows $BD=CR',$ which implies the midpoint of $DR'$ is also the midpoint of $BC,$ and therefore by the homothety the midpoint of $S'R$ is also the midpoint of $PQ,$ which happens to be $M.$ Thus, $MS'=MR.$ Now notice since $D,I,R$ are collinear it follows $R$ and $D$ are antipodes. Thus $\angle DSR=90^\circ.$

Note then $\angle RSS'=90^\circ,$ so the circumcircle of $SS'R$ is centered at $M,$ and thus $MS'=MR=MS,$ so $MS$ is tangent to $\omega,$ and therefore $TM$ is tangent to $\omega,$ as required.

Let $N$ the midpoint of $BC$, and let the tangents from $N$ to $\omega$ be $D,K$ where $D$ lies on $BC$, let $G$ the point where $PQ$ is tangent to $\omega$ (so $I,D,G$ are colinear), let $AM$ intersect $\omega$ at $X,Y$ and let $AD \cap \omega=L$. Now since its known that $A,G,K$ are colinear we project and take polars w.r.t. $\omega$
$$-1=(X,Y; D, K) \overset{A}{=} (Y, X; L, G) \implies \mathcal P_M=LG \implies M \in \mathcal P_L$$
But since $\mathcal P_L=TL$ we have that $T,L,M$ are colinear hence $TM$ is tangent to $\omega$ as desired.

Let $D$ be the contact point of $\omega$ with $BC$, and let $R$ be the antipode of $D$ in $\omega$. Let $S\neq R$ in $\omega$ such that $MS$ tangent to $\omega$. Finally, let $D'=AD\cap PQ$.
By homothety $D'$ is the contact point of the incircle of $\triangle APQ$ with side $PQ$, and so $MD'=MR=MS$, so $RS\perp SD'$.
Also, $DR$ diameter in $\omega$, so $RS\perp SD$, and so $A-D'-S-D$ collinear. Finally,
$T\in \mathcal{P}(A) \Longrightarrow AD=\mathcal{P}(T) \Longrightarrow S\in \mathcal{P}(T) \Longrightarrow T\in \mathcal{P}(S)$,
and so $TM$ is tangent to $\omega$ at $S$.

Let $\omega$ touch $BC$ at $D$, $AD$ meet $\omega$ again at $R$ and intersect $PQ$ at $Y$, $PQ$ touch $\omega$ at $X$, and $AX \cap BC = X_1$.

Since $DERF$ is harmonic, we know $TR$ is tangent to $\omega$. Hence, it suffices to show $MR$ is tangent to $\omega$.

The Diameter of the Incircle Lemma implies $BD = CX_1$. Now, since $APQ$ and $ABC$ are homothetic, it follows that $MX = MY$.

Because $DX$ is a diameter of $\omega$, we have $$\angle XRY = 180^{\circ} - \angle XRD = 90^{\circ}$$ which means $M$ is the circumcenter of $XRY$. This yields $MR = MX$, as required. $\blacksquare$

[b]Remark:[/b] The solution is quite simple. At first, however, I sort of overcomplicated this problem by adding $AM \cap DX \cap EF$ and attempting to use harmonic bundles.

Let the incircle be the unit circle, and set $d = -i$. Then, we have
$$p = \frac{2fi}{f+i} \text{ and } q = \frac{2ei}{e+i},$$so we can compute$$m = \frac{2ief-e-f}{(e+i)(f+i)}.$$Next, using a Cartesian coordinate system, we can compute$$t = \frac{f+\overline f}2 - \frac{i(e+\overline e - f  - \overline f)}{e-\overline e - f + \overline f}\left(\frac{f-\overline f}{2i} + 1\right) - i = \frac{-i(e+i)(f+i)}{ef+1} - i.$$Then, the reflection $d'$ of $D$ over $\overline{TO}$ is$$d' = \frac{t \cdot i}{\overline t} = \frac{2ef+i(e+f)}{e+f+2i}$$by the complex reflection formula; we have $\overline{TD"}$ is tangent to the unit circle. Then, it suffices to show that $M$ is the tangent intersection of the tangent at $i$ and the tangent at $d'$. In other words, we can simply check that$$\frac{2i\left(\frac{2ef+i(e+f)}{e+f+2i}\right)}{\frac{2ef+i(e+f)}{e+f+2i} + i} = \frac{2i ef-e-f}{(e+i)(f+i)} = m,$$so it follows that $\overline{TD'}$ passes through $M$, and thus $\overline{TM}$ is tangent to the incircle.

Let $D$ be the $A$-contact point, $D'$ be the tangency point of $\overline{PQ}$ with $\omega$, and $D'',X$ respectively be the intersections of $\overline{AD}$ with $\overline{PQ}$ and $\omega$. Then $D'$ and $D''$ are reflections of each other over $M$ and $D'$ and $D$ are reflections of each other over the incenter $I$.

Because $T$ lies on the polar of $A$ wrt $\omega$, $A$ lies on the polar of $T$, which passes through $D$ as well. Hence it passes through $X$, so $\overline{TX}$ is tangent to $\omega$. It then suffices to show that $\overline{MX}$ is also tangent to $\omega$, i.e. $MX=MD'$. Since $\angle D''XD'=\angle DXD'=90^\circ$ and $M$ is the midpoint of $\overline{D'D''}$, it follows that it is the circumcenter of right triangle $D'XD''$, hence $MX=MD'=MD''$. $\blacksquare$

Take the circle the unit circle. Let the circle touch $BC$ at $D$ .We denote the complex co-ordinate of a point $X$ by $x$ .
We set  $d=1$  it is clear $PQ$ toiuches the unit circle at $-1$  .By standard formulae $b=\frac {2f}{1+f}$ and $c=\frac{2e}{1+e}$ and $p=\frac{-2f}{f-1}$ and $q=\frac{-2e}{e-1}$. Note that $ \Re{(t)} = 1$ and the equation of chord $EF$ is $z=e+f-ef\overline{z}$  and $t+\overline{t} = 2 \implies \overline t = 2-t$ and putting this in the equation of $EF$ we get $t = \frac{e+f-2ef}{1-ef}$ and $ m= \frac{e+f-2ef}{(e-1)(f-1)}$.let $o'$ be the complex co-ordinate of the reflection of origin wrt the line $MT$   $$\frac{\overline{m}}{\overline t - \overline m } = \frac{o' - m}{t-m}$$. let the mid-point of $O$ and $O'$ is $P$. then $p=\frac{o'+0}{2} = -\frac{e+f-2ef}{e+f-2}$(Stepped the calculation part ) Now note that $ \overline {p} = -\frac{e+f-2}{e+f-2ef}$.So , $p\overline p  = 1$ and so $| p| = 1$ and so $P$ lies on the unit circle and so $OP\perp TM$ and hence we are done .

dkaoskoas

hi because

$I$ is the incenter
$N=AD\cap \omega$
$Y$ is midpoint of $BC$
$D'=DI\cap \omega$
$X=DI\cap AM\cap EF$ (well-known to exist)
$Z=AD\cap EF$

Since $T$ lies on the polar of $A$ so $A$ lies on the polar of $T$ hence $AD$ is the polar hence $TN$ is tangent to $\omega$ hence it suffices to show that $M,N,T$ are collinear. If we can show that $(A,X;M,Y)=-1$ we will be done since we also have $(A,Z;N,D)=-1$ and we could finish by Prism Lemma (aka phantom points).

Now let $\infty_{BC}$ be the point at infinity along line $BC$ and project onto $\overline{DID'}$, suppose $A\to A'$. It suffices to show that $(A',X;D',D)=-1$ which is trivial: realize that $A'\in (AEIF)$ hence by Shooting Lemma
\[IX\cdot IA'=IE^2=IF^2=ID^2=ID'^2\]
done. Yay!

Let $\omega$ intersect $\overline{BC}$ and $\overline{PQ}$ at $D$ and $G$, and define $K = \overline{AD} \cap \omega, L = \overline{AD} \cap \overline{PQ}$. First, we note that $KDFE$ is harmonic, so $\overline{TK}$ is tangent to $\omega$. Thus, it suffices to show that $\overline{MK}$ is tangent to $\omega$ too.

However, since a homothety at $A$ takes $\overline{PQ}$ to $\overline{BC}$ and $G$ to the $A$-extouch point, we have $LM = MG$. But $\overline{GD}$ is a diameter of $\omega$ because $\overline{PQ} \parallel \overline{BC}$, so $\overline{GK} \perp \overline{LKD}$, and $M$ is the circumcenter of right $\triangle LKG$. Therefore $MK = MG$, and since $MG$ is tangent to $\omega$, $MK$ is as well.

Let $D$ and $D'$ be the intersections of $\overline{BC}$ and $\overline{PQ}$ with $\omega$, respectively – let $D''$ be the intersection between $\overline{AD}$ and $\overline{PQ}$, and lastly, let $U$ be the second intersection of $\overline{AD}$ with $\omega$. Because $DD'$ is a diameter of $\omega$, $U$ is the foot from $D'$ to $\overline{AD}$.

We claim that $\overline{MU}$ and $\overline{TU}$ are both tangent to $\omega$, which implies the problem.
[list=1]
[*]To prove that $\overline{MU}$ is tangent to $\omega$: a homothety of factor $\tfrac{1}{2}$ centered at $D'$ sends $D$ to $I$ and sends $D''$ to $M$. Thus, $MI$ is the perpendicular bisector of $\overline{UD'}$. Since $MD'$ is obviously tangent to $\omega$, it follows that $MU$ is tangent to $\omega$ as well.
[*]To prove that $\overline{TU}$ is tangent to $\omega$: let $G$ be the foot from $I$ to $\overline{AD}$. Note that $AFGIE$ is cyclic with diameter $AI$. Now, radical axis theorem on $(AFGIE)$, $(GID)$ and $\omega$ means that $T$, $G$ and $I$ are collinear. So, $TI$ is the perpendicular bisector of $\overline{UD}$, so $\overline{TU}$ is tangent to $\omega$.
[/list]


modular arithmetic
number theory
divisibility tests
algebra unsolved
algebra

Let $ M$ and $ N$ be two nine-digit positive integers with the property that if ANY one digit of $ M$ is replaced by the digit of $ N$ in the corresponding place (e.g. the "tens" digit of $ M$ replaced by the "tens" digit of $ N$) then the resulting integer is a multiple of $ 7$.

Prove that any number obtained by replacing a digit of $ N$ by the corresponding digit of $ M$ is also a multiple of $ 7$.

Find an integer $ d > 9$ such that the above result concerning divisibility by $ 7$ remains true when $ M$ and $ N$ are two $ d\minus{}$digit positive integers.

[quote="AndrewTom"]Let $ M$ and $ N$ be two nine-digit positive integers with the property that if ANY one digit of $ M$ is replaced by the digit of $ N$ in the corresponding place (e.g. the "tens" digit of $ M$ replaced by the "tens" digit of $ N$) then the resulting integer is a multiple of $ 7$.

Prove that any number obtained by replacing a digit of $ N$ by the corresponding digit of $ M$ is also a multiple of $ 7$.

Find an integer $ d > 9$ such that the above result concerning divisibility by $ 7$ remains true when $ M$ and $ N$ are two $ d \minus{}$digit positive integers.[/quote]

Let $ d_p(x)$, for $ p\ge 0$ and $ x\in\mathbb N$ be the digit in $ (p\plus{}1)^{th}$ position from the right in decimal representation of $ x$
Let $ \Delta_p\equal{}d_p(N)\minus{}d_p(M)$

The property may be written $ M\plus{}\Delta_p10^p\equal{}0\pmod 7$ $ \forall p\in\{0,1,...,8\}$

From this we deduce $ \Delta_p10^p\equal{}\minus{}M\pmod 7$ and so $ N\equal{}M\plus{}\sum_p\Delta_p10^p\equal{}M\minus{}9M\pmod 7$ and so $ M\plus{}N\equal{}0\pmod 7$

So $ N\minus{}\Delta_p10^p\equal{}0\pmod 7$ and the property true for $ (M,N)$ is symetrical and true for $ (N,M)$

And this property is obviously true for any $ d\equal{}2\pmod 7$ (in order to have $ M\minus{}dM\equal{}\minus{}M\pmod 7$) and so, for example, for $ d\equal{}16$


$ a\plus{}b\plus{}c\equal{}1$, $ a,b,c$ Real positive. Proof :
$ \frac{ab\plus{}c}{a\plus{}b}\plus{}\frac{ac\plus{}b}{a\plus{}c}\plus{}\frac{bc\plus{}a}{b\plus{}c}\geq 2$

The inequality is equivalent to \[ \sum\frac {ab \plus{} 1\minus{}a\minus{}b}{a \plus{} b}\ge 2\iff\sum\frac {(1 \minus{} a)(1 \minus{} b)}{a \plus{} b}\ge 2\]
So we need only prove that
\[ \sum\frac {(b \plus{} c)(c \plus{} a)}{a \plus{} b}\ge 2
\]
Let $ a \plus{} b \equal{} x,b \plus{} c \equal{} y,c \plus{} a \equal{} z$. Then we have $ x \plus{} y \plus{} z \equal{} 2$ and we need to prove that

$ \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y}\ge 2\iff x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2\ge xyz(x \plus{} y \plus{} z)$

$ \iff \sum_{cyc}(xy \minus{} yz)^2\ge 0$, obviously true.

The following stronger one is also true for $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$
\[ \frac {ab \plus{} c}{b \plus{} c} \plus{} \frac {bc \plus{} a}{c \plus{} a} \plus{} \frac {ca \plus{} b}{a \plus{} b}\ge 2
\]

EDIT: This must be an equality, sorry for not noticing it earlier. Then we get another proof of the original inequality by rearrangement.

[quote="nayel"]The following stronger one is also true for $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$
\[ \frac {ab \plus{} c}{b \plus{} c} \plus{} \frac {bc \plus{} a}{c \plus{} a} \plus{} \frac {ca \plus{} b}{a \plus{} b}\ge 2
\]
[/quote]

$ b\plus{}c\equal{}x , c\plus{}a\equal{}y , a\plus{}b\equal{}z$ ----> $ a\equal{}1\minus{}x,b\equal{}1\minus{}y,c\equal{}1\minus{}z$ ----> $ \frac {ab \plus{} c}{b \plus{} c} \plus{} \frac {bc \plus{} a}{c \plus{} a} \plus{} \frac {ca \plus{} b}{a \plus{} b}\equal{}x\plus{}y\plus{}z\equal{}2$

There is an other solution ,too:
$ ab \plus{} c \equal{} ab \plus{} c(a \plus{} b \plus{} c) \equal{} (b \plus{} c)(c \plus{} a)$
Then continue by the same way of solution no.1  :spam:
Put $ b \plus{} c \equal{} x;c \plus{} a \equal{} b;a \plus{} b \equal{} z$
We have:
$ \sum{ab \plus{} c}{a \plus{} b} \equal{} \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y}$
We will prove that: $ \sum{xy}{z}\geq x \plus{} y \plus{} z(1)$
That right!
By AM-GM:
$ \frac {xy}{z} \plus{} \frac {yz}{x}\geq 2y$
$ \frac {yz}{x} \plus{} \frac {zx}{y}\geq 2z$
$ \frac {xy}{z} \plus{} \frac {zx}{y}\geq 2x$
$ \Rightarrow (1) right$ :D
$ \Rightarrow \sum{ab \plus{} c}{a \plus{} b}\geq x \plus{} y \plus{} z \equal{} 2(a \plus{} b \plus{} c) \equal{} 2$
That inequalities solved


What is the main difference between mathematics and physics?

I would like an answer in simple terms.

Thanks

Physics is an attempt to understand the real world using mathematical models.

The central difference, then, is that the axioms of physics are assumptions about the real world that might change based on new experimental evidence, whereas the axioms of mathematics are arbitrary and change whenever it is useful for them to change.  (Well, perhaps "change" is the wrong word.  I mean that we can introduce new axioms that produce new systems without regard for the real world - non-Euclidean geometry, for example.)

So, a mathematician would not be able to solve physics problems and vice-versa?

Why do colleges have a physics department and a math department?

If BOTH teach numbers, why not include BOTH into one department?

I am looking for many views.

[quote="Interval"]So, a mathematician would not be able to solve physics problems and vice-versa?[/quote]

I'm not sure you understood what I meant.

[b]Physics is an application of mathematics.[/b]  Thinking of them as two separate fields on equal ground with each other ignores the universality of mathematics.

Anything which studies patterns and structure in any kind of logical environment can be called mathematics.  All theoretical work in physics is mathematics.

In fact, one of the most fascinating things I ever learned about the world was that some mathematical systems which appeared to have no corollary in the real world were found later to have usefulness as physical models.

Much of modern physics - quantum mechanics and so forth - is, in its technical details, heavily mathematical.  A mathematician would excel at such theoretical work, and a physicist would have to be a good mathematician.

[quote="Interval"]Why do colleges have a physics department and a math department?[/quote]

The studies have different focuses (again, I'm not sure you understood what I meant).  Physics has the additional focus of [b]experiment.[/b]  Physicists have to be scientists, and the main goal of science is to both postdict and predict observable phenomena.

Pure mathematics is free of such constraints, and is free to work in any system that mathematics deem interesting and worthy of study.

In practice, much of the mathematics you will learn at a college level is a prerequisite for much of the physics you will be studying - just because the departments are distinct doesn't mean they have nothing to do with each other, yet again, they are distinct fields of study.

[quote="Interval"]If BOTH teach numbers, why not include BOTH into one department?[/quote]

You might as well throw the Business department in there too, since business is about numbers.

First off, as I have said repeatedly, [b]mathematics is not just about numbers.[/b]  There are entire fields of mathematics in which concrete numbers hardly ever appear, and mathematicians are instead studying things more bizarre - topology, for example, or set theory.

Second, [b]physics only employs numbers as a tool.[/b]  Quantifying physical phenomena make it easier to suggest a particular model which produces quantifiable and testable predictions.

Your reply is well-understood.

Good job!

I think theres another way that u can look at it.

physics is a system in which in reality, there are an infinite amount of variables(air resistance, pressure,etc.) therefore, in physics, each problem can only focus on a single principle or law. For example, if trying to calculate the amount of pressure on a hydraulic system through pascals principle(easy physics) then there are a million variables, etc. fluid, area, air pressure, fluid pressure, etc. In mathematics you can create a world that is completely arbitrary. in mathematics, you control everything, including the formulas, equation, etc. though there are limitations, like basic axioms, etc. but youn stil  can do anything you want. In physics you are limited my gravity, compressibility of fluids, gas laws, etc.

hope you like my view :surf:

mathematics is a tool in which to further the work of physics, (as well as every other science).

The main difference is that mathematics is fixed and independent of all sciences. That is, if a statement in mathematics is a theorem for us, it will also be a theorem for any alien, even if the alien lives in a universe where the laws of physics are completely different than those of our own universe.

Physics makes calculation to answer the question: What and why is happening?
Mathematics learns the question what the calculation itself is and how the calculations are happening. As a result it provides a good calculational systems for the various sites of the reality.
So, mathematics is a tool for physics. And that is why there were times when physics has caused the development of mathematics putting up new challenges for it.

Physics necessarily uses maths but maths does not requires physics, that's the main difference

thank you

[quote=QuantumMech]If math and physics are so related, why does AoPS have ChemWOOT instead of something akin to 'PhysWOOT'?[/quote]

I think levans said something about where they didn't have they correct people/it takes a long time to develop the initial curriculum.

It would be kinda cool if we wrote the curriculum, kinda like just posting big LaTeX documents on the Physics Forum.

physics=awesome
math=not as awesome(notice i didn't say math isnt awesome)

Someone may say the reverse, you cannot say straightforward that maths is not as awesome....

[quote=RPFeynman110]It would be awesome if AOPS had physics classes. Proposed curriculum:

Week 1: SI Units and Vectors
Week 2: Particle and collision mechanics
Week 3: Rotational motion
Week 4: Fluid mechanics
Week 5: Kinetic gas theory
Week 6: First and Second Law of thermodynamics
Week 7: Electrostatic and Gravitational fields (I grouped them together, since they are so similar.)
Week 8: Circuit analysis
Week 9: Magnetostatics
Week 10: Induction, AC circuits and Faraday`s Law
Week 11: Maxwell`s Equations, EM waves, and Optics
Week 12: Special Relativity
Week 13: Quantum Phenomena
Week 14: Special techniques like Lagrangian and Hamiltonians[/quote]

Hold on.... You mean to say you will cover all of physics in 14 weeks?! I was also thinking about this and there should be at least 3 courses ~18 weeks each. Mechanics, E&M, Misc topics (thermo, optics, modern, etc.).

Proposed Curriculum for mech:

week 1: Math review (calculus, vectors, dimensions, SI units, etc.)
week 2: 1D kinematics
week 3: 2D and 3D kinematics
week 4: Introduction to Newton's laws
week 5: Intermediate newton's laws, friction
week 6: Hard problems in Newton's Laws
week 6: Work and energy
week 7: Problem solving with the conservation of energy
week 8: Momentum
week 9: Collisions
week 10: Rotational kinematics, angular KE
week 11: Torque and Angular momentum
week 12: Hard problems in rotation
week 13: Introduction to gravitation
week 14: More gravitation, Kepler's laws
week 15: Fluid statics
week 16: Fluid Dynamics
week 17: Challenging Problems I
week 18: Challenging Problems II


Find the value of:
$ A\equal{}sin\frac{\pi}{2^n}$

$ B\equal{}cos\frac{\pi}{2^n}$
$ n\in N$

Kyoshiro, I can't quite understand your question - whose value we have to find?
Could you explain it to me a little better?

He wants closed-form expressions for $ \sin \frac{\pi}{2^n}$ and $ \cos \frac{\pi}{2^n}$

It's cleaner for cosine, because you have $ \cos 2x \equal{} 2 \cos ^2 x \minus{} 1 \implies \cos x \equal{} \frac{1}{2} \sqrt{2x\plus{}2}$, so using that recursively its just:

$ \cos \frac{\pi}{2^n} \equal{} \frac{1}{2} \sqrt{2 \plus{} \sqrt{2\plus{} \sqrt{2\plus{} \ldots \sqrt{2 \plus{} \sqrt{2}}}}}$, where the total number of surd signs is equal to $ n\minus{}1$ for $ n\geq 2$

Then you can just use pythagorean on that to get the formula for sine...


Let cevians $AD, BE$, and $CF$ of $\triangle ABC$ meet at an interior point $P$.  If $AB = 3\cdot FB$ and $AC = 4\cdot AE$, then compute the value of $\frac{PA\cdot PB\cdot PC}{PD\cdot PE\cdot PF}$.

I think there are two things that jump out: Ceva's Theorem and calculating ratios by calculating area ratios.  No time to solve it right now though.

[hide]By Ceva $\frac{BD}{DC}=\frac {1}{6}

Now use Menelaus to finish. (e.g, start by going down along a Cevian)[/hide]

[quote="blahblahblah"]If they use my solution, make them prove Ceva and Menelaus  :D[/quote]

most junior students in my school don't know these two theorems yet, so they must prove Ceva and menelaus :D

That's a lot of green smilies.

 :D  :D  :D

[quote="ASquare"]<ignorance>
What's a cevian?[/quote]

A segment connecting a vertex of a triangle to a point on the opposite side.

[quote="ThAzN1"][quote="siuhochung"]dear thazn1,
may I put this question in the training notes for lower form students in my school?
Thanks!  :D[/quote]

Heh, sure.  :D  My solution actually uses just Ceva, and a lot of area ratios with it (which ends up to be basically the same thing).  :)[/quote]

Thank you!! :D (green smiles again)


function
integration
calculus
calculus computations

Let $ 0\le k\le1.$ Find all continuous functions $ f: [0,1] \to \mathbb{R}_ +   \cup \{ 0\}$ that satisfy:

\begin{eqnarray*}\int_0^1 {f(u)\,du} & =& 1\\\int_0^1 {uf(u)\,du}  &=& k\\\int_0^1 {u^2 f(u)\,du}  &=& k^2 .\end{eqnarray*}

The equations imply $ \int_0^1 (u\minus{}k)^2f(u)\ du\equal{}0$. Since $ f$ continuous and nonnegative, so is $ (u\minus{}k)^2f(u)$. This requires $ (u\minus{}k)^2f(u)\equal{}0$ for all $ u\in [0,1]$, which means $ f(u)\equal{}0$ for all $ u\in [0,1]$ (except possibly at $ k$, but then continuity shows that $ f(k)\equal{}0$ as well). But that would make $ \int_0^1 f(u)\ du\equal{}0$. So there is no solution.


integration
calculus
logarithms
limit
real analysis
real analysis unsolved

$a(n)=\int_{0}^{1}\frac{1}{\sqrt{1+(nx^{n-1})^2}}dx$

asymptotic expansion with two terms when n tends +oo ?

Observe, first of all, that $\int_1^\infty \frac 1{\sqrt{1+(nx^{n-1})^2}}\,dx\le \int_1^{\infty}\frac 1{nx^{n-1}}\,dx=\frac 1{n(n-2)}=O(n^{-2})$. So, we can expand the interval of integration to $[0,+\infty)$ if we can neglect terms of order $\frac 1{n^2}$. Let $x_n=n^{-\frac 1{n-1}}=1-\frac{\ln n}{n}+O(n^{-2}\ln^2 n)$. Making change of variable $x=x_ny$, we can write the integral as $x_n\int_{0}^{\infty}\frac 1{\sqrt{1+y^{2n-2}}}\,dy$. Now substitute $y=e^{\frac t{n-1}}$. We get $\int_{0}^{\infty}\frac 1{\sqrt{1+y^{2n-2}}}\,dy=\frac 1{n-1} \int_{-\infty}^\infty \frac{e^{t/(n-1)}}{\sqrt{1+e^{2t}}}\,dt=1+\frac1{n-1}\int_{-\infty}^\infty e^{t/(n-1)}g(t)\,dt$ where $g(t)=\left\{\begin{aligned} \frac 1{\sqrt{1+e^{2t}}},&\quad t\ge 0\\ \frac 1{\sqrt{1+e^{2t}}}-1,&\quad t<0\end{aligned}\right.$. Since $g(t)=O(e^{-|t|})$, we see that the last integral can be written as a sum of inverse powers of $n-1$ with some coefficients (just decompose $e^{t/(n-1)}$ into its Taylor series). We are interested in the free term $c$ of that series, which is $c=\int_{-\infty}^{\infty}g(t)\,dt$. If we find it, we can conclude that the original integral is $\left(1-\frac{\ln n}{n}+O(n^{-2}\ln^2n\right)\left(1+\frac cn+O(n^{-2})\right)=1-\frac{\ln n}n+\frac cn +O(n^{-2}\ln^2 n)$. Now, making change of variable $y=e^t$, we see that the integral expressing $c$ can be written as $\lim_{\delta\to 0+}\int_\delta^\infty \frac 1{y\sqrt{1+y^2}}\,dy-\ln \frac 1\delta$. The first integral can be computed using the substitution $z=\sqrt{y^2+1}$ and we get $c=\ln 2$. Thus $\int_0^1 \frac 1{\sqrt{1+(nx^{n-1})^2}}\,dx=1-\frac{\ln n}n+\frac{\ln 2}n +O(n^{-2}\ln^2 n)$. (I hope I haven't made a stupid mistake somewhere... :) )


In a telephone directory for a certain city in South Dakota, there are 213 people named Smith and 327 people named Johnson. What is the ratio of the number of people named Smith to the number of people named Johnson? Express your answer as a common fraction.

[quote="AceOfDiamonds"]Factor out a 3 so you get
$ \frac{71}{109}$
and both these numbers are prime.[/quote]
They don't have to be both prime, they just have to be relatively prime meaning they have no common factors other than 1.


darkquantum wrote:if:
2a+b+c+d+e=5
a+2b+c+d+e=15
a+b+2c+d+e=30
a+b+c+2d+e=20
a+b+c+d+2e=50

Find abcde.

[hide]

6(a+b+c+d+e) = 120

(a+b+c+d+e) = 20

so...

subtrcting that from each of the equations we find:

a = -15

b = -5

c = 10

d = 0

from here  you know the product is going to be ... 0[/hide]

yea
actaully
u don't even need to solve for a,b,c,e

all u need to see is that a+b+c+d+e=a+b+c+2d+e.
This means d=0.
So ur done


search
combinatorics unsolved
combinatorics

Find a closed form of the following sum
\[ f(i, j) \equal{} \sum_{k \equal{} 0}^i \binom{2 i \minus{} 2k}{i \minus{} k} \binom{k \plus{} j}{k}.
\]

i dont think it has a good close form but we know that:
$ f(i,j)$ = cofficients of $ x^i$  in $ ( \frac {1}{1\minus{}x})^{1 \plus{} j}\frac {1}{\sqrt{1 \minus{} 4x}}$

do we know $ f(i,1)$ in close form?

Similar problems were posted by me here:
http://www.mathlinks.ro/viewtopic.php?search_id=1368030362&t=171763
Probably that approach works here either.

Even $ f(n,0) \equal{} \sum_{k \equal{} 0}^n \binom{2k}{k}$ doesn't have a nice closed form, at least as far as the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] knows (it's sequence [url=http://www.research.att.com/~njas/sequences/A006134]A006134[/url]).  $ f(n,1)$ isn't listed in the OEIS.


Let $A,B \in M_n \left( \mathbb{C} \right)$  matrixs, such that  $A^3  = A^2 $  and  $A + B = I_n $

Prove that  $\exists X \in M_n \left( \mathbb{C} \right)$ such that: $X \cdot \left( {AB + I_n } \right) = \left( {AB + I_n } \right) \cdot X = I_n $

We have to show that $I_n+A-A^2$ is invertible, and this follows from the fact that $(A-A^2)^2=O_n$, so $(I_n+A-A^2)(I_n-A+A^2)=I_n-(A-A^2)^2=I_n$.

Metru, this post should be in the [url=http://www.mathlinks.ro/Forum/index.php?f=8]Superior Algebra[/url] forums. I've moved it there, and please post all college-related problems there. Thanks!


:D  ya que mi novia no entra al msn colgare problemas, que tal?

[color=blue][b]Problema 1. [/b]Dado el triangulo $ABC$ recto en $C,$ sea $D$ un punto en $AB$ tal que los triangulo $ACD$ y $DCB$ tienen el mismo inradio, demostrar que $[ABC]=CD^2$ ($[ABC]$ significa area dle triangulo $ABC$)

[b]Problema 2.[/b] Dado un triangulo, demostrar que sus puntos de Fermat, el centro de su circunferencia de Euler y el circuncentro estan en una circunferencia. [/color]

(vean en:
$\square$ [url]http://http://centros5.pntic.mec.es/~marque12/matem/fermat.htm[/url],
$\square$ http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fermat+%5C%5Ctriangle&t=47955,
 $\square$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=triangles+constructed++the+outside+%28on+the+inside&t=26139]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=triangles+constructed++the+outside+%28on+the+inside&t=26139[/url])

[color=blue][b]Problema 3. [/b]En un trapecio se conoce que la suma de ambas diagonales es 4 y que el área es 2. Según lo anterior se pide calcular la altura.

[b]Problema 4.[/b] Sea $ABCDE$ un pentágono convexo de área $1$ (las diagonales quedan dentro del pentágono). Sean $P,$ $Q,$ $R$ y $S$ los baricentros de los triángulos $ABE,$ $BCE,$ $CDE$ y $DAE,$ respectivamente. Indicar la naturaleza del cuadrilátero $PQRS$ y dar su área.

[b]Problema 5.[/b] La circunferencia inscrita en $\triangle ABC$ tiene centro $O$ y es tangente a los lados $BC,$ $AC$ y $AB$ en los puntos $X,$ $Y$ y $Z,$ respectivamente. Las rectas $BO$ y $CO$ intersecan a la recta $YZ$ en los puntos $P$ y $Q,$ respectivamente. Si  $XP$ y $XQ$ tienen la misma longitud, Indicar la naturaleza del triangulo $ABC.$

[b]Problema 6.[/b] Los vértices de un triangulo $ABC$ pertenecen a la hipérbola $xy=1.$ Demostrar que el ortocentro también pertenece a la hipérbola.[/color]

(Sug. Usar tangentes)

[color=blue][b]Problema 7.[/b] En $ABC,$ sean $AD,BE,CF$ cevianas concurrentes en $M.$ Si los triángulos $BDM,$ $CME$ y $AMF$ tienen la misma área y el mismo perímetro, entonces $ABC$ es equilátero.[/color]

[color=red][b]Problema 8.[/b] (Dedicado a Tipe) Los números enteros del 1 al 2002, ambos inclusive, se escriben en una pizarra en orden creciente 1, 2, . . . , 2001, 2002. Luego, se borran los que ocupan el primer lugar, cuarto lugar, séptimo lugar, etc., es decir, los que ocupan los lugares de la forma 3k + 1. En la nueva lista se borran los números que están en los lugares de la forma 3k +1. Se repite este proceso hasta que se borran todos los números de la lista. ¿Cuál fue el último número que se borró?  :dry: $(detesto$ $la$ $aritmetica)$
Espero manden sus soluciones prontito  ;) [/color]

Cuidense

$Jose^/ Carlos$

[color=red][b]Problema 3. [/b]En un trapecio se conoce que la suma de ambas diagonales es $4$ y que el área es $2$. Según lo anterior se pide calcular la altura.[/color]

[color=blue][b]Lema.[/b] Dado un trapecio, sean $d_1$ y $d_2$ las longitudes de las diagonales y sea $x$ el angulo que forman, entonces el area del trapecio es $(d_1.d_2.senx)/2$.[/color]

Entonces tenemos que $(d_1.d_2.senx)/2=2$, de ahi que $d_1.d_2.senx=4$, es decir $d_1.d_2\geq 4$, por la desigualdad de las medias tenemos que $\frac{d_1+d_2} {2}\geq \sqrt{d_1.d_2}\geq 2$, asi se tiene que $2\geq \sqrt{d_1.d_2}\geq 2$, de ahi que $d_1=d_2=2$.

Ahora bien, como su area es $2$ entonces si $b_1, b_2,h$ son su base mayor, base menor y altura respectivamente, $2=\frac{h(b_1+b_2)} {2}$ y tambien, tenemos que $d_1^2=h^2+(b_1+b_2)^2$ y de ahi que $h^4+4=4h^2$ y se sigue que $h=\sqrt{2}$. $\square$

[quote="Pascual2005"]jeje, veo que tu novia volvio a irse... :D[/quote]

algo asi, jejeje... me gustaria que $CUELGUEN$ $SOLUCIONES!!!$


For each of the series below, decide which of the descriptions:

(i) Does not converge;
(ii) Converges, but not absolutely;
(iii) Converges absolutely

applies, and prove your answer.

1. [sum] from n=1 to oo sin(n*[pi] + (1/n))

2. [sum] from n=1 to oo (e - (1 + (1/n^{2})^{n^{2}})

3. [sum] from n=1 to oo (e - (1 + ((1/n))^{n} / (1 - (1/2n))))

$\sum_{n=1}^{\infty}\left(\sin(n\pi)+\frac{1}{n}\right)$ diverges. note that $\sin(n\pi)$ is 0 whenever n is an integer so this reduces to just

$\sum_{n=1}^{\infty}\frac{1}{n}$ which is the harmonic series and is known to diverge.

[quote="Spasty"]I think that series is $\sum_{n=1}^{\infty}\sin(n\pi+1/n)$.
Now, $\sin(n\pi+1/n)=\sin(n\pi)\cos(1/n)+\cos(n\pi)\sin(1/n)=(-1)^{n}\sin(1/n)$ for $n=1,2,...$, hence the answer is (ii).[/quote]
o woops guess i missed that set of parentheses.  :blush:
question, what do u mean by "hence the answer is (ii)"?

The series converges for Leibniz, but not absolutely.

2) Using Taylor's formula, we can show that
$\lim_{x\to+\infty}x(e-(1+\frac{1}{x})^{x})=\frac{e}{2}$.
Hence, put $x=n^{2}$, we get
$\lim_{n\to+\infty}n^{2}(e-(1+\frac{1}{n^{2}}){}^{n}{}^{2})=\lim_{n\to+\infty}\frac{(e-(1+\frac{1}{n^{2}}){}^{n}{}^{2})}{1/n^{2}}=\frac{e}{2}$.
This means that the series has the same character of $\frac{1}{n^{2}}$ which converges.


How many squares having all 4 vertices on a dot are possible in a 5 by 5 geoboard, like:

. . . . .
. . . . .
. . . . .
. . . . .
. . . . .

--question: Is there an extremely efficient formula or way to count squares rather than drawing them all out?--

This problem is much more fun for rectangles...

Anyway, yes, there is an extremely efficient way to do this.  Count all the possible squares that are $k \times k$ for $k = 1, 2, 3, 4$, and add all that up.

[hide="Generalization"] The number of $k \times k$ squares in an $n \times n$ lattice is the number of possible placements of, say, the top left corner of that square.  Take a moment to see why the possible positions of this corner are a $(n-k) \times (n-k)$ square subset of the lattice.  Hence our total number of squares is

$\sum_{k=1}^{n}(n-k)^{2}= \sum_{k=1}^{n-1}k^{2}= \frac{(n-1)(n)(2n-1)}{6}$

For $n = 5$, we have

$\frac{4(5)(9)}{6}= \boxed{ 30 }$ squares.

(Edited.  Then we need to add in the diagonal squares.) [/hide]

[quote="t0rajir0u"]This problem is much more fun for rectangles...[/quote]
Here are two ways to find all rectangles in an $m\times n$ grid:

[hide="First way"]$x=\binom{m}{2}\binom{n}{2}$[/hide]
[hide="Second way"]$x=\frac{mn(m+1)(n+1)}{4}$[/hide]

[quote="t0rajir0u"]The second one is written wrong, and both formulas (in the correct form) are equivalent, so why did you write them separately?[/quote]
How is the second one wrong?

They are separate because although they are equivalent, they look completely different.

The question does not exclude squares with diagonal sides.

[hide="Casework"]There are the $\frac{n(n+1)(2n+1)}{6}=30$ squares already mentioned.

$9$ with sidelength $\sqrt{2}$.

$8$ with sidelength $\sqrt{5}$.

$2$ with sidelength $\sqrt{10}$.

$1$ with sidelength $2\sqrt{2}$.

So I think it is $30+9+8+2+1=50$, I think  :maybe: .[/hide]

[quote="i_like_pie"]How is the second one wrong?[/quote]

It should be ${m \choose 2}{n \choose 2}= \frac{mn(m-1)(n-1)}{4}$.

[quote="pianoforte"]The question does not exclude squares with diagonal sides.[/quote]

Hmm, fair enough.  We'll have to take those into consideration too.  (It's less generalizable, though :( )  [hide="A little trickier..."] We'll have to count a few different kinds.  First, the $\sqrt{2}\times \sqrt{2}$ squares.  The set of possible locations of their topmost corners is a $3 \times 3$ square, so there are $9$ of these.

Then the $2 \sqrt{2}\times 2 \sqrt{2}$ squares.  We can only fit $1$ of these in.

Next, the $\sqrt{5}\times \sqrt{5}$ squares.  These can come in two orientations, and there are $4$ positions of the topmost corner for both, so there are $8$ of these.

Next, the $\sqrt{10}\times \sqrt{10}$ squares.  Two orientations, and we can only fit $1$ of each, so $2$ of these.

In total, we have added $9+1+8+2 = 20$ squares to our nice generalizable answer, making our final total... $\boxed{ 50 }$? [/hide]

[quote="t0rajir0u"][quote="i_like_pie"]How is the second one wrong?[/quote]
It should be ${m \choose 2}{n \choose 2}= \frac{mn(m-1)(n-1)}{2}$.[/quote]
Try $m=2,n=3$.

$\frac{mn(m-1)(n-1)}{2}$ yields $6$. However, the correct answer, obtained with $\frac{mn(m+1)(n+1)}{4}$, is $18$.

[quote="t0rajir0u"][quote="i_like_pie"]$\frac{mn(m-1)(n-1)}{2}$ yields $6$. However, the correct answer, obtained with $\frac{mn(m+1)(n+1)}{4}$, is $18$.[/quote]

In either case either your first expression is wrong or your second one is.

[quote="pianoforte"]I think t0rajir0u meant $\frac{mn(m-1)(n-1)}{4}$, while thinking of a grid of m horizontal lines, and n vertical lines, while you were thinking of a grid with m rows and n columns.[/quote]

Yes.  Usually, when I speak of an $m \times n$ lattice grid, I expect there to be $m \times n$ lattice points in it.[/quote]
I, too, was talking about a grid with $m\times n$ points. I still don't see how mine is wrong as it gives results that correspond with reality. Try several cases and compare the results that $\frac{mn(m+1)(n+1)}{4}$ and $\frac{mn(m-1)(n-1)}{2}$ give with the real answer.

Did I write that over two?  Sorry, it should be

${m \choose 2}{n \choose 2}= \frac{mn(m-1)(n-1)}{4}$

Consider the degenerate case $m = n = 1$.  This is a point.  A point is not a rectangle (are you counting degenerate rectangles?  I'm not).

This problem is from 2003 Mathcounts, the 2nd problem on the Team round, to be precise. The answer key says that it is [hide] 50[/hide]

Aha.  I was counting the degenerate case before, so subtract $5^{2}= 25$ from my answer and it's correct.  I see.

Edit:  In any case, i_like_pie, I don't see how you can't see that your expressions aren't equivalent.  If you accept that

${n \choose 2}= \frac{n(n-1)}{2}$

Then clearly

${n \choose 2}{m \choose 2}= \frac{mn(m-1)(n-1)}{4}$

Which is not your second expression.

can you explain why you subtract 5^2 and what is "degenerate?" It would help me understand. Thanks.

I was counting points as squares without realizing it.  So I counted all $25$ points already in the square in my original argument...

(A point is a "degenerate" square because it's like a square of side length $0$.)

oh. thanks. it makes much more sense now. i thought that you meant degenerate, as in not a square originally, and did not take it as points.


Do there exist a circle and an infinite set of points on it such that the distance between any two of the points is rational ?


What is the molar mass of barium nitrate, Ba(NO3)2?

1)261 g/mole

2)199 g/mole

3)213 g/mole

4)167 g/mole

5)none of these


[quote="rcv"]What is the tens digit of $12+3^4\times 67^5+89$?[/quote]
[hide="solution"]
All you need are the last two digits of each number.
12 + 81 + 89 + 67:^5:
182 + 67:^5:
82 + 07
89
[b]8[/b]
[/hide]

Treething:  Your method should work, but your order of operations are a little messed up.  Carefully compare your formula with the original problem.


AMC
AIME
USA(J)MO
USAMO
geometry
3D geometry
tetrahedron

I have something I was wondering if people related to/ could give advice about.

I took practice AMC, AIME, and USAMO tests this summer.

On the AMCs, I tended to score in the 120-135 range.

On the AIMEs, I scored in the 8-12 range.

On the USAMO, I would get about 1 completely right. Sometimes 2.

I am just worried because for me, the AMC seems so much tougher to improve in than the AIME or USAMO. My score does not budge and I feel too pressed for time and make so many silly mistakes. Does anyone else find the AIME an easier test taking experience than the AMCs? Is there anything I can do other than practice to help me in timing during the AMC? What do all of you perfect scorers out there do to get a perfect score?

this is irrelevant, but I think that scores take a LOT of work to go up, especially some kinds. Geometry scores on Vestavia practices go up really slowly. I've been practicing forever, as in 3 weeks lol. But once, it did go up a lot.

I am just a green geometer, but I know that practice works and it's the only way as well as studying.

Haha that probably didn't help just like that tetrahedron thing I posted formulas for a long tiem ago. I know....  :blush:

Don't worry about it; if you qualify for USAMO, who cares what your index was, and if you can qualify comfortably, why obsess over slightly better scores when you can be doing olympiad stuff?  Work harder problems and the easier ones take care of themselves.

(By the way, just so you know, don't be too sure of how well you're really doing on the USAMO—look at some of the threads [url=http://www.artofproblemsolving.com/Forum/index.php?f=133&ppage=50&sort=lastpost&order=DESC&start=185]here[/url], especially [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85142]this one[/url]. :wink: )

I had my teacher at school grade it. He is pretty good. Ph.D from Cornell.

I mean there are those questions where they ask "find all possible solutions" and I do that and I show that there are no other possible solutions, or the geometry problems where it says to prove something, etc.


I'm reading paper at http://www.emis.de/journals/EJC/Volume_10/PDF/v10i1r8.pdf . In figure 2, there is that the best known packing has side length 3.8772 and tilt angle 40.182 degrees. How can I verify those values, as another source (Which Way did the Bicycle Go) says that the side length is about 3.877083?

Anyone? This irritates me as I don't know which source I can trust.


How many ways can you color the squares of a $ 2 \times 2008$ grid in 3 colors such that no two squares of the same color share an edge?

[hide="A Beginning..."]Let the colors be red, blue, and green.

The top row can be colored in $ 3\cdot2^{2007}$ ways.

It remains to determine in how many ways the bottom row can be colored. :ninja:[/hide]

[hide="You want to think the other way, since 2008 is arbitrary  :wink: "]
There are $ 3 \cdot 2 = 6$ choices for the leftmost two squares, without loss of generality say $ \textcolor{red}{R}\textcolor{green}{G}$. For the next column, there are $ 3$ possibilities, $ \textcolor{green}{G}\textcolor{red}{R}, \textcolor{blue}{B}\textcolor{red}{R}, \textcolor{green}{G}\textcolor{blue}{B}$, and so forth for the rest of the columns.
\begin{tabular}{|r\parallel{}r|r|}
\hline
\cdots &\textcolor{red}{R}&\textcolor{green}{G}\\
\hline
\cdots &\textcolor{green}{G}&\textcolor{red}{R}\\
\hline
\end{tabular}, \quad
\begin{tabular}{|r\parallel{}r|r|}
\hline
\cdots &\textcolor{red}{R}&\textcolor{blue}{B}\\
\hline
\cdots &\textcolor{green}{G}&\textcolor{red}{R}\\
\hline
\end{tabular}, \quad
\begin{tabular}{|r\parallel{}r|r|}
\hline
\cdots &\textcolor{red}{R}&\textcolor{green}{G}\\
\hline
\cdots &\textcolor{green}{G}&\textcolor{blue}{B}\\
\hline
\end{tabular}
Thus, the answer is $ 6 \cdot 3^{2007} = \boxed{2 \cdot 3^{2008}}$.
[/hide]

Nice solution; I was obviously thinking differently. :D

[hide="LaTeX Tip"]You can use [b]\color{<color>}<text>[/b] instead of [b]\textcolor{<color>}{<text>}[/b].

To be able to continue in black afterwards without explicitly saying [b]\color{black}[/b], use [b]{\color{<color>}<text>}[/b].

Also note that [b]\textcolor{<color>}{<text>}[/b] affects all parts of [b]<text>,[/b] not just actual text. Example: $ \textcolor{blue}{\begin{tabular}{|r|r|}\hline A & B \\
\hline C & D \\
\hline \end{tabular}}$[/hide]


Doesn't this have to do with that Guest viewing profile thing, about how the thing isn't updated in real-time so it delays it a bit?? (idk).  lol...I'm looking at the guest at the bottom who is apparently in the Missouri forum... :rotfl:  :rotfl:  :rotfl: