Hi,

I'm a Junior in High School and I just scored a 6 on the AIME (I made some retarded mistakes).  This is my last year to report my scores to colleges and makeing USAMO would be so incredibly helpful.  Does anyone know If I can take the test on the alternate date, or if that is only for people who didn't take the first one?

You can only take one or the other, not both.

Thanks,

I guess I'm just screwed then.  There's always next year, but then it wont count for college.

Qualifying for USAMO senior year, while it won't help with college admissions, is still something to be proud of, and something you might want to attain if you enjoy math  :D

I was sick on AIME day, sick enough that i couldnt go. I was hoping to have a shot at usamo for college admissions. Now im not sure if my teacher will let me take the AIME II, guess i have to look around more. Next year i will still try to get usamo, but not for college admissions, for my own sake, and because i like math.

It won't help for admissions, but if you get waitlisted, it definitely improves your chances.  Just make sure to call in or write them a letter stating your outstanding achievement, and who knows, you might just get in.

Thats all true,

I'll prepare and make it last year.  I only wish it had dawned on me to prepare for these thigns earlier.  You hear that kids, start NOW.

Well actually, amazingly enough, sometimes you can randomly take the AIME II too.

Reporting next year's AIME score may be too late for college admissions, but a 6 is noteworthy. Contrary to what the poll on this forum may have lead you to believe (crazy nerds -.- *mutters* ) that is quite an achievement. I'd put it down when you're applying to college next year.


hi, im new to problem solving and just took the AMC 10 A. Just wondering how u do the following problems:

10. Coin A is flipped three times and coin B is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?

13. At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

17. Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?

Also I just started reading the AoPs books, im wondering what chapters or topics should I read in order to be able to do these problems?

[quote="02"]hi, im new to problem solving and just took the AMC 10 A. Just wondering how u do the following problems:

10. Coin A is flipped three times and coin B is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?
[/quote]

You break it up into cases.

Case 1:
3 heads from coin A and from B

for coin A it would just be 1/2^3 x 3C3=1/8
coin B=1/2^4 x 4C3=1/4

multiply the probabilities (independent events) to get 1/32.

Case 2:
2 heads from coin A and from B

p(A)=probability of A in this case (same for p(B))

p(A)=1/2^3 x 3C2=3/8
p(B)=1/2^4 x 4C2=3/8

p(A) x p(B)=9/64

Case 3:
1 head from coin A and coin B

p(A)=1/8 x 3C1=3/8
p(B)=1/16 x 4C1=1/4

p(A) x p(B)=3/32

Case 4:
0 heads from coin A and coin B

p(a)=1/8
p(b)=1/16

p(A) x p(B)=1/128

now you add up all the probabilites

1/32+9/64+3/32+1/128=35/128

[quote="02"]
13. At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?
[/quote]

This was basically an inverse propotion.

m=men
w=women

mw=12x3=36

You're given that each woman danced with 2 men, so 2w=36

Thus, there were 18 women.

[quote="02"]

17. Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in [/quote]

I used logic on this problem.  When Brenda and Sally meet, they are obviously at the same point.  So the next time they meet is when Brenda's Distance (B) and Sally's Distance (S) covers the whole track.

So if in the whole track, Sally ran 150 meters, then in half the track she would run 75 meters.

Since Brenda ran 100 meters in half of the track, the distance of half of the track is 100+75=175.  Thus, the length of the whole track is 2(175)=350.

Quote:13. At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

To further explain why Neal multiplied 12 and 3 together: (highlight the text below to view it)

[hide]

Each man danced with three women, and there were 12 men, so there were a total of 36 man-woman pairs.  Since each woman participated in exactly two of those pairs, there were 36/2 = 18 women at the party.

Alternatively, since each man had 3/2 as many partners as each woman did, only 2/3 as many men as women were necessary.  Thus, if w is the number of women, we have 2/3*w = 12, so w = 18.[/hide]

I missed #17 on 10A...:-(

17. Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?

I used algebra.

Brenda's rate --- b
Sally's rate --- s
length of the track --- l
l/2 = bt + st = 100 + st
l = 200 + 2st
l = 2bt + 150
2l = 2(bt+st) + 350
2l = l + 350
l = 350

I missed 17 on the test also - along with 8 others...

What if some men danced with the same woman? Would the answer change?


logarithms
function
complex numbers
absolute value

What is a^z defined as, where a is a real number and z is a complex number. And better yet, what is z_1^z_2, where they're both complex numbers?

Well, those definitions are slightly problematic, because you end up with multiple-valued things.  If we take a positive real number to an imaginary power, we need the following definition of complex numbers:  z = r*e^(i:theta:), where r is the absolute value of z and :theta: is the angle it makes with the positive x axis.  (This is another way of writing the polar form of complex numbers.)  If we just look at numbers z in the upper-half-plane, we can talk about a^(x+iy), where y is positive, we see that it can be written as a^x*e^(i*ln(y)) (where ln is the logarithm base e for real numbers).  So that's nicely defined.  I'm going to let Simon say anything else about this, because he knows it better than I do and I'd just be guessing.

We start by defining e^z. I feel somewhat bad saying this since it turns some of Euler's most amazing work into mere definition, but I'll do it anyway. If z=x+iy, then e^z=e^x*e^(iy), and e^(iy)=cos(y)+i*sin(y).

Now we can turn a^z into a number of the form e^z'. Write a=e^ln(a). (I should note that ln is a real-valued real function only, taking positive reals to reals. ln(-1) and ln(i) are not defined.) a^z is therefore equal to e^(z*ln(a)), or e^(x*ln(a))*e^(iy*ln(a)), or a^x*(cos(y*ln(a))+i*sin(y*ln(a))).

Now a bit more about logarithms: we should write log(z) for the logarithm of a complex number, where it is understood that the base is e. Unfortunately, the complex logarithm function is not uniquely defined. In fact, it has infinitely many values. (To see this, recall that e^(2*k*i*pi)=1 for any integer k.) Thus log(z) is referring the set of values whose antilog is z. We can make matters a bit better defining Log(z) (note the capitalization) to be the value of log(z) with -pi<arg(z)<pi. Note also that this makes logarithms of negative reals undefined. That's about the best we can do.


Hi, I was wondering how one could set of equations to their own line and have each numbered.

For example, if I had $a^2+b^2=c^2$, how could I center that in the page and have a (1) on the right side of it.  I know the double dollar sign can center it, but what about numbering it?

\[ a^2+b^2=c^2 \]

use

\begin{equation}
\label{pythagore}
a^2+b^2=c^2
\end{equation}

which gives (in $\text{\LaTeX}e$ it will be centered):

$\begin{equation} \label{pythagore} a^2+b^2=c^2 \end{equation}$

The \label command is optional but the point of numbering equation is to refer to it.  So according to$~\ref{pythagore}$, we know that...bla bla

~\ref{pythagore}

it will work to in $\text{\LaTeX}e$

The ~ before to command is use instead the space.  This tells latex that it cannot cut there to avoid stuff like that:

bla bla............................................................................and with equation
(3)

for more equation environnement, use this document:

[url=http://www.ctan.org/tex-archive/macros/latex/required/amslatex/math/amsldoc.pdf]www.ctan.org/tex-archive/macros/latex/required/amslatex/math/amsldoc.pdf[/url]

As an added bonus, you can easily tell $\LaTeXe$ how you want the equation labelled (if you for some reason don't like the automatic numbering). For example, if you wanted equation $\clubsuit$ instead of equation (1), use

\begin{equation}
\tag{$\$$\clubsuit$\$$}
\label{whatever}
a^2+b^2=c^2
\end{equation}.

Just try not to overuse this feature: too many consipicuous symbols floating around the page can get in the way. Also, DONT use this feature to number the equations; let $\LaTeXe$ do that for you. Almost all of the time it is best to let $\LaTeXe$ label the equations. This is just an amusing feature.


I have 2 solutions for a problem. Can I send both?

Please don't, for two reasons.  First, if one solution has an error and the other doesn't, it's a problem.  Second, grading is a very time-consuming process, and we frankly have enough to read without trying to read an "extra" solution.

This of course is not to say that you shouldn't try to solve a problem in more than one way!  Not only is it a great way to check your answer, but it's also a really good way to improve your problem-solving skills.

But for the final submission, pick the solution that you like best and submit that one.


Show that $\Sigma_{n=1}^{\infty}\frac{1}{a_{n}}$ converges, where $a_{1}, a_{2}, ...$ is the Fibonacci sequence $1, 1, 2, 3, 5, 8,...$

Since $\lim_{n \to \infty}\frac{1/a_{n+1}}{1/a_{n}}= \lim_{n \to \infty}\frac{F_{n}}{F_{n+1}}= \frac{1}{\alpha}< 1$, the series converges by d'Alembert's test.
($\alpha$ is golden number)


Problema 2 si 5 m-ar interesa....are cineva vreo idee?
Mercic mult!

[u]Solutia problemei 5 (la nivelul clasei a XI - a !).[/u]

$a(x)\equiv \frac{x^x-\sin ^{x}x}{x^3 \ln x}=e^{x\ln (\sin x)}\cdot \frac{e^{x\ln \frac{x}{\sin x}}-1}{x\ln \frac{x}{\sin x}}\cdot$$\frac{\ln \frac{x}{\sin x}}{\frac{x}{\sin x}-1}\cdot \frac{x-\sin x}{x^3}\cdot \frac{x}{\sin x}\cdot \frac{1}{\ln x}$

$\Longrightarrow L_a\equiv\lim_{x\searrow 0} a(x)=1\cdot 1\cdot 1\cdot \frac 16\cdot 1\cdot 0=0.$

$b(x)\equiv \frac{\sin ^x x-\sin ^{\sin x} x}{x^3 \ln x}=e^{\sin x \ln (\sin x)}\cdot \frac{e^{(x-\sin x)\ln (\sin x)}-1}{(x-\sin x)\ln (\sin x)}\cdot$ $\frac{x-\sin x}{x^3}\cdot \frac{\ln (\sin x)}{\ln x}$

$\Longrightarrow L_b\equiv \lim_{x\searrow 0}b(x)=1\cdot 1\cdot \frac 16\cdot 1=\frac 16.$

$c(x)\equiv \frac{x^x-\sin ^{\sin x} x}{x^3\ln x}=a(x)+b(x)$

$\Longrightarrow L_c\equiv\lim_{x\searrow 0} c(x)$ $=\lim_{x\searrow 0} \frac{x^x-\sin^{\sin x} x}{x^3\ln x}=L_a+L_b=0+\frac 16=\frac 16.$

$f(x)\equiv\frac{x^x-\sin ^{\sin x}x}{x^p}=\frac{x^3\ln x}{x^p}\cdot c(x)\Longrightarrow L\equiv\lim_{x\searrow 0}f(x)=\frac 16\cdot \lim_{x\searrow 0}\frac{x^3\ln x}{x^p}\Longrightarrow$

$L=\left\{ \begin {array}{cc} 0&\ if\ \ \ \ p\le 2\\ -\infty&\ if\ \ \ \ p\ge 3 \end {array}\right. .$


Why don't you go look in the Michigan forum?

[quote="mathgeniuse^ln(x)"][quote="yunbo"]Just saying "hi" is enough. I want to make sure I'm not a loner.[/quote]

And whom might you be?[/quote]
Yun Bo?


Of course the reason why these products are formes  :wink:

In $ \ce{XeF_4}$ Xe is in +4 state so the hyrolysis leads to a disproportionation reaction.

But in $ \ce{XeF6}$ Xe is already in the +6 state and hence can't disproportionate and hence undergoes normal hydrolysis to give $ \ce{XeO3}$. In some cases partial hydrolysis can take place to give $ \ce{XeOF_4}$ and $ \ce{XeO_2 F_2}$.

Side questions:

1) What are the geometries of the Xenon species $ \ce{XeF2}, \ce{XeF4}$, $ \ce{XeF5 \minus{} }$, $ \ce{XeF6}$, and $ \ce{XeO6^{4\minus{}}}$?

2) Can $ \ce{XeF6}$ be stored in a glass container?

1. $ {XeF_2}$ ----Linear
    $ {XeF_4}$-----Square planar
    $ {XeF_6}$-----Distorted Octahedron
    $ {XeO_6^4 \minus{} }$--square bipyramidal

    i think $ {XeF_5^ \minus{} }$ has square pyramidal.. structure...

2.I don think so cause $ {XeF_6}$ is a powerful flourinating agent and so the flourine will etch the glass...

1)	XeF2-Linear, XeF4-square planar, XeF5-probably planar( pentagonal with two lone pairs occupying the axial positions of the pentagonal bipyramid) ,XeF6-distorted octahedron ,XeO6(4-)octahedral
2)	Nope, it is stored in nickel or monel containers, as XeF6 reacts with SiO2 to give SiF4 and XeOF4

sorry for double posting
[quote="Euclidean Geometer"]Could some one explain the following two reactions:
 6XeF4  + 12 H2O -----------> 4 Xe -+ 2 XeO3 + 24 HF + 3 O2.

XeF6 + 3 H2O  ------------------> XeO3 + 6 HF[/quote]
Probably all compounds of Xe in the lower oxidation are unstable. And this makes them better oxidizing agents, than Xef6, and in almost all of their reactions, XeF4 and XeF2 are either reduced or disproportionate, whereas almost all reactions of xe in VI state( reactions of XeF6) are not redox reactions.
P.S: only guessing
 :D

chemrock's answer was more appropriate Varun....

oh, first of all i hadnt seen his post. but his post does not explain why XeF6 doesnt get reduced, or why XeF2 doesnt disproportionate and only gets reduced. :D

Could smone temme partial equations if possible for these reactions?


[color=darkred]Let $ ABC$ be a triangle for which $ AB\perp AC$ and $ B < 60^{\circ}\ .$ Prove that exists $ D\in (AB)$ so that $ 2\cdot AD = BC\ .$

Consider the point $ E\in (AB)$ for which $ DB = DE\ .$ Prove that $ EB = EC\Longleftrightarrow B = 36^{\circ}\ .$

[b][u]Another equivalent enunciation.[/u][/b] Let $ ABC$ be a triangle for which $ A=90^{\circ}$ and $ B<60^{\circ}$.

For a point $ D\in (AB)$ consider the point $ E\in (AD)$ for which $ DB=DE$. Prove that if two from

the following three sentencies are truly, than and the another sentence is truly : $ \{\begin{array}{cc}
1\blacktriangleright & 2\cdot AD=BC\\\\
2\blacktriangleright & EB=EC\\\\
3\blacktriangleright & B=36^{\circ}\end{array}$[/color]

because B < 60 so $ cosB\equal{}\frac{AB}{AC} > cos60 \equal{}1/2$
then 2AB > BC so there exist point D satisfying 2AD=BC

I is the midpoint of EC we have ID = BC/2 = AD so angle AID = angle IAD , furthermore angle IAD = angle IEA = 2B , then angle AID = 2 B , however we already have angle EID = angle ECB = B , so angle AIE=B
now look at triangle IAE , angle IAE + angle AIE +angle IEA =180 , in other words , 5 B =180 ; B=36

Let's note $m(\hat B)=x$; take $M$ - the midpoint of $BC$. Then $AM=AD=MB$. This implies $MD\parallel CE$ and $MD=BD$. Consequently $m(\widehat{AMC})=2x, m(\widehat{BMD})=x$ and $\widehat{AMD}=\widehat{ADM}$. But $\widehat{AMD}=180^\circ-3x$ and $\widehat{ADM}=2x$, so $180^\circ-3x=2x\iff x=36^\circ$.

Best regards,
sunken rock


Putnam
linear algebra
matrix
integration

Problem $ B_5$) Let $ A$ be the $ n$x$ n$ matrix with $ a_{ij} = cos((i + j)\phi)$ ,$ \phi = \frac {2\pi}{n}$, $ n \in \mathbb{N},n > 2$
Find the determinant of $ I + A$
 - I learn method from this solution .It is very nice , It can find eigenvalues and eigenvectors but I try general with $ \forall \phi \in \mathbb{R}$ do not work ! :(  But can for solution with $ B$ be the $ n$x$ n$ matrix with $ b_{ij} = sin((i + j)\phi)$ then can proof that : $ \det(I + A) = \det(I + B) = 1 - \frac {n^2}{4}$
 General It to see more nice problem !!!
Lemma 1) Let $ x_1,x_2,..,x_n,y_1,y_2,...,y_n$ are real numbers . Put :
  $ A_1 = \begin{bmatrix}sin(x_1 + y_1) & sin(x_1 + y_2) & .... & sin(x_1 + y_n) \\
sin(x_2 + y_1) & sin(x_2 + y_2) & .... & sin(x_2 + y_n) \\
... & ... & ... & ... & ... \\
sin(x_n + y_1) & sin(x_n + y_2) & .... & sin(x_n + y_n)\end{bmatrix}$
$ A_2 = \begin{bmatrix}cos(x_1 + y_1) & cos(x_1 + y_2) & .... & cos(x_1 + y_n) \\
cos(x_2 + y_1) & cos(x_2 + y_2) & .... & cos(x_2 + y_n) \\
... & ... & ... & ... & ... \\
cos(x_n + y_1) & cos(x_n + y_2) & .... & cos(x_n + y_n)\end{bmatrix}$
Then $ rank(A_1) = rank(A_2) \leq 2$
 Proof lemma 1)For $ A_1$ are similary for $ A_2$
Put  $ C_{j} = \begin{bmatrix}sin(x_1 + y_j) \\
sin(x_2 + y_j) \\
... \\
sin(x_n + y_j)\end{bmatrix} = cos(y_j)\begin{bmatrix}sin(x_1) \\
sin(x_2) \\
... \\
sin(x_n)\end{bmatrix} + sin(y_j)\begin{bmatrix}cos(x_1) \\
cos(x_2) \\
... \\
cos(x_n)\end{bmatrix} = cos(y_j)\alpha + sin(y_j)\beta \in Vect(\alpha,\beta)$
 easy see $ Vect(C_1,C_2,..,C_n) \subset Vect(\alpha,\beta)$
 Then $ rank(A_1) = \dim(Vect(C_1,C_2,..,C_n)) \leq \dim(Vect(\alpha,\beta)) \leq 2$
Similary we have $ rank(A_2) \leq 2$
 Lemma 2) (It is very interesting with calculate $ \int$ ,....)
  $ \forall x \neq k\pi ,k \in \mathbb{Z}$  then
 $ (\frac {sin(nx)}{sinx})^2 = n^2 - 4\sum_{1 \leq i < j \leq n}{sin^{2}(j - i)x}$ !
 Proof lemma 2)
 Put $ M = \sum_{k = 1}^{n}{sin(2kx)}$,$ N = \sum_{k = 1}^{n}{cos(2kx)}$
We have :
  $ 2sin(x)M = \sum_{k = 1}^{n}{2sin(2kx)sin(x)} = \sum_{k = 1}^{n}{(cos(2k - 1)x - cos(2k + 1)x)} = cos(x) - cos(2n + 1)x = 2sin(nx)sin(n + 1)x$
  Hence $ M = \frac {2sin(nx)sin(n + 1)x}{sinx}$
the same :
  $ 2sin(x)N = \sum_{k = 1}^{n}{2cos(2kx)sin(x)} = \sum_{k = 1}^{n}{(sin(2k + 1)x - sin(2k - 1)x)} = sin(2n + 1)x - sin(x) = 2sin(nx)cos(n + 1)x$
 Hence : $ N = \frac {2sin(nx)cos(n + 1)x}{sinx}$
Then :
$ (\frac {sin(nx)}{sinx})^2 = (\frac {2sin(nx)sin(n + 1)x}{sinx})^2 + (\frac {2sin(nx)cos(n + 1)x}{sinx})^2 = M^2 + N^2 = (\sum_{k = 1}^{n}{sin(2kx)})^2 + (\sum_{k = 1}^{n}{cos(2kx)})^2 = n + \sum_{1 \leq i < j \leq n}{(2sin(ix)sin(jx) + 2cos(ix)cos(jx)) = n + 2\sum_{1 \leq i < j \leq n}{cos2(j - i)x} = n^2 - 4\sum_{1 \leq i < j \leq n}{sin^{2}(j - i)x}}$
 Proof complete lemma2 )
Now become problem for lemma 1 ) $ x_i = y_i = i \phi$ , $ \phi = \frac {2p\pi}{n},\frac {2p}{n} \notin \mathbb{Z},p \in \mathbb{Z}$ (putnam 1999 $ p = 1$)
then by $ rank(A_1) = rank(A_2) \leq 2$ Hence example $ A_1$
 easy $ M = N = 0$ and $ \sum_{1 \leq i < j \leq n}{sin^{2}(j - i)\phi} = \frac {n^2}{4}$
We have :
 $ \det(I + A_1) = 1 + Tr(A_1) + \sum_{1 \leq i < j \leq n}{det(\begin{bmatrix}a_{ii} & a_{ij} \\
a_{ji} & a_{jj}\end{bmatrix})}$
 We have $ Tr(A_1) = M = 0$ and
$ \sum_{1 \leq i < j \leq n}{det(\begin{bmatrix}a_{ii} & a_{ij} \\
a_{ji} & a_{jj}\end{bmatrix})} = \sum_{1 \leq i < j \leq n}{(sin(2i \phi})sin(2j\phi) - sin^{2}((i + j)\phi)) = \frac {1}{2}\sum_{1 \leq i < j \leq n}{cos(2(j - i)\phi) - cos(2(i + j)\phi) - 1 + cos(2(i + j)\phi)} = - \sum_{1 \leq i < j \leq n}{sin^{2}(j - i)\phi} = - \frac {n^2}{4}$
 easy with $ A_2$ (similary a result)
 and $ \det(I + A_1) = \det(I + A_2) = 1 - \frac {n^2}{4}$
Now for $ \phi = \frac {\pi}{n + 1}$, ..,.... Very interesting !
 More : To more  interesting You can calculate determinant $ I + K$ Here  $ K = [k_{ij}]_{1 \leq i,j \leq n}$
 Here $ k_{ij} = C_{n}^{i}cos((i + j)\phi)$ , to similary $ k_{ij} = C_{n}^{i}sin((i + j)\phi)$ , $ \phi = \frac {2\pi}{n},n > 2 , n \in \mathbb{N}$
 You can proof that :
1) $ rank(K) \leq 2$
2) $ 4\sum_{1\leq i < j\leq n} {C_{n}^{i}C_{n}^{j}sin^{2}((i - j)x)} = (\sum_{k = 1}^{n}{C_{n}^{k}})^2 - 4^ncos^{2n}{(x)} + 2^{n + 1}cos^{n}{(x)}cos(nx) - 1$
 You can check It  It is don't dificult  :P  ......

What was the point of this rambling? Please ask clear questions.

[quote="jmerry"]What was the point of this rambling? [/quote]
 :)  I will thinks !
But I find a some different point can be constant .....
 Example : Here $ (\frac {sin(nx)}{sinx})^{2} \equal{} n^{2} \minus{} 4\sum_{1\leq i < j\leq n}{sin^{2}(j \minus{} i)x}$
 We have $ cos(ix) \equal{} ch(x)$ and $ sh(x) \equal{} \minus{} isin(ix)$ I find :
$ M_1 \equal{} \sum_{k \equal{} 1}^{n}{sh(2kx)} \equal{} \frac {2sh(nx)sh(n \plus{} 1)x}{shx}$,$ N_1 \equal{} \sum_{k \equal{} 1}^{n}{ch(2kx)} \equal{} \frac {2sh(nx)ch(n \plus{} 1)x}{shx}$
  $ (\frac {sh(nx)}{shx})^{2} \equal{} n^{2} \plus{} 4\sum_{1\leq i < j\leq n}{sh^{2}(j \minus{} i)x}$ ,$ x \neq 0$
Point $ x$ to $ \frac {sh(nx)}{shx} \equal{} ?$ .....


A sequence of letters is formed by writing 1 A, 2 B's, 3 C's, and so forth, increasing the number of letters written by one each time the next letter of the alphabet is written. What is the 200th letter in the sequence?

200 is between the 19th and 20th triangular numbers (190 and 200)

so our answer is the 20th letter of the alphabet, T.


the day before my english essay was due, i had already finished it and brought it to school. during lunch, i let my friend read it, and i made him promise me that he wouldnt plagiarize it. but a few days after the essay was due, i asked him to see his essay and the whole intro was almost exactly like mine, and so was the rest of the essay. is this considered plagiarism on my part?

If I understand your story correctly, it would be plagiarism on his part. Also, why are you friends with people that would do that?

Plagiarism (by you)?  Probably not, although your school (or even your English department) might have a formal definition you can read.  Ethically questionable?  Absolutely, especially if this was even remotely predictable (which it sounds like it was).  Grounds for finding a new friend?  That's up to you, but I might consider it.

Well i don't think it's plagiarism because you didn't "copyright" it? or something along those lines...for example if two people accross the world wrote a similar essay they cannot call it plagarism because they

a) did not see the other's essay (not online or in a book, etc.) so they could not copy it because it was not available to them
b) they do not even know each other

however i believe your friend was simply out of ideas...

No, not plagiarism because it wasn't really published work (even if "published work" now means Wikipedia).  However, it was certainly cheating.  If he really needed help, you should have helped him, not let him read your paper.  Unless he edited it for you, there was little to no benefit to you, and in return you gave him your paper.  It was cheating because you said that he blatantly copied it.  I would talk to him about it, because you could both get in serious trouble if your teacher notices.

Let's be clear here: plagiarism is copying without attribution.  It doesn't matter whether the work is published, copywrited, or spray-painted on a subway car: if you copy work without attribution, that's plagiarism.  Your friend is unquestionably guilty of plagiarism.  In fact, some kinds of plagiarism are worse than others, and your friend is unquestionably guilty of a "bad" kind.

If your friend needed help, he should have talked to the teacher, done his own research and written his own paper.  Regardless of whether you benefited from it, whether your teachers caught it, whether your friend was "out of ideas," he acted in a dishonest and ethically unsound way, and you helped him.

funcia, if you can explain the meaning of that comment, I won't delete it.

I think funcia means there is a different rule in New Zealand about plagiarism maybe?

I know several classmates who got a zero for plagiarism on a historical fiction critical analysis book report.

They copied, and even though it was not published, just a rough draft, still a big fat plump 0.

JBL is correct about the defintition of plagiarism. Copying someone else's work is plagiarism if you don't give the first author credit. When doing a school paper you should always do some of your own thinking and own writing.

Copyright belongs in a work as soon as it is created, under modern copyright law. You don't need to register the copyright nor do you need to put a copyright notice on the work (although both of those steps give you additional legal protections in court). If someone else wrote something, and you copy it, you may also be infrininging a copyright in addition to committing plagiarism.

[quote="ProtestanT"][quote="now a ranger"]Hopefully, your english teacher won't read it and will just give you credit for doing them.

I wouldn't let ANY of my friends copy my paper almost exactly if I had one.  :blush:[/quote]
wait do u mean if u had one friend or if u had one piece of paper?[/quote]

If I had a paper to write, i meant.

sorry for being ambiguous

[quote="tokenadult"]Copyright belongs in a work as soon as it is created, under modern copyright law. You don't need to register the copyright nor do you need to put a copyright notice on the work (although both of those steps give you additional legal protections in court). If someone else wrote something, and you copy it, you may also be infringing a copyright in addition to committing plagiarism.[/quote]

Even in the US? That's something I truly didn't know if so.

Oh and yes that is what I meant with NZ.

[quote="7h3.D3m0n.117"][quote="Treething"]Hmmm.

Why does it matter so much?[/quote]

Because you can get it REALLY big trouble.[/quote]
Missed the sarcasm tag? (I hope)

[quote="JBL"]Because it's intellectually lazy and dishonest.

[url=http://timpanogos.wordpress.com/2007/02/19/a-little-plagiarism-a-little-book/]Here[/url] is a blog post I just recently ran across about plagiarism and what's wrong with it.[/quote]

...reading over this article, I'd say some are simply too lazy to care. If the sin of plagiarism lies in morals and ethics, where does the stereotypical lazy, amoral, Asian honors student lie?


geometry
3D geometry
sphere
Locus
Locus problems
IMO
IMO 1963

Point $A$ and segment $BC$ are given. Determine the locus of points in space which are vertices of right angles with one side passing through $A$, and the other side intersecting segment $BC$.

Let F be the foot of a normal from the point A to the line BC and P an arbitrary point on the segment BC. If the point A does not lie on the line BC, the locus of the vertices V of the right angles $\angle AVP$ is a sphere with a diameter AP and passing through the point F. This sphere is centered on the midline M, N of the triangle $\triangle ABC$ parallel to the side BC, where M, N are the midpoints of the sides AB, AC. If the point A lies on the line BC, then the points A, F and the locus of the vertices V of the right angles $\angle AVP$ is a sphere centered on segment M, N, where M, N are the midpoints of the segments AB, AC. As the point P moves on the segment BC, the locus of the vertices V becomes a portion of the pencil of spheres centered on the segment MN and passing through the points A, F. If the point A does not lie on the line BC, the pencil of spheres is elliptic (the spheres intersect in a fixed circle with the diameter AF and perpendicular to the line BC). If the point A lies on the line BC, the pencil of spheres is parabolic (the spheres touch the plane perpendicular to the line BC at the point A).


We will use Lagrange multipliers to prove this ineq $ 0<a,b,c,d<1$ and $ f(a,b,c,d)\equal{}a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}8(1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)$ and $ g(a,b,c,d)\equal{}a^2\plus{}b^2\plus{}c^2\plus{}d^2$ $ grad(g)$ not 0 and $ grad(g)$ and $ grad(f)$ is continious then we must  decide system
$ 3a^2\minus{}8(1\minus{}b)(1\minus{}c)(1\minus{}d)\minus{}x2a\equal{}0$
$ 3b^2\minus{}8(1\minus{}a)(1\minus{}c)(1\minus{}d)\minus{}x2b\equal{}0$
$ 3c^2\minus{}8(1\minus{}b)(1\minus{}a)(1\minus{}d)\minus{}x2c\equal{}0$
$ 3d^2\minus{}8(1\minus{}b)(1\minus{}c)(1\minus{}a)\minus{}x2d\equal{}0$
$ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$
from first and second $ 3a\minus{}\frac{8(1\minus{}b)(1\minus{}c)(1\minus{}d)}{a}\equal{}3b\minus{}\frac{8(1\minus{}a)(1\minus{}c)(1\minus{}d)}{b}$ or $ 3(a\minus{}b)\equal{}\frac{8(a\minus{}b)(a\plus{}b\minus{}1)(1\minus{}c)(1\minus{}d)}{ab}$ then cyclic alternate numbers and we will find that $ a\equal{}b\equal{}c\equal{}d$ so by $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$
$ a\equal{}b\equal{}c\equal{}d\equal{}1/2$ is solution of this system then $ f(a,b,c,d)\geq f(1/2,1/2,1/2,1/2)\equal{}1$


function
algebra
domain
trigonometry
algebra unsolved

Find all f functions for which
f(2x)= (f(x)-1)^2 - 1 and f(1) =3.

Isn't there any other restriction? Is $ \mathbb{R}$ the range and the domain of the function?

If so, there are much functions satisfying this equation... Where "much" means, as the grammar says, a non countable number. $ f$ can be random in $ [1, 2[$ (except that $ f(1) \equal{} 3$ and $ f(x) \ge \minus{} 1$).

well there were nothing about f(0) we just know that f(x)=2^x + 1 satisfies the given condition

Let $ g(x)\equal{}f(x)\minus{}1$. We have $ g(2x)\equal{}g(x)^2\minus{}2$. If we let $ a\equal{}g(0)$. We have $ a^2\minus{}2\equal{}a$. So ffao is correct.
Obviously $ g(x)\equal{}2^x$ is not a solution. So fallinlovewithmaths is wrong.

feliz is right in that there are a lot of solutions.

Define $ A(x)\equal{}\{y|y\equal{}2^nx, n\in\mathbb{Z}\}$ for $ x\ne0$.
For any $ x\in A(x_0)$, we can have
either
$ g(x)\equal{}e^{\alpha(x_0)x}\plus{}e^{\minus{}\alpha(x_0)x}$
or
$ g(x)\equal{}2\cos(\theta(x_0)x)$.
Here $ \alpha(\cdot)$ and $ \theta(\cdot)$ are arbitrary real-valued functions. In particular, $ \alpha(1)\equal{}0$ and $ \theta(1)\equal{}0$.
Notice that $ \alpha(a)\equal{}\alpha(b)$ if $ a,b\in A(x_0)$. Same for $ \theta(\cdot)$.


Each post you are to spend 3 points deducting or adding points to your preference.
You can also add your username.
Also, you can buy upgrades with your own username.
1 point-light bomb-take off 3 extra from a username.
3 points-heavy bomb-take off 10 extra.
5 points-shield-cannot be attacked for precisely 24 hours.

List:
ternary0210-10 points +3=13 points


inequalities
trigonometry
geometry proposed
geometry

Let $ m_a,m_b,m_c$ be length of medians of acute triangles $ ABC$ prove that
\[ a^2 \plus{} b^2 \plus{} c^2\ge\frac {m_bm_c}{\cos\frac {B}{2}\cos\frac {C}{2}} \plus{} \frac {m_cm_a}{\cos\frac {C}{2}\cos\frac {A}{2}} \plus{} \frac {m_am_b}{\cos\frac {A}{2}\cos\frac {B}{2}}
\]
(standard notations).

[color=red]Sorry for my spam but I only can prove your inequality in the case the triangle is acute.
  Let we have a lemma : If the triangle is acute then $ m_a \le\sqrt{2(b^2\plus{}c^2)}.\cos\frac{A}{2} (*)$ and similarly for $ m_b,m_c$.
  Indeed by squaring and some small calculation we have $ (*)$ is equivalent to: $ (b\minus{}c)^2(b^2\plus{}c^2\minus{}a^2)\ge 0$ which is clearly true because the triangle is acute.
  Thus we only need to show that $ 2(a^2\plus{}b^2\plus{}c^2)\ge \sqrt{a^2\plus{}b^2}.\sqrt{b^2\plus{}c^2}\plus{} \sqrt{b^2\plus{}c^2}.\sqrt{c^2\plus{}a^2}\plus{}\sqrt{a^2\plus{}b^2}.\sqrt{c^2\plus{}a^2}$
  Which is true by AM-GM inequality.So we are done. :) .But how can we solve the problem in the case the triangle is obtuse  :?: [/color]

[quote="encyclopedia"]Let $ m_a,m_b,m_c$ be length of medians of [b]acute triangles[/b] $ ABC$ prove that
\[ a^2 + b^2 + c^2\ge\frac {m_bm_c}{\cos\frac {B}{2}\cos\frac {C}{2}} + \frac {m_cm_a}{\cos\frac {C}{2}\cos\frac {A}{2}} + \frac {m_am_b}{\cos\frac {A}{2}\cos\frac {B}{2}}
\]
(standard notations).[/quote]

My problem is for acute triangle, the solution is base on nice indentity of triangle
\[ 2(b^2 + c^2)\cos^2\frac {A}{2} = 4m_a^2 + (b - c)^2\cos A\ge 4m_a^2
\]
So
\[ \sqrt {\frac {b^2 + c^2}{2}}{\ge\frac {m_a}{\cos\frac {A}{2}}
}\]
Similarly and use AM-GM we get this result.

On the same occasion

[quote="Arsenaler"][color=red]Sorry for my spam but I only can prove your inequality in the case the triangle is acute.
  Let we have a lemma : If the triangle is acute then $ m_a \le\sqrt {2(b^2 + c^2)}.\cos\frac {A}{2} (*)$ and similarly for $ m_b,m_c$.
[/color][/quote]

The above is wrong, it should be
 $ m_a \le\sqrt {\frac {(b^2 + c^2)}{2}}.\cos\frac {A}{2}$

[color=red]Ok I have some mistakes in my post but anyway the database of my solution is correct.
  And the last here is the lower bound it's true for arbitrary triangle: $ m_a\ge\frac{(b\plus{}c)\cos\frac{A}{2}}{2}$[/color]


linear algebra
matrix
combinatorics unsolved
combinatorics

We call a society ideal if for each two distinct members of that society there exists a person familiar with exactly
with one of them. Prove that for any ideal society of $n$ members there exist a $n-1$-member ideal society included in that.
(You can consider this as a simple graph)

In terms of adjacency matrices, the problem becomes:

Given an $n\times n$ symmetric binary matrix with distinct rows, prove that for some $i\in\overline{1,n}$, after we eliminate the $i$'th row and column, we're left with an $(n-1)\times(n-1)$ matrix with the same property.

Suppose the contrary. For each $i$, after we eliminate the $i$'th row and column, we're left with a matrix having at least one pair of identical rows. Consider the following graph: the vertices are the rows of the initial matix, and two vertices are connected iff the corresponding rows differ in precisely one component. By our assumption, we see that this graph on $n$ vertices has at least $n$ edges, hence it must have a cycle $C=(r_1,r_2,\ldots,r_k,r_1)$ such that the components where $r_j,r_{j+1}$ differ are distinct as $j$ varies in $\overline{1,k}$. This, however, is clearly impossible: if we assume WLOG that the first component of $r_1$ is $0$ and the first component of $r_2$ is $0$, at some point, in order to get back to $r_1$, some $r_j$ will have to differ from $r_{j+1}$ in the first component.


algorithm
number theory
relatively prime

if [b]a,n,m[/b]are natural number and a>1 [b]prove that:[/b]
[b]((a^m) -1,(a^n) -1)=a^(m,n) -1[/b]

Ah...have done this before, but too lazy to redo it.  If you let $m=x(m, n)$ and $n=y(m, n)$, then you just have to prove that $a^{(x-1)(m, n)}+a^{(x-2)(m, n)}+...+a^{(m, n)}+1$ and $a^{(y-1)(m, n)}+a^{(y-2)(m, n)}+...+a^{(m, n)}+1$ are relatively prime, which follows from repeatedly using the division algorithm, because x and y are relatively prime...

What?  Come on, Krishanu!  Just WOP it!  :D

Let $d = (a^{m}-1, a^{n}-1)$.  Then

$a^{m}\equiv a^{n}\equiv 1 \bmod d$
$\implies a^{(m, n)}\equiv 1 \bmod d$
$\implies d | a^{(m, n)}-1$

And of course we can show $a^{(m, n)}-1 | d$.  Hence the two are equal :D


linear algebra
matrix
combinatorics solved
combinatorics

Let $x_1$, $x_2$, ..., $x_n$ and $y_1$, $y_2$, ..., $y_n$ be real numbers. Let $A$ be a $n \times n$ matrix such that $a_{ij}=1$ if $x_i \geq y_j$, and $a_{ij}=0$ otherwise. Let $B$ be a $n \times n$ matrix made of $0$'s and $1$'s such that the sum of any row of $A$ equals the sum of the corresponding row of $B$, and the same for the columns. Prove that $A=B$.

Is there a crucial difference with that one?
http://www.mathlinks.ro/viewtopic.php?t=5758

Pierre.

It's the same, just take $-y_i$ instead of $y_i$.   ;)

It is proposed by Iran in IMO shortlist 2003


geometry
AMC
AIME
analytic geometry
calculus
integration
trigonometry

I'm in this online course that helps you prepare for the AIME, and even our teacher had problems solving this one... :wallbash_red:
I don't get this that well, either, but you guys are geniuses (improper grammar, i know :rotfl: )
so good luck!!   :thumbup:

[img]http://i17.tinypic.com/62sh4zk.jpg[/img]

lol i thought about coordinates :rotfl:

[hide]
By Power of a Point, $ DA^{2}= 3*1 \Rightarrow DA = \sqrt{3}$ (Because DA = A(extension of AD onto circle). Similarly, $ CE=\sqrt{3}$. Hence, $ \angle DOA = 60$, so $ \angle DOC = 30$. Now, the area of the sector is the area of sector $ ODE$ - 2*$ [DBO]$ = $ \frac{\pi}{3}-(\sqrt{3}-1)$ so yeah
[/hide]

Umm >_>

[hide="Daesunbash"]
...so $ BD = \sqrt3-1$, meaning that the shaded area is slightly more than $ \frac{(\sqrt3-1)^{2}}{2}\approx 0.268$. Choice A is the closest.
[/hide]

I think this is the reason that calculators are now banned from the AMC.

mkay...This takes like 5 seconds with calculus if you know your basic integrals.

$ \int_{1}^{\sqrt{3}}\sqrt{4-x^{2}}-1 \, dx=2\sin^{-1}\frac{x}{2}+\frac{x}{2}*\sqrt{4-x^{2}}-x |_{1}^{\sqrt{3}}=\frac{\pi}{3}+1-\sqrt{3}$

I am sure that trig substituions are part of the AP calculus BC curriculum. I guess that is what I mean by basic.

here's a simple way to do it, pretty sure it works

[hide]draw a line from O to D and O to E, now OE and OD are both 2 since they are the radii, We know CO and AB is 1, so we can now find the lenth of CE and DA, which is $ \sqrt{3}$, so you can see the triangle is a 30-60-90 triangle, so angle DOA is 60 degrees, that means COD is 30 degrees, and from the other triangle, we can see that COE is 60 degrees, that means EOA is 30, so COA-COD-EOA=30
degrees, now find the area of region DOE, it is $ \frac{30}{360}*2^{2}\pi$ now you can find the lenth of BE and BD by subtracting CB and AB from CE and AD, so BE and BD is $ \sqrt{3}-1$ now draw a line from BO, and we can prove
triangle DOB and OBE is equal by sss, to find the area of OBE, we know the height which is CO=1, and we know the base, BE=$ \sqrt{3}-1$, and we can find the area of OBE, just remember to multiply by two to get the area of triangle ODB also, now subtract the area of the two triangles from the region DOE
$ \frac{\pi}{3}-(\sqrt{3}-1)$ it equals answer choice A[/hide]

thanks!
but the integral part confused me, cuz i didn't take calculus.. i know just a teeny bit of derivatives, and that's it

[quote="Elemennop"]yea, it's definitely not a "hard" integral to solve, but it's not exactly a quick integral[/quote]

When you look at an integral, first you go throught the obvious set of techniques

i) direct use of an integral rule
ii) u substition
iii) integration by parts
iv) trig substitution

it is obvious that iv) is what you do here...the problem is solved by the 'AP calculus' algorithm, hence I would consider it to be a mechanical, easy problem when you use the calculus approach.

[hide]Split the region bound by C, O, E, and the arc. It is formed by a sector of area $ r\theta = \pi/3$ and a triangle of area $ \sqrt{3}/2.$ (Note: you know these are 30,60,90 triangles because the hypotenuse/radius 2, is twice the leg 1) The total area is $ \pi/3+\sqrt{3}/2.$

The area of the region bound by C, O, A and D has same area. Hence, the total area of everything in the quarter circle excluding the desired region is
\[ 2\left(\pi/3+\sqrt{3}/2\right)-1, \]
where 1 is the area of the square.

Subtract from $ \pi$ to get A as answer.[/hide]


a little bit hard for me
how about you?
share your experience^^ :lol:

yeh....
sorry...I mean Rubik's cube :P

Try the $3 \times 3 \times 3 \times 3$ one ;)

By 3x3x3 magic cube, do you mean a 3x3x3 cube filled with the numbers 1-27 such that the sum of the numbers in any row, column, or diagonal is constant?

Or do you mean some sort of Rubik-based puzzle?

I think she means the rubix cube.  In chinese, it literally translates into magic cube.  I don't know about other places though.

In Germany, it's called both "Rubix' Cube" and "Zauberwürfel" (="magic cube").

I used to know how to do it, its not too complicated but you have to learn how.  Its nearly impossible without knowledge of exactly how to do it in my opinion.

[quote="math92"]my best time is like 15 minutes, :blush:[/quote]

well, if you did it without memorizing the algorithms, then thats pretty good...
i used to be impressed until i learned how, and found that it was just applying these methods in various cases...

anyways, try doing it in the 27 turns or so which is calculatedto be the minimum number of turns in order to solve the cube in any position... o.0

Hm...the first time i did it took like a week of on and off thinking.
After that, I structured my method more, made it more methodical.  Then I sped it up, adopted things from other methods.  It turned out my method was very similar to Lars Petrus's...at the start at least.
Recently, I took an average around 19 seconds.

[quote="K81o7"][quote="kyyuanmathcount"]I know someone who takes about 8 or 9[/quote]

Really?  Average?  ...
If that were so, that person would hold the world record by a bit.[/quote]

agreed.
im not sure its even possible to solve a well mixed cube in 8 seconds...

[quote="kyyuanmathcount"]I know someone who takes about 8 or 9[/quote]

That's not true - the official world record is just over 12.

Maybe he means 8 or 9 minutes? :D

I do the layer by layer method because I just got a cube for Christmas.

I got my time down to one minute.

But I think I need to use a better method if I want to break 30 seconds.

By the way, I think some algorithm proved that it takes at most 20 moves to solve any given cube, given that you make the right moves of course!

ok, 2X moves to solve then =D

how would you prove such a thing anyways? :?

my own algorithm gets me about 30 sec, which is not bad. i guess it also depends on the "turnability" of the cube, and the speed of your turns...

Wow, you people are so good...my record irs 51.18 seconds.

Does anyone have suggestions for lubricating a cube with common household materials (I don't have silicon/silicone (spelling)).

You can prove it by developing a very complex set of algorithms.

The reason people don't use such an available algorithm is because there are too many cases to consider while cubing.

sometimes some lubricants make the cube messy...i could also suggest taking your cube apart and just wiping everything, if that makes sense...

of course, using the jelly would also require taking the cube apart:
for those of you who dont know, taking the cube apart is quite simple. all you have to do is turn one face to a 45 degree angle and pop out an edge piece with your finger if your cube is extremely loose, or a pen. then the rest of the cube sort of falls apart.

I use Vaseline and it seems to work. Yeah, a 3x3 is easy to put back together but a 5x5 is not  :lol:

About the 8 or 9 seconds, the world record is 7 isnt it?

erm, no, i think its around 12, 11.75 to be exact...

Best time : 55 seconds after gotten lucky

The current world record for 3x3x3 two-handed is 11.75 for a single solve, held by Jean Pons from France. Average of 5 is 14.59, held by Shotaro Makisumi ([url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=2889]litoxe2718281828[/url]). One-handed single-solve is 22.05, held by Ryan Patricio, and one-handed average of 3 is 26.73, also held by Ryan Patricio. Blindfolded is 1:46.47, by Leyan Lo.

As for my personal records, I have ~27 seconds single solve (not lucky), average is somewhere in the low 30's, and my one-handed is ~1:06. As you can see, I haven't accurately timed myself very much. :P Also, I'm pretty sure I've done better than these at times, but never timed them, so I don't know exactly how fast they were.

Also, the unofficial WR for fewest moves to solve a cube is 21, and I think the theoretical minimum is 20, although I could be wrong. Official WR is 28. My best is 48 (that I counted for).

The fastest lucky solve is difficult to determine. Some have done it in under 5 seconds, but often because the cube wasn't scrambled enough. The fastest I could find that I was sure was scrambled properly and simply lucky (unofficial, though) was Macky's 7.40 second solve, where he basically skipped half of the last layer. :P The fastest unofficial non-lucky solve is 9.54 seconds.

Fastest solve: 33.95 seconds
Fastest average of 10 (12, but with the fastest and slowest taken out): 39.18 seconds

Cubing is my current addiction.  :)

My best average is 18.76, best solve without luck is 14.40 and best solve with luck is 13.90. It's been a while though so I'm probably out of shape.

[b]Don't[/b] lubricate with vaseline. It contains petroleum and will wear down your cube eventually. Use a silicon based lubricant instead. Preferably some kind of spray. (Any professional cuber will tell you this :)).

Have any of you solved blindfolded? I can do it in about 15 minutes. It's really frustrating until you do it right for the first time. When you do it wrong, of course, you have little clue of _where_ you went wrong, so getting it right for the first time is difficult. But when it happens - ohh man. :D

For lubrication, I first used Snap silicone spray. Works quite well. However, I haven't bothered to reapply it, and that was something like 10 months ago. While I'm sure my cube would spin better if I did reapply, at the moment it works just fine through wearing down inside and being lubricated by cube dust. :D

I really hate to be an Admin. but if you didn't notice there is a whole forum on rubik's cube, and similar math based games! ;)  ;)  ;).

For those who still have not found it: :D

[url=http://www.artofproblemsolving.com/Forum/index.php?f=360]www.artofproblemsolving.com/Forum/index.php?f=360[/url]

But, personally, i have seen people use almost any kind of lubriant, everything from Baby oil to olive oil (don't ask). :huh:

my record is like somewhere around 40 or 50 seconds. and btw... its all about silicon lube. thats teh best stuff out there for cubes. some ppl i kno r super fast meaning less than 7 seconds as a record.

Fire, it's not your duty to be an administrator, and the capital font could be misinterpreted as shouting, which by all means isn't the message you [i]want[/i] to be communicating.  :P

Speed depends on

1. How easily the springs and sides turn (some people call it the smoothness)
2. How well the springs stretch (the margin of error)
And of course, the cuber.  But really, one needs a balance between weak springs and strong springs.  Too strong means harder to turn, almost no margin of error.  Too weak means it'll literally fall apart in your hands.  I had a cube which was extremely easy to turn and the margin of error was about 60% (literally).  As long as you didn't go crazy on it, it didn't pop (however, you could rotate corners in place with easy, and pop out pieces very easily as well.  You could also put corners directly in).  However, I lost that...yet it also has to do with wearing the cube in, so my other cube has some potential.  It is a bit harder to turn and has less margin of error, but then again, it's sturdier, and I haven't used it as long.

it takes me about 1 second to rotate once.....

lol..

Do you use your fingers rather than your whole hand to turn a side? If you learn to just flick it with your finger, it'll be much faster. However, your cube needs to be fairly loose to do that. Just lubricate it with something good, preferably silicone spray, and work it around a little, and you'll be good to go. Make sure to work with it for about an hour after lubricating to get the excess out, washing your hands every 5 to 10 minutes, since you don't want to keep that stuff on your hands. ;)

Yeah, you ideally combine flicking and turning to get really really fast (you want smooth transitions) but ultimately you really only touch the cube with your fingertips (pretty much, and you mostly use thumb and index fingers, some ring finger).

I forget my best time, but I haven't solved a Rubik's Cube since MP2005, my best time then was... under a minute I think...

I remember I solved one while submerged underwater, so it couldn't have been over one and a half minutes.

Once you know how, you can scramble and unscramble the cube in less than 5 min. I can!

i can generally solve the cube in 2-3 minutes, i did recently time myself but i forgot what the time was (that's effective  :lol: ).  the 5x5 takes me about 25 minutes (is this extremely long?) and the 4x4 somewhere in between there.

as for lubricating them, my old 3x3 turned horribly, but my new one (25th anniversary cube) i got at christmas, it turns really well.  the 5x5 is sometimes a pain, though.  i actually haven't ever lubricated/taken apart any of my cubes, strangely enough, but i assume it will happen inevitably...

I can do it using the internet.  :D Does that count??

hm... my record was 2 min (im pretty bad yes)

WHAT???!!!

That's the record I'll never break!

*fanfare*

I think my best time was 1:16, but I had a sucky cube..
I don't speedcube anymore, I find it a waste of time.


Find all the numbers n such that n!+1 is a perfect square.
Cheers!

This seems to be a very serious conflict. I have just one question: do you have solutions for these problems? I may be stupid, but they seem very hard. If they really are, they I do not understand your point. But I'm not interested, so I let Valentin to think about them.

As far I as know this problem is not only difficult but this also is an open problem.
Thus, I do not see any interest to your way of asking problems even if they are not given to me....

You will learn more about the problem at :
http://mathworld.wolfram.com/BrocardsProblem.html

Pierre.


Just to make sure: $y_n$ on RHS [i]should[/i] be in the index?

[quote="Beat"]It is given the sequence $y_1=y_2=1, y_{n+2}=(4k-5)y_{n+1-y_n}+4-2k, n\ge 1$.
Find all integer numbers numbers $k$, for which all the terms of the sequence are perfect squares. (Enjoy)[/quote]
It would be:
[quote]It is given the sequence $y_1=y_2=1, y_{n+2}=(4k-5)y_{n+1}-y_n+4-2k, n\ge 1$.
Find all integer numbers numbers $k$, for which all the terms of the sequence are perfect squares.(Enjoy)[/quote]

My answer is: there's no integer value of k satisties the problem's condition.
Is it true?, or false? :)

Maybe I had a mistake in calculating. Can you post your solution? :(

I solved this very beautiful problem .yes the only solution for $k$ is $1$ & $3$ .you can solved it ,only take sequence mod  $k-2$ ,and prove
$k-2$ must be $1$ or $-1$ so this very very very good problem can solved. :D  :D

[quote="dna"]only take sequence mod  $k-2$ ,and prove
$k-2$ must be $1$ or $-1$ so this very very very good problem can solved. :D  :D[/quote]

How :?:

And further...


Kyle's science project grade has two parts. The oral presentation is worth 30 points, and the written report is worth 70 points. Kyle earned $ 84\%$ of the possible points for his written report. To earn at least $ 87\%$ of the possible points for the entire project, what percent of the possible points for the oral presentation must he earn?

If he earned 84 percent on the oral presentation, then he earned $ 0.84 \times 70 \equal{} 58.8$ points. He wants a total of 87 points, so he needs $ 87 \minus{} 58.8 \equal{} 28.2 \div 30 \equal{} \boxed{94}$ percent.


topology
function
advanced fields
advanced fields unsolved

Let $A$ and $B$ be the [b]topological space [/b]that both have fix point property and $A\cap B=\bar{A}\cap B=A\cap\bar{B}=\{c\}$ then $A\cup B$ also have [b]fix point property[/b].
The set have fix point property respect to its topology if any function from this set to itself have fix point.

Saman [b]Gharib[/b]

it is challenging problem so it's nice to thinking.

Saman Gharib

Of course, "any function" must mean any continuous function. And I'm not sure what topological separation axioms we might need - I think we need points to be closed.

Define the function $p_A: A\cup B\mapsto A$ by $p_A(x)=\{\begin{array}{cc}x&\text{ if }x\in A\\c&\text{ if }x\in B\end{array}.$

$p_A$ is a continuous function. We see that by taking a closed $F\subset A\cup B.$ We have that either $p_A^{-1}(F)=F\cap A$ or $p_A^{-1}(F)=(F\cap A)\cup \{c\}.$ Since $A$ and $B$ are both closed in $A\cup B,$, this is closed.

Similarly, define $p_B(x)=\{\begin{array}{cc}x&\text{ if }x\in B\\c&\text{ if }x\in A\end{array}.$

Now, suppose $f: A\cup B\mapsto A\cup B$ is continuous. Define a new function $g$ by $g=p_A\circ f,$ restricted to $A.$ $g: A\mapsto A$ is continuous, hence by hypothesis has a fixed point. If this fixed point is in $A\setminus\{c\},$ then we have a fixed point of $f$ and are done. The alternative is that $f(c)\in B.$

Repeat this argument with $h=p_B\circ f,$ restricted to $B.$ As before, either we find a fixed point of $f$ in $B,$ or $f(c)\in A.$ But if both $f(c)\in B$ and $f(c)\in A,$ then $f(c)=c$ and $c$ is a fixed point.


[quote="LawOfSigns"]It's one month till state for me, I and I got these books because they were the top 3 on the mathcounts list and I didn't think I would have time for 4 books.  [b]Anyone who hasn't done the 2007 state yet[/b], what chapters should I read?  (I'm in geometry honors, but I got the geometry book because its challenge problems never come near to being covered at school)  I'm worst at # theory, poor at prob, and ok at geometry; (by mathcounts standards anyways) which chapters are most helpful, from your experience with 2006 & earlier competition scores?[/quote]Because of the boldface in the original posters' question, I don't feel comfortable leaving this discussion open.  (Basically, there's no way to police this.)  So although this is a good question, I feel like I need to lock this thread.

Please refrain from asking or suggesting particular topics to study until the conclusion of all the state contests.

I'm sorry I have to do this, but I feel it is necessary to protect the integrity of the contest.


Hi. I have a friend who started a new forum for school. But he's currently planning to put a math section into the forum. He was wondering how you guys put LaTeX into the forum so we could do stuff like $\frac{x}{y}$, $\sqrt[z]{x}$, or $x^{2}-4=0$ and stuff like that. His forum is php or IPB(I forget which). Please help my friend, that would mean a lot to him.
Thanks.

I installed latexrender and set up the paths ..etc

BUT it doesn't render any .gif files.

When I hit the render button I get a dos window that appears and disappears very quickly, so that I can't read it.

<Q?> where does latexrender place its output files??

And what is the problem/fix?

TIA

Please don't post in two threads. I can answer your problem with the detail you give in the other thread.

My friend used the LaTeX renderer on his forum, but the forum required didn't enable the html as well as one of the other incompatible choices. How do you enable html well without an ipb forum?

I would need a lot more detail before being able to make any suggestions. How about posting a link to the forum so I could see what you are referring to?

I try to download LATEX but I always end up downloading RIBBIT. Does anybody know why this happens and what's RIBBIT's difference with LATEX.

Ribbit is a piece of software that uses LaTeX to add equations into Word. So it isn't LaTeX at all.

Have you followed the instructions and links at [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php]AoPS Downloads for LaTeX[/url] which gets you to download MiKTeX - a LaTeX distribution for Windows?

Well, I think so :blush: Anyway, thanks for the information. I'll try again... :wink:


Find all functions $ f:\mathbb{R}\to\mathbb{R}$ such that
\[ f(x\plus{}y\plus{}z)\equal{}f(x)f(1\minus{}y)\plus{}f(y)f(1\minus{}z)\plus{}f(z)f(1\minus{}x)\] for all $ x,y,z\in\mathbb{R}$.

[quote="Pirkuliyev Rovsen"]Find all functions $ f\colon \mathbb{R}\to\mathbb{R}$ such that $ f(x \plus{} y \plus{} z) \equal{} f(x)f(1 \minus{} y) \plus{} f(y)f(1 \minus{} z) \plus{} f(z)f(1 \minus{} x)$ for all $ x,y,z\in\mathbb{R}$[/quote]

Let $ P(x,y,z)$ be the assertion $ f(x \plus{} y \plus{} z) \equal{} f(x)f(1 \minus{} y) \plus{} f(y)f(1 \minus{} z) \plus{} f(z)f(1 \minus{} x)$

$ P(x,1 \minus{} y,y)$ $ \implies$ $ f(x \plus{} 1) \equal{} (f(x) \plus{} f(1 \minus{} x))f(y) \plus{} f(1 \minus{} y)^2$
$ P(x,1 \minus{} z,z)$ $ \implies$ $ f(x \plus{} 1) \equal{} (f(x) \plus{} f(1 \minus{} x))f(z) \plus{} f(1 \minus{} z)^2$

Subtracting these two lines, we get : $ (f(x) \plus{} f(1 \minus{} x))(f(y) \minus{} f(z)) \equal{} f(1 \minus{} z)^2 \minus{} f(1 \minus{} y)^2$ and so :

If $ f(x)$ is not constant ($ \exists f(y)\ne f(z)$), then $ f(x) \plus{} f(1 \minus{} x) \equal{} a$ but then $ P(x,1 \minus{} y,y)$ becomes $ f(x \plus{} 1) \equal{} af(y) \plus{} f(1 \minus{} y)^2$ constant hence contradiction.
So $ f(x) \equal{} c$ constant. Plugging back in the original equation, we get $ 3c^2 \equal{} c$ and $ c\in\{0,\frac 13\}$

Hence the only two solutions :
$ f(x) \equal{} 0$ $ \forall x$

$ f(x) \equal{} \frac 13$ $ \forall x$


modular arithmetic
quadratics
number theory

Um can someone prove this question?
"Prove that when you take an odd number and square it followed by subtracting 1, it can always be divided by 8." Ex: 5 squared is 25, then minus one is 24 which is divisible by 8. :lol:  :D  :P

$ 2n\plus{}1$ will always be an odd number because $ Even\plus{}1\equal{}Odd$
Therefore,
$ (2n \plus{} 1)^{2} \minus{} 1 \equal{} 4n^2 \plus{} 4n$
Now we will assume that that is divisible by 8
$ 4n^2 \plus{} 4n \equal{} 8k$
$ n^2 \plus{} n \equal{} 2k$
Now this means that as long as any number squared, added to the original number, is divisible by 2,then any odd number squared subtracted by 1 will be divisible by 8.
And since any number squared, and then added to the original number will be even, because $ Even \plus{} Even \equal{} Even$ and $ Odd \plus{} Odd \equal{} Even$, then any off number squared subtracted by 1 will be divisible by 8.
I suck at proofs.
 :wink:

Make the last statement stronger. Other than that it is ok.

[quote="ZhangPeijin"]$ 2n \plus{} 1$ will always be an odd number because $ Even \plus{} 1 \equal{} Odd$
Therefore,
$ (2n \plus{} 1)^{2} \minus{} 1 \equal{} 4n^2 \plus{} 4n$
Now we will assume that that is divisible by 8
$ 4n^2 \plus{} 4n \equal{} 8k$
$ n^2 \plus{} n \equal{} 2k$
Now this means that as long as any number squared, added to the original number, is divisible by 2,then any odd number squared subtracted by 1 will be divisible by 8.
And since any number squared, and then added to the original number will be even, because $ Even \plus{} Even \equal{} Even$ and $ Odd \plus{} Odd \equal{} Even$, then any off number squared subtracted by 1 will be divisible by 8.
I suck at proofs.
 :wink:[/quote]The ideas contained in here are sufficient enough to solve the problem but the way this proof is written, it is not logically correct. First, you do not assume the conclusion, ie 'Now we will assume that that is divisible by 8' and then prove it??? Brainstorming $ \neq$ proving. Here is a typical way to improve your proof and shorten it:

$ 2n \plus{} 1$ will always be an odd number because $ Even \plus{} 1 \equal{} Odd$

Therefore,
$ (2n \plus{} 1)^{2} \minus{} 1 \equal{} 4n^2 \plus{} 4n\equal{}8\frac{n(n\plus{}1)}2$.

$ n(n\plus{}1)$ is an even integer because either $ n$ or $ n\plus{}1$ is even.

So, $ (2n \plus{} 1)^{2} \minus{} 1 \equal{}8\frac{n(n\plus{}1)}2\equal{}8k$, where $ k\equal{}\frac{n(n\plus{}1)}2$ is some integer.

[quote="hunter34"]Thats not a proof. You just showed for one case. You have to prove [b]ALL[/b] cases.[/quote]
I know. I just wanted to show an example

Sorry. Did not see the EX in front of it.

[hide="if n is any odd integer, then"]
$ n\pmod{8}\equiv \{1,3,5,7\}\equiv \{1,3, \minus{} 3, \minus{} 1\}$ which leads to: $ n^2\pmod{8}\equiv \{1\}\Rightarrow (n^2 \minus{} 1)\equiv 0\pmod{8}$.
The latter statement verifies the equation $ n^2 \minus{} 1 \equal{} 8k$. An exception to LHS divisibility by eight is $ k \equal{} 0$ which implies $ n\neq\pm1$.

Therefore $ 8\mid{(n^2 \minus{} 1)}$ if and only if $ |n|\neq1$.[/hide]
I liked that. :)

EDIT: Modular arithmetic is clearly a powerful tool here, but only because we so easily defined n as any odd integer. This problem is related to what is known as [url=http://mathworld.wolfram.com/QuadraticCongruenceEquation.html]quadratic congruency[/url], and that can become messy in a quick way if you don't know what you're doing. AOPS volume II has a nice section on it, as does Introduction to Number Theory by Mathew Crawford (I think).

[quote]
Therefore $ 8\mid{(n^2 \minus{} 1)}$ if and only if $ |n|\neq1$.[/quote]

are you sure?

if $ n\equal{}2$
$ 8\not|(2^2\minus{}1)$
is it ?

[quote="igiul"]An exception to LHS divisibility by eight is $ k \equal{} 0$ which implies $ n\neq\pm1$.\[/quote]

Zero divides nothing but everything (integers except zero) divides zero. That is, $ z|0$ for all integers $ z\neq0$ (where we throw out zero since $ \frac {0}{0}$ is indeterminate), so there is no exception to the result you got. Nice proof, though!

[quote="binomial_4eva"][quote]
Therefore $ 8\mid{(n^2 \minus{} 1)}$ if and only if $ |n|\neq1$.[/quote]

are you sure?[/quote]

For odd $ n$, yes. What is the problem?

An alternate solution:

We want to prove that $ 8\mid (n^2\minus{}1)$ for all odd $ n$. But, $ n^2\minus{}1\equal{}(n\plus{}1)(n\minus{}1)$ For odd $ n$, both of $ n\plus{}1$,$ n\minus{}1$ are even, and since of every two consecutive evens one is a multiple of 4, we have that their product is a multiple of $ 2\cdot 4\equal{}8$.

Or this could be slightly changed by letting $ n\equal{}2k\plus{}1$ so that $ (2k\plus{}1)^2\minus{}1\equal{}(2k\plus{}2)(2k)\equal{}4(k\plus{}1)k$ and exactly one of $ k$, $ k\plus{}1$ is even.


Suppose you need to use at least 1 dollar worth of stamps to mail a package. You have as many 3 cent stamps as you need, but only four 32 cent stamps. How many of each stamp can you use?

What systems of inequalities would describe this situation?

[hide="Solution?"]Let $ x$ be the number of $ 3$-cent stamps, and $ y$ be the number of $ 32$-cent stamps. We have the following three inequalities: $ 3x \plus{} 32y \ge 100$ and $ x \ge 0$ and $ 0 \le y \le 4$.[/hide]

If you're just looking for systems of inequality, then I think darkdieuguerre has the right idea. As for how many you can use, that would be infinite, I assume, cause you have an unlimited number of 3 cent stamps.....


Call an integer $m$ [b]good[/b] if $|m|$ is not a square. Determine all integers $m$ with the following property :::

$m$ can be represented , in infinitely many ways, as a sum of three distinct [b]good[/b] integers whose product is the square of an odd integer.

Posted before:

IMO Shortlist 2003, problem N5 (or N6)...


integration
function
limit
algebra unsolved
algebra

Let f and g be two continious, distinct funtions from $[0, 1]\rightarrow R$ such that:
$\int\limits_{0}^{1}f(x)d(x)=\int\limits_{0}^{1}g(x)d(x)$.
Let ${y_n=\int\limits_{0}^{1}\frac{f^{n+1}(x)dx}{g^n(x)}}$, for ${n\geq0}$, natural.
Prove that  $y_n$ is an increasing and divergent sequence

$f^{n+1}$ is $f\circ f\circ f\circ \ldots \circ f$ ?

No, it isn't. i remember that i posted a solution to this nice problem before. Anyway, here it goes again :D :

 [color=blue] [b] Problem. [/b]

    Let $f, g: [0,1]\to (0,\infty)$ two continous functions such that \[ \int_{0}^{1} f(x)dx=\int_{0}^{1} g(x)dx.  \]
    Let us define the sequence \[ x_{n}=\int_{0}^{1} \frac{f^{n+1} (x)}{g^{n} (x)} dx.  \]
    Prove that $(x_{n})_{n\in\mathbb{N^{*}}}$ is increasing and divergent.
 [/color]

  [i] Solution. [/i]

     It is easy to see that there is $c\in[0,1]$ such that $f(c) >g(c)$ if $f\neq g$. From here we deduce that there is
     a compact interval $I$ such that $f(x)> g(x)$. So we have

                              \[ \int_{0}^{1} \frac{f^{n+1} (x)}{g^{n} (x)}dx=\int_{0}^{1} f(x)\cdot\frac{f^{n} (x)}{g^{n} (x)}dx\geq (\inf_{x\in I} \frac{f(x)}{g(x)})^{n}\cdot\int_{I} f(x)dx.  \]

      Since $\lim_{n\to\infty}(\inf_{x\in I} \frac{f(x)}{g(x)})^{n}=\infty$ by the sandwich theorem the conclusion follows.

And by the way, this isn't an algebra problem for sure. I think that this should be moved to College Playground-Analisys.... ;)

Wow, I forgot to prove that the sequence is increasing. Here it goes:

 \[ x_{n}-x_{n-1}=\int_{0}^{1} \frac{(f^{n}(x)-g^{n}(x))(f(x)-g(x))}{g^{n}(x)} dx\geq 0. \]


I want to change the font of the page number that is automatically generated on the first page of a new chapter by the \chapter{} function.

Also is it possible to change the alignment of this page number? i.e. from center to outer?

many thanks in advance

....ok, so now I have an identical problem with the Index page.... i.e. different font and alignment for the page number compared to teh rest on the doc.

The only reason that I can see for a different font for the page numbers is that you have changed the font for parts of the document but without a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] it is impossible to tell.

The best way to deal with footer (and header) layout is to use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=fancyhdr]fancyhdr package[/url]; \fancyfoot[LE,RO]{\thepage} for example puts the page number on the left on even pages and on the right on odd pages.

Ok, sorry for not including any code. I know about fancyhdr and am using it but when you invoke a new chapter it produces a new page format of its own automatically; i.e. it suppresses the header and footer rules and text that I have setup in fancyhdr and produces a blank header and footer without rules and with a centered page number in a serif font.

I don't mind the blank headers and footers but would like the page number font to match and aslo possibly to align it at 'outer' like the rest of my two-side book. The page number revets back to my settings after the first page of the chapter without me having to do anything. As I say, I have the same issue with my index pages which are also producing a centered serif font page number.

Any ideas?

At the start of the doc I use;

\fancyfoot[LE,RO]{\sffamily{\thepage}}

to setup my page numbers.

A typical chapter start looks like;

\cleardoublepage
\mainmatter
\chapter{Introduction}

\fancyhead[LE,RO]{\sffamily\textbf{Introduction}}
\renewcommand{\headrulewidth}{1.0pt}      %Underlines the header. (Set to 0pt if not required).
\renewcommand{\footrulewidth}{1.0pt}      %Underlines the footer. (Set to 0pt if not required).

many thanks

Unfortunately it's impossible to tell exactly what you are doing as you haven't provided a [b]complete[/b] minimal example with the preamble etc, so I am left with guessing what is happening.
\chapter redefines the page style to plain, so I suspect that you haven't redefined the plain pagestyle. Section 7 of the fancyhdr manual shows you how to do this.

Ok, that was all I needed. I have redefined the plain pagestyle and that has solved the problem.

Thanks very much - your help is really appreciated


probability
geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope

here is my question:
Three points are chosen randomly and independently on a circle. What is the probability that all 3 pairwise distances between the points are less than the radius of the circle?

Of course it's possible. Set it so that the two outer points have less than the distance of the radius between them.

I remember seeing this problem before, but I'm not completely sure how to solve it. Obviously, the placement of the first point is irrelevant. After that, the second point has a $\frac{1}{3}$ chance of being within the possible area. However, I get stuck after this. Essentially, if the second point is placed (an arbitrarily small distance under) one radial length from the first point, then the third point must be placed between them, so we have a $\frac{1}{6}$ chance of meeting the conditions. However, the closer the second point is placed to the first, the larger the chance is that the third point will be placed so that the conditions are met. How am I supposed to approach this?

Woo... awesome problem...

[hide][img]http://img127.imageshack.us/img127/7399/amc2003b254qe.png[/img]

Look at the picture (circle has circumference $1$). Call the smallest distance between two points $x$. Call the left point the "center," that is, both the other points are closer to the "center" than each other. The third point is located on the arc of length $1-x$.

(1) Then the distance between the left point and the third point is smaller than $\frac{1-x}{2}$ (to maintain the "center" status) but greater than $x$ (so that $x$ is still the smallest distance). However, we also note that $x$ has to be less than $\frac{1-x}{2}$ or the third point will become the center, so $x < \frac{1-x}{2} \Rightarrow x < \frac{1}{3}$.

(2) Similarly apply this argument for the third point (reflect the graph as shown below), and we get a graph that looks like this ($x$-axis is the distance from the center to the right point, $y$-axis is the distance from the third point to the center).

[img]http://img423.imageshack.us/img423/9702/amc2003b2514ci.png[/img]

The red indicates the area described by (1) and the blue is (2). The yellow is the area such that the sum of the arcs is less than $\frac{1}{6}$, or equivalently the distance is less than one radius (note that by picking a "center" we guarantee this is the total area, because the sum of the distances is always greater than the independent distances).

Hence our answer is $\frac{\frac{1}{72}}{\frac{1}{9}+2\left(\frac{1}{36}\right)} = \frac{\frac{1}{72}}{\frac{1}{6}} = \frac{1}{12}$. Whew.[/hide]

[quote="WarpedKlown1335"]paladin8, how'd you get from x<1/3 to your graphs which show x<1/2?

Sorry I'm confused.[/quote]

So for part (1), you get the red portion, where $x < \frac{1}{3}$ and (call the other arc length $y$ for now) $y < \frac{1-x}{2}$, $y > x$ (explanation in previous post).

Then in part (2), flip the graph over the line $y=x$ and apply the same argument, to fill in the blue portion.

Essentially, it is necessary to create the $x < \frac{1}{3}$ condition because as you can see the slope of the graph changes as it gets to $\frac{1}{3}$ (also to maintain the center condition).

For a revamped solution, see [url=http://wangsblog.com/jeffrey/?p=16]here[/url]. The same idea, hopefully better explained.


geometry
circumcircle
ratio
function
trigonometry
rectangle
cyclic quadrilateral

Let $K$ is the point of intersection the diagonals of cyclic quadrilateral $ABCD$. In the triangle
$AKD$ exists the point $P$, such that $\angle APC=\angle ADC+90^\circ$ and $\angle BPD=\angle BAD+90^\circ$.  Prove that the diagonals of
the convex quadrilateral, formed the foots of perpendiculars from $P$ on sides of $ABCD$, are perpendicular.
(Sorry for my English).

Let the tangents to the quadrilateral circumcircle (O) at the vertices A, C meet at a point U and let the tangents at the vertices B, D meet at a point V. The circle (U) with radius UA = UC meets the circle (V) with radius VB = VD at points P, Q, the point P lying inside the quadrilateral ABCD. As long as $\angle D < 90^\circ$ and $\angle A < 90^\circ$, the point P lies inside the triangle $\triangle AKD$ and obviously, $\angle APC = \angle D + 90^\circ$ and $\angle BPD = \angle A + 90^\circ$. PQ is the radical axis of the circles (U), (V) and the circumcircle (O) is perpendicular to both of them, hence, it is centered on their radical axis, $O \in PQ$. The diagonal AC is the radical axis of the circles (O), (U) and the diagonal BD is the radical axis of the circles (O), (V). The diagonal intersection K is then the radical center of the circles (O), (U), (V) lying on the radical axis PQ of (U), (V).

Let P' be an arbitrary point on PQ and E', F', G', H' the feet of normals from P' to the quadrilateral sides AB, BC, CD, DA. When the point P' coincides with the circumcenter O, then E', F', G', H' are the midpoints of AB, BC, CD, DA, so that $\frac{AE'}{AB} = \frac{DG'}{DC}$ and $\frac{BF'}{BC} = \frac{AH'}{AD}$. When the point P' coincides with the diagonal intersection K, then E', F', G', H' are the feet of the K-altitudes of the triangles $\triangle ABK, \triangle BCK, \triangle DCK, \triangle ADK$. Since the triangle pairs $\triangle ABK \sim \triangle DCK, \triangle BCK \sim \triangle ADK$ are similar, again, $\frac{AE'}{AB} = \frac{DG'}{DC}$ and $\frac{BF'}{BC} = \frac{AH'}{AD}$. Since these ratios are linar functions of the position of the point $P' \in PQ$ and since they are equal for 2 positions of this point, they are equal for all positions of $P' \in PQ$. In particular, they are equal when P' coincides with the point P. Thus if we denote E, F, G, H the feet of normals to AB, BC, CD, DA from P, we have $\frac{AE}{AB} = \frac{DG}{DC}$ and $\frac{BF}{BC} = \frac{AH}{AD}$.

Let the lines AB, CD meet at X and the lines BC, DA at Y. Since $\angle A < 90^\circ, \angle D < 90^\circ$, the point X lies on the opposite side of the line BC than the quadrilateral vertices A, D. Point U is the pole of the diagonal AC and point V is the pole of the diagonal BD with respect to the circumcircle (O). Hence, UV is a polar of the diagonal intersection K, which is identical with the line XY, i.e., the points U, V, X, Y are collinear. Let (X) be a circle centered at the point X and with radius XP. Using the sine theorem for the triangles $\triangle AUX, \triangle BVX$,

$\frac{UA}{UX} = \frac{\sin \widehat{UXA}}{\sin \widehat{UAX}},\ \frac{VB}{VX} = \frac{\sin \widehat{VXB}}{\sin \widehat{VBX}}$

$\frac{UX}{VX} = \frac{UA}{VB} \cdot \frac{\sin \widehat{UAX}}{\sin \widehat{VBX}}$

because $\sin \widehat{UXA} = \sin \widehat{VXB}.$ But

$\angle UAX = \angle BDA = \angle BCA = 180^\circ - (\angle B + \angle CAB) =$

$= 180^\circ - (\angle B + \angle VBC) = \angle VBX$

and consequently,

$\frac{UX}{VX} = \frac{UA}{VB} = \frac{UP}{VP}$

Thus PX bisects the angle $\angle UPV$ in the triangle $\triangle UVP$. But since $(U) \perp UP, (V) \perp VP, (X) \perp XP$, the circle (X) bisects the angle formed by the circles (U), (V).

Invert the quadrilateral ABCD with the circumcircle (O) and the circles (U), (V), (X) in a circle (P) with radius PX (same as radius of the circle (X)). The pencil of circles (U), (V), (X) (with collinear centers and all passing through the point P) is carried into a pencil of concurrent lines u, v, x, such that the line x bisects the angle formed by the lines u, v, because inversion preserves angles between curves. The circle (O) centered on the radical axis PQ of this pencil and perpendicular to 2 circles (U), (V) of this pencil (i.e., perpendicular to all circles of this pencil) is carried into a circle (O') centered at the concurrency point O' of the lines u, v, x. The quadrilateral vertices A, B, C, D are carried into the intersections A', B', C', D' of the lines u, v with the circle (O'), i.e., the quadrilateral ABCD is carried into a rectangle A'B'C'D'. Since the circle (X) is congruent to the inversion circle (P), the line x (midline of the rectangle A'B'C'D') is the perpendicular bisector of the segment PX. Therefore, the triangles $\triangle D'PA' \cong \triangle C'XB'$ are congruent. But the triangles $\triangle APD \sim \triangle D'PA'$ with the common angle at the vertex P are similar by the basic properties of inversion $PA \cdot PA' = PD \cdot PD' = PX^2$. It follows that the triangles $\triangle APD \sim \triangle D'PA' \cong \triangle C'XB'$ are similar. The triangle $\triangle C'XB'$ is the inverted triangle $\triangle CXB$ in the circle (P) and the triangle $\triangle EFG$ is the pedal triangle of the triangle $\triangle CXB$ with respect to the point P. This means that the triangles $\triangle C'XB' \sim \triangle EFG$ are also similar and consequently, the triangles $\triangle APD \sim \triangle D'PA' \cong \triangle C'XB' \sim \triangle EFG$ are all similar. In entirely the same way, we can show that the triangles $\triangle BPC \sim \triangle EHG$ are also similar.

If the triangles $\triangle APD, \triangle BPC$ are carried into the triangles $\triangle EFG, \triangle EHG$ with the common side EG by similarity transformations, then using the relation $\frac{BF}{BC} = \frac{AH}{AD}$ proved previously, this means that the points $H \in AD, F \in BC$ are both carried into the same point $L \in EG$. Since $FL \perp EG, HL \perp EG$, the points F, L, H are collinear, i.e., L is the diagonal intersection of the quadrilateral EFGH and $FH \perp EG$.   :cool:

[hide="Here are some interesting ideas."][color=darkred][b]Lemma 1.[/b] Let $ABCD$ be a quadrilateral $(AB=a\ ,$ $BC=b\ ,$ $CD=d\ ,$ $DA=d\ ,$ $AC=e\ ,$ $BD=f)\ .$ Then $A+C\in \{90^{\circ},270^{\circ}\}$ if and only if $e^2f^2=a^2c^2+b^2d^2\ .$[/color]

[color=darkblue][b]Lemma 2.[/b] Let $ABCD$ be a cyclic quadrilateral and let $I$ be a point of the its interior. Denote the projections $M$, $N$, $P$, $R$ of the point $I$ on the sides $[AB]$, $[BC]$, $[CD]$, $[DA]$ respectively. Then the quadrilateral $MNPR$ is orthodiagonally $(MP\perp NR)$ if and only if $\frac{IA^2+IC^2}{IB^2+ID^2}=\left(\frac ef\right)^2\ .$[/color][/hide]I will come back soon. Here time is 2.51 AM !


Since we have parallel lines, we know $ \angle ABC \equal{} 124^\circ$. $ \triangle ABC$ is isosceles, so that means $ \angle BAC \equal{} \angle BCA$, so they are $ \frac{180\minus{}124}{2} \equal{} 28$. Therefore, $ x \equal{} 124 \minus{} 28 \equal{} 96$.

Actually, $ x\equal{}180\minus{}124\minus{}28\equal{}28$, or alternatively we have $ \angle BCA \equal{} x$ because the lines are parallel.


MATHCOUNTS LaTeX ARML USAMTS search AMC AIME

I need really a lot practice.. So, here 's what great community memebers can do to me:
1. Post the question.....
2. Answer the question..... (with explanation)
I hope questions will in middle level becaus I'm attending MATHCOUNTS for state and I think those questions are pretty high level... So, here' s my question..

The sum of three numbers a, b, and c is 77. If a decreased by 5 is equal to b increased by 5 and also equal to c multiplied by 5, what is the value of b?

[hide]
There are several points that I want to add in my application.

1. In 8th grade, I was the one who received Jay Ernest Award for Science.
2. Under School Extracurricular Activities, Gifted club includes activities like Model UN, Scholastic Scrimmage, Stock Market Game, and several math competitions like AMC (American Mathematics Competition) and Bucknell University Gold Exam. I had to take an exam in junior high with psychologist and passed it to become part of gifted.
3. In addition for School Extracurricular Activities: Scholastic J 9-12 and Varsity J 10-12.
4. Under Community Activities, I brought up the idea of volunteering as an assistant coach for junior high math competition, Mathcounts, which I entered in my junior high years and became the only person from our school to qualify to states.  I had to receive permissions from both junior high and high school, and tried to go there almost once in a week to teach the kids in junior high gifted.
5. Art of Problem Solving is an internationally respected forum devoted for learning math, both for classroom and competition.  This forum is run by very talented mathematicians who have all qualified to national or international level in math competition.  I was asked personally by one of these respected administrators of the forum.  I have been a moderator more than four years, and in the process, I learned a mathematical language for computer called LaTeX to help more effectively.
6. I participated in numerous math competitions, including ARML (American Regional Math League - a national competition for highly exceptional high school students from each state), USAMTS (USA Mathematical Talent Search - A national competition that requires students to write full-written proofs for five problems on four rounds), Lehigh University Math Contest for High School Students, Bucknell Gold Exam (first place in my school all participating years; second place in the competition in sophomore year), AMC/AIME (American Mathematics Competition/American Invitational Mathematics Examination - national level math competition for high school students; AIME only picked certain high scorers from AMC).
7. With the work activities, I want to add that although I do not have any work experience, it was because I wanted to devote my time fully into academics and community services.  Academics has been my number one priority during high school, and after that is community service.  Because of these two factors, I did not have much time to find a job because I'm either volunteering in one place or working on my academics. [/hide]

[quote="Silverfalcon"]I need really a lot practice.. So, here 's what great community memebers can do to me:
1. Post the question.....
2. Answer the question..... (with explanation)
I hope questions will in middle level becaus I'm attending MATHCOUNTS for state and I think those questions are pretty high level... So, here' s my question..

The sum of three numbers a, b, and c is 77. If a decreased by 5 is equal to b increased by 5 and also equal to c multiplied by 5, what is the value of b?[/quote]

we'll write this in terms of b.
b+(b+10)+(b+5)/5=77.
2b+10+b/5+1=77
2b+b/5=66
10b/5+b/5=66
11b/5=66
11b=330.
b=30

heres my question:
What is the remainder when 5^1993 is divided by 7?

5^1 mod 7 = 5
5^2 mod 7 = 4
5^3 mod 7 = 6
5^4 mod 7 = 2
5^5 mod 7 = 3
5^6 mod 7 = 1
5^7 mod 7 = 5 repeats from here,  so 5^1993 mod 7= 5^(1993 mod 6) mod 7=5 mod 7 = 5

what is the greatest possible value of g in p(x)=x^8-6x^7+ax^6+bx^5+cx^4+dx^3+ex^2+fx+g
if all roots of the polynomial are positive

I hope I don't mean anything by this but I meant that I want the question to be in middle level, not like this 8th power polynomial questions.. I know some of polynomial questoins but this is more like to be from Math Olympiad.. So, here's my question from state competition in MATHCOUNTS...

The surface area of a cube is numerically equal to twice its volume. Find the length of a diagonal of the cube.

Scrambled's question was already fairly hard compared to the original. Demon's is way harder. He's not getting any help if you just keep on feeding him really hard questions that he doesn't want.

Anyway, to answer yours:

[hide]

6 sides of a cube.

6d^2 = 2d^3

d^3 - 3d^2 = 0

d^2(d-3) = 0

d=3, 0

0 doesn't work, so d=3. If you want the diagonal, that would be sqrt(3^2 + (3 :rt2: )^2) which is  :sqrt: 27 or 3 :rt3: .

[/hide]

Here's a geometry problem to mix it up:

You have a bunch of identical circles. How many circles can you place around a central circle so that all circles are tangent to one another without overlapping? Justify.

Thanks to you, rlee for understanding that I don't truly get the hard questoins.. But what's those signs you put? Are they the answer for my question? If you are saying root 3 or root 2 are answer, it is incorrect.. And what does your question mean? Can you post the question with drawing or more explanation?

Oh.. I didn't understand the picture..
Yes, it is 3  :sqrt: 3...
Then what does his question mean?

I don't get it...
I have algebra II but I don't get what you meant by that...
What's that formula for?

The formula is the sum of the roots of a polynomial to the n degree.

For the circle question you don't really need a picture.  The problem is just that you don't understand the question.

The question says that you have a circle.  Around that circle you draw identical circles that are tangent to it( tangent means that they touch it at one and only one point).  How many of these tangent circles can you draw around it so that none of them overlap each other?

Solution:[hide]

This question is a lot easier to explain with a picture but I don't have any software for that so I will try my best to do it in words.

I started with a circle of radius r ( you can pick a value for the radius but ultimately it doesn't matter because they all have the same radius).  Then I drew a circle tangent to this one and connected the centers with a line.  Tangent to both of these circles I drew another circle which means that now I have the original and two circles tangent to it.  Connecting the centers of all the circles I was left with a triangle.  This triangle was equilateral because each side was equal to 2r.  This means that the angles were all 60.  Since a circle has 360 degrees and every circle I draw tangent will make a 60 degree angle I can conclude that there will be 6 of these angles and thus 6 circles can be drawn around the first circle.[/hide]

Sorry. I accidently posted the same thing twice.  The first time I did it my computer froze while I was looking over it so I thought it didn't get posted but it did.

[quote="Demon"]Mine wasn't really a polynomial question,  if you know the formula sum and product of the roots,  you get sum=6 and product=g. the maximum value of the product is achieved when all the roots are equal 6/8=3/4 so all the roots are 3/4 the product in this case is just (3/4)^8[/quote]

It's an interesting problem, but are you sure that it responds to the request of the person who originally posted the thread, and that it really fits on the Getting Started board? I rather think that a problem solution stated in those terms is into the intermediate level.

BTW, joml88 has a great verbal explanation, written in spoiler  :)  to the question posed by rlee. That's great stuff to have in this forum.

(After edit:) I will be deleting posts liberally in this forum, particularly in this thread, so that the original poster can readily find the answers to the very on-topic and on-level questions he is posing.

[quote="Tare"][quote="Demon"]Mine wasn't really a polynomial question,  if you know the formula sum and product of the roots,  you get sum=6 and product=g. the maximum value of the product is achieved when all the roots are equal 6/8=3/4 so all the roots are 3/4 the product in this case is just (3/4)^8[/quote]
Arg not Viete's...I [b]still[/b] don't get how to use that thingy, can someone explain that in plain english?[/quote]

Lets say you have a polynomial such as x^3 - 34x^2 + 12x - 9. I just made that up so it probably doesn't have 3 real roots but forget that. Lets say its roots are a, b, and c. Then we can write it as (x-a)(x-b)(x-c). If you expand that, you get x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc.
So we must have a+b+c = 34, ab+bc+ca = 12, and abc=9.
And of course you can do that for any polynomial you want.

Marathon is not ended...
Since we spent a lot time in really hard question, let's go with little bit nerves-off questions....

There is this octahedron. When two faces met, the edge is colored orange. How many orange edges are there?

[quote="demji"][quote="beta"]
Let f be a function for which f(x/3) = x :^2:  + x + 1. Find the sum of all values of z for which f(3z) = 7.[/quote]
[/quote]

My answer seems to disagree with Silverfalcon's answer.

But, then again, maybe someone can tell me where I went wrong.

You have f(x/3) = x^2 + x + 1, and you want z for f(3z) = 7.

So let f(x/3) = g(y).

g(y) = (3y)^2 + 3y + 1.

So, we have f(3z), which by our ealier function, is (3(3z))^2 + 3(3z) + 1 = 81z^2 + 9z + 1 = 7.

So, 81z^2 + 9z - 6 = 0.

And 3(27z^2 + 3z - 2) = 0.

So, 3(9z-2)(3z+1) = 0.

So, the solutions are 2/9 and -1/3, which summed equals-1/9.

Discrepency... yikes.

[quote="white_horse_king88"]Question: How many positive integers less than 2002 are divisible by either 3, 4, and/or 5?[/quote]

I'm not sure if this is right, but...

[hide]Number divisible by 3 - 667
divisible by 4, but not 12 -  333
divisible by 5, but not 15 or 20 - 240

total - 1240[/hide]

If ${f(x) = \frac{x!}{f(x-1)}}$ and $f(1) = 1$, what is $f(5)$?

i have a nice question but don't know if it belongs here so i'll put it down.
you have 2 parallel chords in a circle.  I line drawn perpendicular to one of the chords intersects the other at a distance of 1.  The outer chord is 6 and the one closer to the center is 8.  What is the diameter of the circle.

Hint[hide] use the Pythagorean Theorem[/hide]
this problem might have been on one of the mock amc's

[quote="sam3.14"]How many zeroes are at the end of $\frac{200!}{5^3}$?[/quote]
[hide="Answer"]
$[200/125]=1$
$[200/25]=8$
$[200/5]=40$
$40+8+1=49$
$49-3=46$[/hide]

[quote="demji"][quote="Rep123Max"]For what value of n does (1^3+2^3+3^3+...n^3)/(1+2+3+...n)=36[/quote][/quote]

[hide="Solution"]$\frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)}{2}}=\frac{n(n+1)}{2}$

$n(n+1)=72$

$n=8 \text{ or } n=-9$[/hide]

What are the number of paths from (0,0) to (6,6) with x >= y at all points. Or if you'd like, y >= x at all points.

RC-7th, you might want to look at your reply with the Latex on the  second one.

It doesn't look so nice.

(it needs a space between \\ and [, or it considers \[ to be starting displaymath).

isn't anyone going to answer the circle problem
extra hint:  use 2 triangles.

[quote="RC-7th"][quote="sam3.14"]How many zeroes are at the end of $\frac{200!}{5^3}$?[/quote]
[hide="Answer"]
$[200/125]=1$
$[200/25]=8$
$[200/5]=40$
$40+8+1=49$
$49-3=46$[/hide]

[quote="demji"][quote="Rep123Max"]For what value of n does (1^3+2^3+3^3+...n^3)/(1+2+3+...n)=36[/quote][/quote]

[hide="Solution"]$\frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)}{2}}=\frac{n(n+1)}{2}$

There's only one solution.  Ignore the negative.
$n(n+1)=72$

$n=8 \text{ or } n=-9$[/hide]

What are the number of paths from (0,0) to (6,6) with x >= y at all points. Or if you'd like, y >= x at all points.[/quote]

ok i'll post the circle solution since no one is trying it.
[hide]
let r be the radius of the circle.  Let s be the distance from the midpoint of the chord of length 8 to the center of the circle.
We are then able to make 2 triangles with 1 leg equal to half each chord.
r 2 =4 2 +s 2
r 2 =3 2 +(s+1) 2
then we can set them equal
16+s 2 =9+s 2 +2s+1
2s=6
s=3
then we use the pythagorean theorem:
3 2 +4 2 =r 2
r 2 =25
r=5

but the question asks for the diameter.
d=2r
d=10
[/hide]

Just in case anyone forgot:

What are the number of paths from (0,0) to (6,6) with x geq y at all points. Or if you'd like, y geq x at all points.

When u say paths do u mean shortest paths as in no overlaps? If so

Solution
[hide]
Here is some basic theory. Suppose that we live in a town whose streets run in two perpendicular directions (as in this case), where all the streets of the town are represented in the form of horizontal and vertical lines. We can number the horizontal lines with number $0, 1, 2, 3, ...$ and do the same for the vertical lines. Then we can denote their interesctions by pairs of coordinates $(m,n)$, where $m$ is the number of the "vertical" street which passes through the intersection and $n$ is the number of the "horizontal street". Suppose that we have to go from a house located at the intersection $(0,0)$ (as in this case) to a house located at the intersection $(m,n)$. There will then be $\displaystyle{m+n\choose n}$  shortest paths joining the two houses, for each of these shortest paths is the $m+n$ blocks long--$m$ blocks in the horizontal direction and $n$ blocks in the vertical direction. Applying this knowledge to our problem at hand we see that the number of paths is

\[\displaystyle{6+6\choose 6}=\displaystyle{12 \choose 6}=924\]

[/hide]

I believe that this is the correct solution for any integer points $(x,y)$ where $0 \leq x,y \leq 6$.

Is that right? It seems a bit big but I think I did it right unless I did something really stupid ;)


[quote="aadil"]what do you mean by the first inequality?[/quote]

the first inequality is equivalent to the second inequality.

in general $ (x_{1},x_{2},...,x_{n})\equal{}\sum_{cyc} a_{1}^{x_{1}} a_{2}^{x_{2}}... a_{n}^{x_{n}}$. :D

[quote="mestav"]yes but not cyclic it's symmetric[/quote]

oh yeah, sorry :oops:

original inequality is $ a^{3}(b \plus{} c \plus{} d) \plus{} b^{3}(a \plus{} c \plus{} d) \plus{} c^3(a \plus{} b \plus{} d) \plus{} d^3(a \plus{} b \plus{} c)\le\frac {3}{4}(a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2})^2$

I multiplied and I have this inequality

is original inequality false? or my operations false?

It's equivalent to $ (a \minus{} b)^4 \plus{} (a \minus{} c)^4 \plus{} (a \minus{} d)^4 \plus{} (b \minus{} c)^4 \plus{} (b \minus{} d)^4 \plus{} (c \minus{} d)^4 \ge 0$, but I think the right side of the inequality should be $ 4(a^3b \plus{} b^3a \plus{} a^3c \plus{} c^3a \plus{} a^3d \plus{} d^3a \plus{} b^3c \plus{} c^3b \plus{} b^3d \plus{} d^3b \plus{} c^3d \plus{} d^3c)$ instead of what you have.

I am sorry I fixed it and thank you dgreenb801

The convention in this forum is to use square brackets for symmetrical sums, like $ [3,1,0,0]$. Note that this sum has always $ n!$ terms, e.g. $ [4,0,0,0] \equal{} 6\sum a^4$, as the symmetrical sum counts [i]all [/i]the permutations.

If one writes $ \sum_{sym}$, it is understood to count each different term only once, e.g. if we deal with 4 variables, $ \sum_{sym}a^3b$ has 12 terms, not 24. And $ [3,1,0,0]\equal{}2\sum_{sym}a^3b$.

This sometimes leads to confusion, let's just be coherent in each line. :)