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number theory unsolved
number theory

Find the least possible value of $ a \plus{} b$ where $ a,b$   are positive integers such that $ 11$ divides $ a \plus{} 13b$ and $ 13$ divides $ a \plus{} 11b$.


The figure shown is called a [i]trefoil[/i] and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length $ 2$?

[asy]pair A = (0,1);
pair F = (1,1);
pair C = (2,1);
pair D = ((1/2),(sqrt(3)/2)+1);
pair E = (D.x + 1, D.y);
pair B = (1,sqrt(3) + 1);
pair G = midpoint(Arc(F,1,180,120));
pair H = midpoint(Arc(E,1,180,120));
pair I = midpoint(Arc(D,1,0,60));
pair J =midpoint(Arc(F,1,0,60));

draw(A--B--C--cycle,linetype("8 8")+linewidth(0.6));
draw(D--E--F--cycle,linetype("8 8")+linewidth(0.6));

draw(Arc(F,1,180,120),linewidth(0.75));
draw(Arc(E,1,180,120),linewidth(0.75));
draw(Arc(D,1,0,60),linewidth(0.75));
draw(Arc(F,1,0,60),linewidth(0.75));

draw(D--I, Arrow);
draw(E--H, Arrow);
draw(F--G, Arrow);
draw(F--J, Arrow);

draw((F.x - 0.05, F.y - 0.1)--(A.x,A.y - 0.1), Arrow);
draw((F.x + 0.05, F.y - 0.1)--(C.x,C.y - 0.1), Arrow);

draw((A.x,A.y - 0.175)--(A.x,A.y - 0.025),linewidth(0.75));
draw((C.x,C.y - 0.175)--(C.x,C.y - 0.025),linewidth(0.75));

label("$2$",F,S);[/asy]

$ \textbf{(A)}\ \frac{1}{3}\pi \plus{} \frac{\sqrt{3}}{2}\qquad
\textbf{(B)}\ \frac{2}{3}\pi\qquad
\textbf{(C)}\ \frac{2}{3}\pi \plus{} \frac{\sqrt{3}}{4}\qquad
\textbf{(D)}\ \frac{2}{3}\pi \plus{} \frac{\sqrt{3}}{3}\qquad
\textbf{(E)}\ \frac{2}{3}\pi \plus{} \frac{\sqrt{3}}{2}$

[hide]eventually the trefoil becomes 4 sectors with center angle 60 degrees.

since the radius of one sector is 1, the area is $ 1*1* \pi *\frac {240}{360} \equal{} \frac {2}{3}\pi$

Answer: B[/hide]

[hide="Click for solution"]
We can arrange the trefoil so it is four equilateral triangles and four circular segments, which can be combined to form four $ 60^\circ$ sectors.  Since the radius of the circle is $ 1$, the area of the trefoil is $ \frac{4}{6}(\pi \times 1^2)\equal{}\frac{2}{3}\pi$, or $ \boxed{\textbf{(B)}}$.
[/hide]


Let $ a,b,c>0$ and $ abc\equal{}1$. Prove that

$ 4(a^2\plus{}b^2\plus{}c^2)\plus{}(a\plus{}b\plus{}c)\geq4(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})\minus{}(ab\plus{}bc\plus{}ca)\plus{}6$

_________________
 :blush:  Panagiote Ligouras

[quote="caubetoanhoc94"][quote="Ligouras"]Let $ a,b,c > 0$ and $ abc \equal{} 1$. Prove that

$ 4(a^2 \plus{} b^2 \plus{} c^2) \plus{} (a \plus{} b \plus{} c)\geq4(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}) \minus{} (ab \plus{} bc \plus{} ca) \plus{} 6$

_________________
 :blush:  Panagiote Ligouras[/quote]

We have:
$ 4(a^2 \plus{} b^2 \plus{} c^2) \ge 4(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})$
$ a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca \ge 6$[/quote]

Hi!
it is easy............
my demonstration is different.
Thank you [b]caubetoanhoc94[/b]


Euler
geometry
incenter
circumcircle
cyclic quadrilateral

Dear Mathlinkers,
let ABC a triangle, (1) the Euler's circle of ABC, A'B'C' the medial triangle of ABC, A"B"C" the orthic triangle of ABC, A+, B+, C+ the meetpoint of B'C' and B"C", of A'C' and A"C", of A'B' and A"B", S" the meetpoint of  B"B+ and C"C+.

It is known that 	1.  S" is the second Schröder's point of ABC
2.  S" lies on (1)
3.  S", A+ and A" are collinear.

Now, let A*B*C* the tangential triangle of A'B'C' and M the midpoint of B*C*.
Prouve synthetically:   M is on A+A".

Sincerely
Jean-Louis

First I will change the names of points to available for my solution.
[b]Lemma 1:[/b] If $ SA_1 \equal{} Max\{SA_1, SB_1,SC_1\}$ then $ SA_1 \equal{} SB_1 \plus{} SC_1$
For the proof you can see theorem 2.5 -"[b]Generalization of the Feuerbach point[/b]" on Darij grinberg's site:
http://www.cip.ifi.lmu.de/~grinberg/
[b]Lemma 2:[/b] Let $ X$ be the [b]Feuerbach[/b] point of triangle $ ABC$ and $ A',B',C'$ are the midpoints of $ BC, CA, AB. A'X \equal{} Max\{A'X,B'Y,C'Z\}$ then $ A'X \equal{} B'Y \plus{} C'Z$
This lemma appeared some times in Mathlinks.
[b]Lemma 3:[/b] Given triangle $ ABC. AA_1, BB_1, CC_1$ are three bisectors. $ A_2, B_2,C_2$ be the midpoints of triangle $ ABC. (I)$ be the incircle. $ A_1A_3$ is the tangency of $ (I) (A_3\not\in BC)$. Similar for $ B_1B_3, C_1C_3$. Then $ A_2A_3, B_2B_3, C_2C_3$ concur at the [b]Feuerbach[/b] point of triangle ABC.
For the proof: See my post at http://www.mathlinks.ro/viewtopic.php?t=283779
[b]Return to problem:[/b]
We have $ A'I\perp B_3C_3$ and $ B_3C_3//BC$ then $ A'I\perp BC$. But $ I$ is the [b]Euler circle's [/b]center of triangle $ ABC$ thus $ A'I$ bisect $ A_1A_3$. Denote $ D$ the intersection of $ A'I$ and $ B'C'$ we obtain $ DA_1$ is the tangency of $ (I)$.
Similar for $ B_1,C_1$. Applying [b]Lemma 3[/b] we have $ MA_1, NB_1, PC_1$ concur at the [b]Feuerbach[/b] point $ S'$ of triangle $ A'B'C'$.
Applying [b]lemma 2[/b] we get $ MS' \equal{} NS' \plus{} PS' (1)$
Two triangles $ MNP$ and $ A_1B_1C_1$ have $ A_1B_1//MN, B_1C_1//NP, A_1C_1//MP (2)$
From $ (1)(2) \Rightarrow A_1S' \equal{} B_1S' \plus{} C_1S'$
On the other side, applying [b]lemma 1[/b] we claim $ A_1S \equal{} B_1S \plus{} C_1S$
We will show that there is one location of $ S$ satisfy $ A_1S' \equal{} B_1S' \plus{} C_1S'$
Put the lengthsides of $ B_1C_1,C_1A_1, A_1B_1$ are $ a,b,c.$
Applying [b]Ptolemy'[/b]s theorem for cyclic quadrilateral $ C_1S'B_1A_1$ we have:
$ A_1S'.B_1C_1 \equal{} B_1S'.A_1C_1 \plus{} C_1S'.A_1B_1$
$ \Leftrightarrow A_1S'.a \equal{} C_1S'.c \plus{} B_1S'.b\Leftrightarrow C_1S'.a \plus{} B_1S'.a \equal{} C_1S'.c \plus{} B_1S'.b$
$ \Leftrightarrow B_1S'(a \minus{} b) \equal{} C_1S'(c \minus{} a)\Leftrightarrow \frac {B_1S'}{C_1S'} \equal{} \frac {c \minus{} a}{a \minus{} b}$ is constant.
So $ (3)$ is true. Which follows that $ S'\equiv S.$
We end the proof. :wink:
[hide="note"] By [b]Pascal[/b]'s theorem, I found an interesting result: $ B_2,A,A',C_2$ are collinear. Similar for $ B',B,A_2,C_2$ and $ B_2,A_2,C',C$[/hide]

Dear Livetolove and Mathlinkers,
in my mail, you have to read "Schroeter".
I have written an article "Les deux points de Schroeter" that you can see on
http://perso.orange;fr/jl.ayme/.
To day, I have extend this article in order to prove synthetically my proposition without calculus. See p. 14-20.
Livetolove, you will see your figure that I discover now.
Sincerely
Jean-Louis

Dear Livetolove and Mathlinkers,
Thank you Livetolove.
I made a typo in my email :

http://perso.orange.fr/jl.ayme    vol. 2   Les deux points de Schroeter

Sincerely
Jean-Louis

[quote="jayme"]Dear Mathlinkers,
now the result is

the second Schroeter's point of a triangle is the Feuerbach's point of the tangential triangle of the median triangle of the initial triangle.

Sincerely
Jean-Louis[/quote]

By Fontene theorem we get the second Schroeter's point of a triangle is the Anti-steiner point of the Euler line WRT its median triangle(i.e. the orthopole of the Euler line WRT the reference triangle), but it's well-known the Feuerbach point of triangle is the Anti-steiner point of the Euler line of the intouch triangle (through the incenter and the circumcenter) WRT its intouch triangle, so the second Schroeter's point ($ X_{125} $ in ETC) is the Feuerbach point of the tangential triangle of the median triangle :).

[quote="TelvCohl"][quote="jayme"]Dear Mathlinkers,
now the result is

the second Schroeter's point of a triangle is the Feuerbach's point of the tangential triangle of the median triangle of the initial triangle.

Sincerely
Jean-Louis[/quote]

By [b]Fontene[/b] theorem we get the [b]second Schroeter's point[/b] of a triangle is the Anti-steiner point of the Euler line with respect to its median triangle(ie. the orthopole of the Euler line with respect to the reference triangle) , but it's well known that the [b]Feuerbach point[/b] of triangle is the Anti-steiner point of the Euler line of contact triangle (pass through the [b]incenter[/b] and the [b]circumcenter[/b]) with respect to its contact triangle, so the [b]second Schroeter's point[/b] ( $ X(125) $ in ETC ) is the [b]Feuerbach point[/b] of the tangential triangle of the median triangle . :)[/quote]

By the way, use similar idea we can prove $ X_{125} $ of $ \triangle ABC $ is $ X_{100} $ of its orthic $ \triangle $ . :)


linear algebra
matrix
number theory
least common multiple
linear algebra unsolved

I'm sorry, I have meant $M_n$ to be $P_n$-permutation matrix $n \times n$.

[quote="eugene"]Let $P_n$ be a permutation matrix i.e. $P_{ij}=\delta_{i\sigma(i)}$, where $\sigma$ is permutation from $S_n$. I'm interested in whether we can find the number of solutions of the equation $(P_n)^k=I_n$ in terms of $k,n$.
For example it is not difficult to find it when $k$ is prime, even more if $k$ is prime and $k>n$ then we haven't solutions(unless i'm wrong with my solution).[/quote]This is really a question about the number of permutations $\sigma \in S_n$ whose order is $k$. By decomposition into cycles, the possible values for $k$ are all numbers of the form $lcm(n_1,n_2,\dots,n_r)$ where $n_1 + \dots + n_r = n$ (and their multiples). $lcm$ = least common multiple.

thanks for the reply, anyway i was more interested in the number of solutions=the number of permutations of order $k$.


what must our score be after the first three rounds to be called to take aime?

I believe that it's been 68 for the past few years.

well the thing is that i dont know if my school is gonna do amc 10 and 12 so the only chance i will have to make aime may be usamts :( otherwise i no its harder i can do amc half asleep :P but i dont think i even got 20 on round one  :what?:

[quote="azsxder"]I wouldn't worry too much about qualifying for the AIME through USAMTS, though. If you can even get close to a 68 on the USAMTS, you're probably closer to qualifying for the USAMO than not making the AIME (besides, you already have 2 chances in the A and B tests).

Also, if you only qualify through the USAMTS, 100 will be used as your AMC score for your USAMO index, making it much harder to qualify for the USAMO than those who did it through the AMC's.[/quote]
I think it's actually easier to qualify for AIME using AMCs.

I thought that he implied the same thing. Anyway, it's different for different people. Maybe some people do horrible under time pressure, and do better with the amount of time and access to resources that USAMTS offers. The question stated by the OP has already been answered though, so unless somebody wants to add something, we should probably stop giving opinions as to which is easier to qualify with.


number theory
relatively prime
probability and stats

A discrete random variable X takes values $ \frac{r}{s}$ , where r and s are mutually prime positive integers.

If Pr[X=$ \frac{r}{s}$] = $ \frac{k}{e^{r+s} -1}$
Find:
1.k
2.E(X)
3.Pr(X$ \geq$ 2)

\[ k = \left(\sum _{r=1} ^\infty {\sum _{(s,r)=1,s\geq 1}{\frac{1}{e^{r+s}-1}\right)^{-1}}}\]
\[ \mathbb {E}(X)=k\sum_{r=1}^\infty{\sum_{(s,r)=1,s\geq 1}\frac{r}{s\left( e^{r+s}-1\right)}}\]
\[ P\left(X\geq 2\right)=k\sum_{r=1}^\infty{\sum_{(s,r)=1,1\leq s \leq \frac{r}{2}}\frac{1}{ e^{r+s}-1}}\]

But I suspect that's not what you were going for  :D.  I expect you probably use $ \frac{1}{e^{r+s}-1}=\sum_{j=1}^\infty {e^{-j(r+s)}}$  to simplify everything, but the summation over just relatively prime integers is throwing things off.  Care to offer a hint?

If $ (r,s)$ are all pairs of relatively prime integers, then $ (jr,js)$ are all pairs of integers ;)


limit
integration
function
calculus
real analysis
real analysis unsolved

Show  $ \displaystyle\lim_{n\to\infty}	\displaystyle\int^n_0 (1 \plus{} 	\dfrac{x}{n})^n e^{\minus{}2x}\,dx\equal{}1$

One hint: let $ \mathbf{1}_{[0,n]}$ be the function that is $ 1$ on the interval $ [0,1]$ and zero elsewhere (the characteristic or indicator function of that interval. Write the limit as

$ \lim_{n\to\infty}\int_0^{\infty}\left(1 \plus{} \frac {x}n\right)^n e^{ \minus{} 2x}\mathbf{1}_{[0,n]}(x)\,dx.$

The point of doing that is so that we can see this as the limit of the integrals of a sequence of functions on the same set.

Clearly, we'd like to take the limit inside the integral - to say that the limit of the integrals is the integral of the limit. So, let's ask two questions:

(1) Is the sequence of (nonnegative) functions monotone increasing in $ n$ so that we can invoke the Monotone Convergence Theorem?

(2) Can we find a useful dominating function that is integrable on $ [0,\infty)$ so that we can invoke the Dominated Convergence Theorem?

(By the way, the answers to the questions are "yes" and "yes".  Your choice for which one you want to use.)


I have one more for you guys :)

Let $\{x_n\}_{n=1}^\infty$ be a sequence in the Banachspace $X$ such that $\sum_{n=1}^{\infty}\|x_n\|$ converges on $\mathbb{R}$.
Let $y_n = x_1 + x_2 + \dots + x_n$. Show that $\{y_n\}_{n=1}^\infty$ is convergent.

=======================

Since I know that all Cauchy sequences converges in X it seems to be a good idea to show that $\{y_n\}$ is Cauchy. This means that

$\|y_n - y_m\| < \epsilon$ when $n,m > N \in \mathbb{N}$

Is it ok to assume that n > m?

$\|y_n - y_m\| = \|x_n + x_{n-1} + \dots + x_{n-m+1}\| < \|x_n\| + \dots + \|x_{n-m+1}\|$

This should be equal to $\|y_n - y_m\| < \sum_{n=1}^{n}\|x_n\| - \sum_{n=1}^{m}\|x_n\|$

That $\sum_{n=1}^{\infty}\|x_n\|$ converges means that $\|\sum_{n=1} \|x_n\| - x \| < \epsilon$ when $n > N$, right?

What is wrong, how do I finish this?

Nothing is wrong.

We know that $\sum_{n=1}^{\infty}\|x_n\|$ is convergent. Apply Cauchy's criterion, for $n>m>N$, we have
$\sum_{n=1}^{n}\|x_n\| - \sum_{n=1}^{m}\|x_n\| < \epsilon$


geometry
trigonometry
integration
analytic geometry
calculus
calculus computations

[b]1.[/b] Find the area of the finite region enclosed by one loop of the curve $ r^2 \equal{} \minus{} \sin 2\theta$. [hide]

I know that the area enclosed by the polar curve $ r \equal{} f(\theta)$ and the rays $ \theta \equal{} \alpha$ and $ \theta \equal{} \beta$ is given by $ \frac {1}{2}\int_{\alpha}^{\beta} r^2 d\theta$. However, the curve given here is of the form $ r^2 \equal{} f(\theta)$. [/hide]

[b]2.[/b] Find the area of the surface generated by revolving $ r^2 \equal{} \sin 2\theta$ about the $ y$-axis in the form $ k \int_{0}^{a} \cos \theta d\theta$. [hide] Stuck on this too just like problem 1 [/hide]

For #1: and why is $ r^2=f(\theta)$ a problem when you want $ r^2?$

But of course we need $ r^2\ge 0$ so we need $ -\sin2\theta\ge 0.$ This happens for $ \pi\le 2\theta\le2\pi$ and $ 3\pi\le2\theta\le4\pi,$ which we can write as $ \frac{\pi}2\le\theta\le\pi$ for one loop. (There's another loop, of course.)

Hence the area is $ \int_{\frac{\pi}2}^{\pi}-\frac12\sin2\theta\,d\theta$

You can figure that yourself.

As for #2:

Again, there are two lobes of this, one for $ 0\le\theta\le\frac{\pi}2.$ I'll compute the area for one lobe; if you want both, just double the result.

For a surface of revolution about the $ y$ axis, the surface area element is $ dA=2\pi x\,ds,$ where $ ds$ is the arclength element. For a curve given in polar coordinates, $ ds=\sqrt{\left(\frac{dr}{d\theta}\right)^2+r^2}\,d\theta.$

We have $ r=\sqrt{\sin2\theta}.$ That means $ \frac{dr}{d\theta}=\frac{\cos2\theta}{\sqrt{\sin2\theta}}.$

So ${ \left(\frac{dr}{d\theta}\right)^2+r^2=\frac{\cos^22\theta}{\sin2\theta}+\sin2\theta}=\frac{\cos^22\theta+\sin^22\theta}{\sin2\theta}=\frac1{\sin2\theta}$

So $ ds=\frac1{\sqrt{\sin2\theta}}\,d\theta.$

We multiply that by $ 2\pi x=2\pi r\cos\theta=2\pi\sqrt{\sin2\theta}\cos\theta.$

So $ dA=2\pi\sqrt{\sin2\theta}\cos\theta\frac1{\sqrt{\sin2\theta}}\,d\theta=2\pi\cos\theta\,d\theta.$

$ A=\int_0^{\frac{\pi}2}2\pi\cos\theta\,d\theta=2\pi.$ Double that for both lobes.

Now: I'm giving a test on precisely this material tomorrow and I haven't written it yet. What questions am I going to use? We'll see ...

Can you recommend any good books or websites which have more interesting problems on this material?


function
inequalities
algebra unsolved
algebra

show that thee is no   function f definedonthe positive real numbes such that :
                 $f(y) > (y-x)f(x)^2$

I'm will be very interested to see the proof to this.

LOL, a function that grows so fast, so fast that it cannot exist!

There must be some $x$ such that $f(x)>2$. Then $f(x+1)>f(x)^2>4$, $f(x+1+1/2)>f(x+1)^2/2>8$, $f(x+1+1/2+1/4)>f(x+1+1/2)^2/4>16$, ... it turns out that $f(x+3)>f(y)^2$ for every $y<x-2$ will be too big to be a real number.

I love this problem.

I realized that  the function's results would be infinite, but I didn't know how to make a mathematically rigorous proof for it.  I liked you use of an inifinite geometric sequence.

But don't you need to prove that there is some value of [i]x[/i] for which \[ f(x)=2  \]

In fact, one of the results I got when working with this problem, is that \[ 0<f(x)<1  \] for every [i]x[/i].

[quote="draperp"]

In fact, one of the results I got when working with this problem, is that \[ 0<f(x)<1  \] for every [i]x[/i].[/quote]

Impossible:  $f(x) > (x-x) f(x) = 0$ for all $x$, and $f(x+k) > k f(x)^2$ for all $x$, so $f(x)$ is bounded below by a linear function and therefore cannot be bounded above by any constant.

(In fact, since there must be some $f(a) > 1$, we get a recursive formula that tells us $f(a + k) > f(a)^{2^{k}}$ for all integers $k$. )

Yesterday I found a slightly shorter proof based on the above idea that the function cannot exist.

$f(x+1) > f(x)^2$
$f(x) > (-1) f(x+1)^2$
$f(x+1) > f(x+1)^4$ implies $f(x) < 1$ for all $x$, which is impossible by the above argument.

[quote="leepakhin"]Is my proof correct?
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=67226]http://www.mathlinks.ro/Forum/viewtopic.php?t=67226[/url][/quote]

Yes, and I like it very much.

(It's too bad for me that is very similar to a proof that I was just about to post.)

And, t0rajir0u, although my inequality is false, that is what I came up with by letting x=0 and y>0 near the end of leepakhin's proof.  Any number of statements could be reasoned since no function [i]f[/i] exists in the first place.


algebra
polynomial
function
calculus
calculus computations

[size=150]Prove that the polynomial f(x)=X^3 - 3X + m never has two roots in the interval [0,1], no matter what m may be (this is a consequence of Rolle's Theorem[/size])

The Cauchy-Bolzano's theorem states that:
if $f: [a,b] \to R$,  is a continous function, and $f(a)*f(b)<0$ then there is $at \;least$ one point c from $(a,b)$ for which $f(c)=0$. Our function is a polynomial one, it is continous, and $f(0)*f(1)=m^2$. There isn't any real number m for which $m^2<0$, and for this reason you can't have a solution in the interval $(a,b)$. Of course for certain m you can find solutions like 1 or 0 (for m=2 you can obtain the solution x=1), but in the interval $(a,b)$ you can not find another one (a distinct one, different form 1). So, no matter what m is, f(x) never has $two$ distinct roots in the interval [0,1].

Using Rolle:
-for m from $(-\infty,-2)$ in the interval $[-1,1]$ we don't have real roots
-for m=-2 in the interval $[-1,1]$ we have the double root -1
-for m from $(-2,2)$ in the interval $[-1,1]$ we have only a real root
-for m=2 in the interval $[-1,1]$ we have the double root 1
-for m from $(2,\infty)$ in the interval $[-1,1]$ we don't have real roots

If $f$ has two roots $\alpha<\beta$ in $[0,1]$, then by Rolle's theorem there is $\gamma\in (\alpha,\beta)$ such that $f'(\gamma)=0$. On the other hand $f'(x)=3x^2-3=0$ if and only if $x=\pm 1$. So it is impossible to have such $\gamma$.


geometry
geometric transformation
reflection
circumcircle
angle bisector
perpendicular bisector
cyclic quadrilateral

Let ABCDE be a pentagon with $ AE \equal{} ED$, $ AB \plus{} CD \equal{} BC$, and $ \angle BAE \plus{} \angle CDE \equal{} 180^{\circ}$. Prove that $ \angle AED \equal{} 2\angle BEC.$

Let $ S$ be the intersection of $ AB$ and $ CD$, because $ \angle BAE\plus{}\angle CDE\equal{}180^\circ$, so we have the quadrilateral $ SAED$ is cyclic. On the other hand, $ AE\equal{}ED$, which implies that $ E$ lies on the angle bisector of the angle $ ASD$ $ (1)$.

Now, let $ E'$ be the center of the excircle of triangle $ SBC$ and denote by $ A'$, $ B'$ the tangency points of $ (E')$ with $ SB$, $ SC$, respectively. Hence, we conclude that $ BA'\plus{}CD'\equal{}BC\equal{}AB\plus{}CD$. With out loss of generality, assume that $ A'$ lies on the segment $ AB$ and $ D$ lies on the segment $ CD'$. Then we can easily to see that $ \triangle E'AA'\equal{}\triangle E'DD'$, so $ E'A\equal{}E'D$, i.e. $ E'$ lies on the perpendicular bisector of the segment $ AD$ $ (2)$.

From $ (1)$ and $ (2)$, we conclude that $ E$ and $ E'$ lie on the angle bisector of the angle $ BSC$ and also lie on the perpendicular bisector of the segment $ AD$, which implies that $ A\equiv A'$, $ D\equiv D'$ and $ E\equiv E'$. And the result follows.

$ \bullet$ I'm not sure about this solution, so please inform me if it was incorrect. Thank you very much.

April.

[quote="April"]

From $ (1)$ and $ (2)$, we conclude that $ E$ and $ E'$ lie on the angle bisector of the angle $ BSC$ and also lie on the perpendicular bisector of the segment $ AD$, which implies that $ A\equiv A'$, $ D\equiv D'$ and $ E\equiv E'$. And the result follows.

[/quote]

I think that your solution is correct. But the problem is what if the triangle SAD is isoscale ? I don't think you can apply the property in such case.

I have an alternative solution.

Let F be a point on ray BA such that $ AF \equal{} CD$. It leads to $ BC\equal{}BF (1)$
(because $ BC\equal{}AB\plus{}CD \equal{} BA\plus{}AF\equal{}BF$)
Since $ \angle{BAE} \plus{} \angle{CDE} \equal{} 180$, $ \angle{FAE} \equal{} \angle{CDE}$
From here, we can easily prove that $ \Delta AFE\equal{} \Delta DCE$
And deduce $ EF \equal{} EC (2)$ and $ \angle{AEF}\equal{}\angle{DEC} \rightarrow \angle{FEC}\equal{}\angle{AED}$
From $ (1)$ and $ (2)$ F is the reflection of C about BE.
$ \rightarrow \angle{FEC} \equal{} 2 \angle{BEC} \rightarrow \angle{AED}\equal{}2\angle{BEC}$
(Q.E.D)

If the triangle SAD is isoscale , we have $ \angle EAS = \angle EDF = 90^{\circ}$.
So, draw $ EG%Error. "botBC" is a bad command.
$ intersect $ BC$ at $ G$
$ \triangle EAB \cong \triangle EGB$ and $ \triangle EGC \cong \triangle ECD$
And the result follows

We denote as $ S,$ the intersection point of $ AB,$ $ CD$ and let $ A',$ $ D'$ be, the orthogonal projections of $ E,$ on the sidelines $ AB,$ $ CD,$ respectively.

So, because of $ \angle BAE \plus{} \angle CDE \equal{} 180^{o},$ we have that the quadrilateral $ SAED,$ is cyclic.

It is easy to show that $ AA' \equal{} DD'$ $ ,(1)$ from the congruent right triangles $ \bigtriangleup A'AE,$ $ \bigtriangleup D'DE$ $ ($ because of $ AD \equal{} AE$ and $ \angle A'AE \equal{} \angle D'DE$ $ ).$

From $ (1),$ and from $ AB \plus{} CD \equal{} BC,$ we have $ A'A \plus{} AB \plus{} CD' \equal{} BC$ $ \Longrightarrow$ $ A'B \plus{} CD' \equal{} BC$ $ ,(2)$

From $ (2),$ we conclude that the point $ E,$ is the $ S \minus{}$excenter of the triangle $ \bigtriangleup SBC$ $ ($ because of only the $ S \minus{}$excenter has the property of $ (2)$ $ ).$

So, it is easy now to show that $ \angle A'ED' \equal{} 2\angle BEC$ $ ,(3)$

But $ \angle AED \equal{} \angle A'ED'$ $ ,(4)$ $ ($ because of $ \angle AEA' \equal{} \angle DED',$ from the congruent right triangles $ \bigtriangleup A'AE,$ $ \bigtriangleup D'DE$ $ ).$

From $ (3),$ $ (4)$ $ \Longrightarrow$ $ \angle AED \equal{} 2\angle BEC$ and the proof is completed.

Kostas Vittas.

PS. In the particular case of $ \angle A \equal{} 90^{o} \equal{} \angle D,$ the cyclic quadrilateral $ SAED$ is orthodiagonal with $ SE,$ as the diameter of its circumcircle $ (O)$ and then, we conclude that $ A'\equiv A$ and $ D'\equiv D,$ as mentioned by [b][size=100]April[/size][/b].


probability
conditional probability
expected value
real analysis
real analysis unsolved

John tosses consecutively a coin. If it appears a head he wins a dollar, if tail he loses a dollar. Game is over when John loses all his money or earns [b]y[/b] dolars (y has previously been defined). At the beginning John has [b]x[/b] dollars. What's the probability John will lose.
pls help!

Let $u(t)$ be John's net winnings at time $t.$ Define it so that $u$ is a integer in the interval $[-x,y].$ Now let $p(u)$ be the conditional probability that John wins given that his current net winnings are $u.$ The boundary condititions are that $p(-x)=0$ and $p(y)=1.$

Consider one play of the game, and use the theorem that the expected value of a random variable is the expected value of its conditional expectations. This should give you $p(u)$ in terms of $p(u-1)$ and $p(u+1).$ You can solve this second-order constant-coefficient linear difference equation to determine the complete formula for $p(u).$ Once you have done that, $p(0)$ is the answer you were looking for.

It should also be possible to use the same conditional expectation idea to determine how long, on average, it will take for ths game to be played to its conclusion.

The problem is well-known, and goes by the name "gambler's ruin." Once you have completely solved this version, try altering the problem to make the game unfair. (Think of it is having a "loaded" coin.)


topology
inequalities
real analysis
real analysis unsolved

Let $(X,\rho)$ be a metric space. For A a nonempty subset of X, and $x\in X$, the distance $\rho(x,A)$ from x to A is defined by $\rho(x,A)=\inf\{\rho(x,a):a\in A\}$. I need to show that $|\rho(x_1,A)-\rho(x_2,A)|\leq \rho(x_1,x_2)$. Can anyone suggest a proof?

Cheers,
Joe

Of course, that should be a capital A to make the expression symmetric. [Fixed by moderator.]

Choose $\epsilon>0$, and let $a\in A$ be such that $\rho(x_1,a)<\rho(x_1,A)+\epsilon$. We have $\rho(x_2,A)\le\rho(x_2,a)\le\rho(x_1,x_2)+\rho(x_1,a)<\rho(x_1,x_2)+\rho(x_1,A)+\epsilon$.
Since this is true for all $\epsilon$, $\rho(x_2,A)\le\rho(x_1,A)+\rho(x_1,x_2)$. The other half of the inequality is true by exchanging labels.


modular arithmetic
number theory unsolved
number theory

The sequence $ (a_n)_{n\equal{}0}^{\infty}$ is defined by $ a_0\equal{}0,a_1\equal{}1$, and $ a_{n\plus{}2}\equal{}2a_{n\plus{}1}\minus{}5a_n$.

Determine whether there exists a natural number $ k$ such that $ a_{3k}\equal{}\minus{}3$.

There are the nice congruences $ a_n\equiv 2^{n\minus{}1}\pmod{5},a_n\equiv n\pmod{4}$ for $ n\geq 1$. (I found out that mod 4 would be useful by "solving" for $ a_n$ and expanding using binomial theorem.)

Now suppose $ a_n\equal{}\minus{}3$. Then $ n\equiv \minus{}3\equiv 1\pmod{4}$. But this gives $ a_n\equiv 1\not\equiv \minus{}3\pmod{5}$. So the sequence never attains the value -3.


geometry
circumcircle
trigonometry
projective geometry
geometry proposed

Let $ ABC$ ba a triangle circumcircle $ (O)$. $ P$ is an arbitrary point. $ A',B',C'$ are the second intersections of $ PA,PB,PC$ with circumcircle $ (O)$ of triangle $ ABC$. $ Q$ is a point on line $ OP$. $ A''B''C''$ is pedal triangle of $ P$. Prove that $ (A'QA''),(B'QB''),(C'QC'')$ have two common points (they are coaxial circles).

Note that, this is generalization of the problem on the post [url=http://www.mathlinks.ro/viewtopic.php?t=316260]Concurrent 4[/url].

[b]Problem:[/b] Let $ABC$ be a triangle with its circumcircle $(O)$. Let $P$ and $Q$ be arbitrary points such that $P, O, Q$ are collinear. $A_1B_1C_1$ is the pedal triangle of $P$ wrt $\Delta ABC, A_2B_2C_2$ is the circumcevian triangle of $Q$ wrt $\Delta ABC$. Show that $(PA_1A_2), (PB_1B_2), (PC_1C_2)$ are coaxal.

[b]Proof:[/b]
Denote $O_a, O_b, O_c$ the centers of $(PA_1A_2), (PB_1B_2), (PC_1C_2)$; $X$ the intersection of the line through $A_2$ and perpendicular to $PA_2$. Similarly we define $Y, Z$.
Note that $O_a$ is the midpoint of $PX$ therefore $(O_a), (O_b), (O_c)$ are coaxal iff $O_a, O_b, O_c$ are collinear iff $X, Y, Z$ are collinear. $(1)$
We have $\frac{XB}{XC}=\frac{S_{XBA_2}}{S_{XCA_2}}=\frac{\sin \angle XA_2B. A_2B}{\sin \angle XA_2C. A_2C}=\frac{\sin \angle XA_2B}{\sin \angle XA_2C}.\frac{\sin \angle BAQ}{\sin \angle CAQ}$
But $\frac{\sin \angle XA_2B}{\sin \angle XA_2C}=\frac{\cos \angle BA_2P}{\cos \angle CA_2P}$.
Let $A_3$ be the intersection of $A_2P$ and $(O), A_4$ be the intersection of $A_3O$ and $(O)$. Similarly we define $B_3, B_4, C_3, C_4$.
We have $\frac{\cos \angle BA_2P}{\cos \angle CA_2P}=\frac{\cos \angle BA_4A_3}{\cos \angle CA_4A_3}=\frac{\sin \angle BA_3A_4}{\sin \angle CA_3A_4}=\frac{\sin \angle BAA_4}{\sin \angle CAA_4}$
Hence $\frac{XB}{XC}=\frac{\sin \angle BAQ}{\sin \angle CAQ}.\frac{\sin \angle BAA_4}{\sin \angle CAA_4}.$
Do the same with $\frac{YC}{YA}, \frac{ZA}{ZB}$ then applying Ceva-sine theorem we obtain $(1)$ $\Leftrightarrow AA_4, BB_4, CC_4$ are concurrent.
On the other side, according to Pascal's theorem for 6 points $A, B, A_3, B_3, A_2, B_2$ we claim $M, P, Q$ are collinear ($M$ is the intersection of $AB_3$ and $BA_3$)
Applying Pascal's theorem again for 6 points $A, B, A_3, B_3, A_4, B_4$ we get the intersection of $AA_4$ and $BB_4$ lies on $PQ$. Similarly the intersection of $CC_4$ and $BB_4$ lies on $PQ$. This means $AA_4, BB_4, CC_4$ are concurrent. We are done.

PS: Another special case:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=386935

My solution is similar to Linh's but I went for different arguments. Keeping the same initial notations of Linh's solution, let lines $PA_2,PB_2$ cut $(O)$ again at $D,E$ and $DO,EO$ cut $(O)$ again at $U,V.$ Define $T \equiv VA_2 \cap UB_2.$ By Pascal theorem for nonconvex cyclic hexagons $VA_2DUB_2E$ and $AUB_2BVA_2,$ it follows that $T,P,O$ and $R,T,Q$ are collinear $\Longrightarrow$ $R \equiv AU \cap BV \cap PQ.$ If we define the point $W$ similarly as $U,V,$ then  $AU,BV,CW$ concur on $PQ.$ Let $UA_2,$ $VB_2$ and $WC_2$ cut $BC,CA,AB$ at $X,Y,Z.$ Then we have

$\frac{XB}{XC}=\frac{A_2B}{A_2C} \cdot \frac{UB}{UC} \ , \ \ \frac{YC}{YA}=\frac{B_2C}{B_2A} \cdot \frac{VC}{VA} \ , \ \ \frac{ZA}{ZB}=\frac{C_2A}{C_2B} \cdot \frac{WA}{WB}$

Since cyclic hexagons $AWBUCV$ and $AC_2BA_2CB_2$ have concurrent diagonals, we get

$\frac{UB}{UC} \cdot \frac{VC}{VA} \cdot \frac{WA}{WB}=1 \ ,  \ \frac{A_2B}{A_2C} \cdot \frac{B_2C}{B_2A} \cdot \frac{C_2A}{C_2B}=1 \Longrightarrow \ \frac{XB}{XC} \cdot \frac{YC}{YA} \cdot \frac{ZA}{ZB}=1$

By the converse of Menelaus theorem, the points $X,Y,Z$ are collinear. But since $\angle PA_2U$ is right, then $PX$ is a diameter of the circle $(O_a) \equiv \odot(PA_1A_2)$ $\Longrightarrow$ $O_a$ is the midpoint of $PX.$ Likewise, centers $O_b,O_c$ of $\odot(PB_1B_2)$ and $\odot(PC_1C_2)$ are the midpoints of $PY,PZ.$ If $X,Y,Z$ are collinear, then $O_a,O_b,O_c$ are collinear $\Longrightarrow$ $(O_a),(O_b),(O_c)$ are coaxal.

Dear Mathlinkers,
this problem can be simply solved with the Jerabek's theorem.

Sincerely
Jean-Louis


What is the number of factors of the product $ 126\times 216$?

$ 126\times216$
$ \equal{}2\times3\times3\times7\times2\times3\times2\times3\times2\times3$

$ 2\plus{}3\plus{}3\plus{}7\plus{}2\plus{}3\plus{}2\plus{}3\plus{}2\plus{}3\equal{}\fbox{30}$.


I just thought of this problem when doing a problem from Challenging Problems in Geometry, but can't seem to solve it without messy equations.  :)

Ok, lets assume that a triangle is inscribed in a circle. That circle can have a triangle circumscribed around it that is similar to the triangle inscribed in the circle. Find the ratio of the perimeter of the inscribed triangle to the perimeter of the circumscribed triangle.

Never mind, got it. You can still post though. :)

$\triangle ABC\sim \triangle A'B'C'\Longrightarrow\rho =\frac aa'=\ldots =\frac{h_a}{ h'_a} =\ldots =\frac{m_a}{m'_a}=\ldots =\frac{l_a}{l'_a}=\ldots =$
$=\frac{R}{R'}=\frac{r}{r'}=\frac{p}{p'}=\frac{r_a}{r'_a}=\ldots =\sqrt{\frac{S}{S'}}=\ldots$

In our case, if the triangle $ABC$ is circumscribed to the circle $i=C(I,r)$ and is inscribed to the circle $e=C(O,R)$,while the triangle $A'B'C'$ is circumscribed to the circle $e$, then the ratio of similarity is $\rho=\frac rR$.


trigonometry geometry ratio inequalities

These are really aimed at those who want a few little training exercises to help them on the way to become Olympiad geometryl masters. Seasonsed veterans out there could probably slaughter these in minutes, so could you keep all solutions in hide tags please.

1. Show there exists a point $\Omega$ inside $\triangle ABC$ such that $\angle \Omega AB = \angle \Omega BC =\angle \Omega CA = \omega$. Deduce there exists another point $\Omega '$ such that $\angle \Omega ' BA = \angle \Omega ' CB =\angle \Omega ' AC = \omega$ and that  $\Omega$ and $\Omega '$ are isogonal conjugates.

2. Show that $\triangle = \frac{1}{4}\tan \omega(a^2 + b^2 + c^2)$ where $\triangle$ is the area of $\triangle ABC$ and deduce that $\cot \omega = \cot A + \cot B + \cot C$.

3. Show that $\omega \leq \frac{\pi}{6}$ and find the equality case.

This is too hard for Intermediate Topics. Can one of the mods of this forum move it to Pre-Olympiad please?

Really? There's nothing too clever required to solve the questions. Oh well, never mind. I'll keep the difficulty level in mind next time.  :)

[hide]
Using the trigonometric form of Ceva's Theorem, such points $\Omega$ and $\Omega^\prime$ will exist if and only if $\sin^3 \omega=\sin(A-\omega) \sin (B-\omega) \sin (C-\omega)=\prod_\sigma (\sin A\cos \omega -\cos A\sin \omega)$.

Writing this in terms of $\cot \omega$, $\prod_\sigma (\sin A\cot \omega -\cos A)=1$. Now substitute in $\cot \omega=\cot A+\cot B+\cot C$ and show that it works.

The $\cos A$ terms will cancel, so $\sin A\cot \omega-\cos A$ becomes $\sin A(\cot B+\cot C)=\sin A\cdot \frac{\cos B\sin C+\cos C\sin B}{\sin B\sin C}=\frac{\sin^2 A}{\sin B\sin C}$.

The other two terms are similar, so $\prod_\sigma (\sin A\cot \omega -\cos A)=\prod_\sigma \frac{\sin^2 A}{\sin B\sin C}=1$, and $\cot \omega =\cot A+\cot B+\cot C$ works.

For the area formula, $\triangle =\frac 12 ab\sin C=\frac{a^2\sin B\sin C}{2\sin A}$. Then $\frac 14 \tan \omega (a^2+b^2+c^2)=\frac 14 \triangle \tan \omega \sum_\sigma \frac{2\sin A}{\sin B\sin C}$.

$\frac{\sin A}{\sin B\sin C}=\frac{\cos B\sin C+\sin B\cos C}{\sin B\sin C}=2(\cot B+\cot C)$. Using this in the sum, $\sum_\sigma \frac{2\sin A}{\sin B\sin C}=4(\cot A+\cot B+\cot C)$ and the area formula follows. There's probably some way to use area ratios to derive this formula.

Part 3 I have no clue how to do, although I can prove $\sum_\sigma \cot \frac A2 \geq 3\sqrt{3}$ if that helps.
[/hide]

For part 3 you could use $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$.

[hide="2."]
A nicer way of doing it is to first derive the area equation.

Let $A\Omega = x$, $B\Omega = y$ and $C\Omega = Z$. Now consider the Cosine Rule on $\triangle AB\Omega$, $\triangle BC\Omega$ and $\triangle CA\Omega$:

$c^2 + x^2 - 2cx\cos\omega = y^2$
$a^2 + y^2 - 2ay\cos\omega = z^2$
$b^2 + x^2 - 2bz\cos\omega = x^2$

Add the three equations to get $\cos\omega = \frac{a^2 + b^2 + c^2}{2(ay + bz + cx)}$. But we have $\triangle = \frac{1}{2}\sin\omega(ay + bz + cx)$. The result follows immediately.

The next equation follows from the fact $\cot A + \cot B + \cot C = \frac{a^2 + b^2 + c^2}{4\triangle}$ which is again easily shown as $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{\sin A(b^2 + c^2 - a^2)}{4\triangle}$.
[/hide]

[hide="3."]
We know $\cot\omega = \frac{a^2 + b^2 + c^2}{4\triangle} \geq \frac{ab + bc + ca}{4\triangle}$.

Therefore $\cot\omega \geq \frac{1}{2}(\csc A + \csc B + \csc C)$ and as $\csc \theta$ is convex from $0 < \theta < \pi$, we have by Jensen's Inequality $\cot\omega \geq \frac{3}{2}\csc\frac{\pi}{3} = \sqrt{3}$. Thus the result follows.
[/hide]

Simpler: \begin{eqnarray*}\cot^2 \omega & = & \cot^2 A + \cot^2 B + \cot^2 C + 2 \cot A \cot B + 2 \cot B \cot C + 2 \cot C \cot A \\ & \geq & 3(\cot A \cot B + \cot B \cot C + \cot C \cot A) \\ & = & 3. \end{eqnarray*} Hence $\cot \omega \geq \sqrt{3}$, which implies the result.


Support
number theory unsolved
number theory

Define $ Supp(m)\equal{}${$ p: p$ is a prime such that $ p$ divides $ m$}. Show that there exists infinitely many pairs of positive integers $ (a,b)$ such that $ a$ and $ b$ are unique and

i) $ Supp(a)\equal{} Supp(b)$
ii) $ Supp(a\plus{}1)\equal{}Supp(b\plus{}1)$

I dont if this problem was posted before but kindly anyone please give me a solution to this!!!

[hide]Let $ m \plus{} 1$ be a power of $ 2$ (at least $ 4$). Then we can set

$ a \equal{} m \minus{} 1$
$ b \equal{} m^2 \minus{} 1$.

We can see that $ Supp(m^2 \minus{} 1) \equal{} Supp((m \plus{} 1)(m \minus{} 1)) \equal{} Supp(m \plus{} 1)\cup Supp(m \minus{} 1)$, but since $ Supp(m \plus{} 1) \equal{} \{2\}$, and $ 2\in Supp(m \minus{} 1)$, we have $ Supp(m \plus{} 1)\cup Supp(m \minus{} 1) \equal{} Supp(m \minus{} 1)$, so $ Supp(a) \equal{} Supp(b)$.

And it is easy to see that $ Supp(m) \equal{} Supp(m^2)$ so $ Supp(a \plus{} 1) \equal{} Supp(b \plus{} 1)$.[/hide]


Alternate with the post above you. Example:

A guy falls out of a plane.

Fortunately, he had a parachute.
Unfortunately, the parachute didn't open.
Fortunately, there was a haystack underneath him.
Unfortunately, there was a pitchfork sticking right out of it.
Fortunately, he missed the pitchfork.
Unfortunately, he missed the haystack.
.
.
.
(Idea from Douglas Hofstadter)

[b]Aliens are spotted in space, and they're heading straight for earth in their flying machine thingy.[/b]

Fortunately, the earth was prepared for the attack
Unfortunately, the aliens countered everyone of earth's futile resistances
Fortunately, Earth had a backup plan to move to Mars
Unfortunately, the aliens already blew up Mars
Fortunately, we still had spaceships to fly away in
Unforunately, it ran out of fuel once we passed the atmosphere

[b]AoPS is being destroyed by a terrible virus[/b]

EDIT: uh, did i do it wrong?
DOUBLE EDIT: if not, continue w/ mine
if so,
[b]Unfortunately, those weren't the aliens flying towards earth[/b]

Fortunately, a group of medieval knights came to fight off the evil aliens.

There was a topic about this with almost the exact same topic title.

Unfortunately, the ebola virus just happens to afflict one major city per country.

Fortunately, ebola gives protection to alien lasers.

Unfortunately, the aliens were armed with AK-47s.  (That's a type of gun, BTW)

Fortunately, the medieval knights had gun proof vests. (Don't ask me how)

Edit: NVM: Pacman got there before me.

Unfortunately, they found bullets in a cave on the earth and decided to use them.

unfortunately, ebola kills almost everyone.

Unfortunately, one alien survived, and that alien happened to have a backpack filled with 1-up mushrooms.

Unfortunately, the aliens had their own Chuck Norris clone.

[quote="pacman2812"][quote="Flame"]Unfortunately pacman2812 was kicked out of heaven because god doesn't allow pwning in heaven so pacman's soul wandered alone for eternity.

you started everything by killing me in three words game unprovoked... now you'll be pwned for eternity :D[/quote]

...
hehe
hahaha
.
.
.
BWAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! :rotfl:  :rotfl:  :rotfl:

Fortunately, Flame was overpwned, got overweight and fell out of the heaven, next to pacman2812's soul, at a place where overpwning was allowed, so pacman2812 could continue pwning Flame forever and forever, and nobody could stop it, and Flame couldn't go out of there, so pacman2812 pwned him FOREVER!!!!!!!!!!!!!!

EDIT: IS THIS FORTUNATELY???
EDIT 2: OBVIOUSLY; BECAUSE Flame WAS PWNED!!! BWAHAHAHAHA!!!!!![/quote]

not only is that very disturbing, SOMEONE has a VERY LARGE ego.

Unfortunately, pacman2812 died. end of story

sigh, can we have something w/o pacman in it?

Bob meets Bob III

rofl pacman2812 you were again pwned

fortunately they became friends

sorry, but I was dead yet, so nothing happened

so you and mz94 got pwned

Unfortunately, Bob II hated both them

not really but fine

fortunately Bob I and III didn't mind that

Well, what about if we finish this stupid war?  :D

Unfortunately, Bob II killed them

well its extremely entertaining to me and I don't see it as a war

Fortunately Bob II was arrested and imprisoned for life

Unfortunately, Bob II broke free and killed everyone called Bob.

Fortunately Bob II was arrested again and this time sentenced to death

Unfortunately, he used a dummy to die in his place while he escaped.

Fortunately that dummy was pacman2812

PWNED AGAIN!!!!!!!!!!!!!!

Unfortunately, Bob II decided to change the dummy because he was very cool, and placed <random.com told me that...> TPAM and TPBM as dummies. Also he decided not to place pacman2812 as dummy never more.

Please... don't try to pwn me if you don't want to die again... :diablo:

omg you just pwnd yourself  :rotfl:
[quote][b]Unfortunately[/b], Bob II decided to change the dummy[/quote]

Fortunately Bob II was caught while trying to change dummies again and was finally killed while dummies survived.

unfortunately, the omnipotent being got so annoyed with this "pwnage war", he decided to smite both pacman2812 and Flame.

THE END!!!!!!!!!!!!!!!!!!!!!!!!!

start a new story!!!

The Earth goes out of it's orbit

hmm... Flame
[quote]omg you just pwnd yourself   :rotfl:

[quote]Unfortunately, Bob II decided to change the dummy [/quote]
[/quote]

uhmmm... it's unfortunately TO YOU!!!!!!!


I don't know why but when I put this in Latex code:

[code]\binom{29}{p}[/code]

It doesn't show up.

But when I used this:

[code]{29 \choose p}[/code]

It works.. Why? :(

\binom needs the amsmath package so requires \usepackage{amsmath} whereas \choose is a native TeX command


Let $ a,b,c \geq 0$
and $ a\plus{}b\plus{}c\equal{}1$
PROVE :$ \sum \frac{a^2\plus{}1}{b^2\plus{}1} \leq 3.5$

[quote="nguyensilva"]Let $ a,b,c \geq 0$
and $ a \plus{} b \plus{} c \equal{} 1$
PROVE :$ \sum \frac {a^2 \plus{} 1}{b^2 \plus{} 1} \leq 3.5$[/quote]

$ \sum \frac {a^2 \plus{} 1}{b^2 \plus{} 1} \leq 3.5$

$ \Leftrightarrow \frac {7}{2} \minus{} \sum{\frac {a^2 \plus{} (a \plus{} b \plus{} c)^2}{b^2 \plus{} (a \plus{} b \plus{} c)^2}} \ge 0$

$ \Leftrightarrow 124a^2b^2c^2 \plus{} \sum_{cyc}{(6a^5b \plus{} 6a^5c \plus{} 23a^4b^2 \plus{} 21a^4c^2 \plus{} 34a^3b^3 \plus{} 38a^4bc \plus{} 82a^3b^2c \plus{} 82ab^2c^3)}\ge$ $ 0$

obvious...

actually, if $ a,b,c\ge0,a\plus{}b\plus{}c\equal{}1$, we have

\[ 3 \leq \sum\frac{a^{2}\plus{}1}{b^{2}\plus{}1}\leq 3.5\]

[quote="nguyensilva"]Let $ a,b,c \geq 0$
and $ a \plus{} b \plus{} c \equal{} 1$
PROVE :$ \sum \frac {a^2 \plus{} 1}{b^2 \plus{} 1} \leq 3.5$[/quote]
This is problem 101, pages 143 in the book "Bat dang thuc va nhung loi giai hay".
Authors: Vo Quoc Ba Can - Tran Quoc Anh, Ha Noi Publishing House 2009.
You can find nice solutions in this book  :)

[quote="wangsacl"][quote="nguyensilva"]Let $ a,b,c \geq 0$
and $ a \plus{} b \plus{} c \equal{} 1$
PROVE :$ \sum \frac {a^2 \plus{} 1}{b^2 \plus{} 1} \leq 3.5$[/quote]wondering: when does the equality hold?[/quote]

$ (a,b,c)\equal{}(0,0,1)$


The Big Telescope Company sells circular mirrors. Their largest mirrors have radii of 5 meters and their smallest mirrors have radii of 1 meter. The cost of every mirror is proportional to the cube of the mirror's radius. What is the ratio of the total cost of 25 of the company's smallest mirrors to the cost of one of the company's largest mirrors? Express your answer as a common fraction.

Each of the small mirrors cost $ 1^3\equal{}1 dollar$ Each largest one cost $ 5^3\equal{}125 dollar$
$ \frac{25}{125}\equal{}\boxed{\frac{1}{5}}$


geometry
Gauss
conics
geometric transformation
reflection
circumcircle
angle bisector

given four lines $g_1,g_2,g_3,g_4$, pairwise nonparallel, let A be the intersection of $g_1,g_2$, A' be the intersection of $g_3,g_4$, B intersection of $g_1,g_3$, B' intersection of $g_2,g_4$, C intersection of $g_1,g_4$, C' intersection of $g_2,g_3$. show

a) the midpoints of $AA'$, $BB'$, $CC'$ are collinear.
b) let P be an arbitrary point. we have $\angle APB+\angle A'PB'=180^{\circ}$ iff $\angle APC+\angle A'PC'=180^{\circ}$ iff $\angle BPC+\angle B'PC'=180^{\circ}$, where angles are oriented mod 180°.

Peter

The line obtained in a) is the well-known Newton-Gauss line of the complete quadrilateral. I'll think about b), which is new to me.

[quote="Peter Scholze"]b) let P be an arbitrary point. we have $\angle APB+\angle A'PB'=180^{\circ}$ iff $\angle APC+\angle A'PC'=180^{\circ}$ iff $\angle BPC+\angle B'PC'=180^{\circ}$, where angles are oriented mod 180°.[/quote]

Hey Peter, you really did the best out of this year's IMO problem 5... and there are indeed several solutions to your problem, depending on what one likes most: pedal circles, isogonal conjugates, inscribed conics, van Rees isoptic cubics, ...

So let me post the most elementary of the solutions:

I will solve part [b]b)[/b] only, for part [b]a)[/b] is well-known.

I will seriously use directed angles modulo 180°, so I will write 0° instead of 180°. My task will be to show that the equations < APB + < A'PB' = 0°, < APC + < A'PC' = 0° and < BPC + < B'PC' = 0° are pairwisely equivalent.

For solving [b]b)[/b], I construct the orthogonal projections $P_1$, $P_2$, $P_3$, $P_4$ of the point P on the lines $g_1$, $g_2$, $g_3$, $g_4$, respectively. I will show that < APC + < A'PC' = 0° holds if and only if the points $P_1$, $P_2$, $P_3$, $P_4$ lie on one circle. Of course, once this will be shown, it will be analogous to prove that each of the conditions < APB + < A'PB' = 0° and < BPC + < B'PC' = 0° also holds if and only if the points $P_1$, $P_2$, $P_3$, $P_4$ lie on one circle, and hence all three equations < APB + < A'PB' = 0°, < APC + < A'PC' = 0° and < BPC + < B'PC' = 0° are equivalent, so that the problem is solved.

Hence it just remains to show that < APC + < A'PC' = 0° holds if and only if the points $P_1$, $P_2$, $P_3$, $P_4$ lie on one circle.

But this is really straightforward: Since $\measuredangle AP_1 P = 90^{\circ}$ and $\measuredangle AP_2 P = 90^{\circ}$, the points $P_1$ and $P_2$ lie on the circle with diameter AP. Thus, $\measuredangle PP_2 P_1 = \measuredangle PAP_1$. In other words, $\measuredangle PP_2 P_1 = \measuredangle PAC$. Similarly, $\measuredangle PP_2 P_3 = \measuredangle PC'A'$. Thus,

  $\measuredangle P_1 P_2 P_3 = \measuredangle PP_2 P_3 - \measuredangle PP_2 P_1 = \measuredangle PC'A' - \measuredangle PAC$.

Similarly,

  $\measuredangle P_1 P_4 P_3 = \measuredangle PA'C' - \measuredangle PCA$.

But the points $P_1$, $P_2$, $P_3$, $P_4$ lie on one circle if and only if $\measuredangle P_1 P_2 P_3 = \measuredangle P_1 P_4 P_3$, what is therefore equivalent to < PC'A' - < PAC = < PA'C' - < PCA. This can be rewritten as < PC'A' - < PA'C' = < PAC - < PCA, or, equivalently, as < PC'A' + < C'A'P = < PAC + < ACP. Now, since the sum of the angles of a triangle is 0°, we have < PC'A' + < C'A'P = - < A'PC' and < PAC + < ACP = - < CPA, and thus our condition is equivalent to - < A'PC' = - < CPA, i. e. to < A'PC' = < CPA, i. e. to < A'PC' = - < APC, what means nothing but < APC + < A'PC' = 0°. After what I have said before, this completes the proof.

Note that we can prove much more: All the following conditions are pairwisely equivalent:

[b]1.[/b] We have < APB + < A'PB' = 0°.
[b]2.[/b] We have < APC + < A'PC' = 0°.
[b]3.[/b] We have < BPC + < B'PC' = 0°.
[b]4.[/b] The orthogonal projections $P_1$, $P_2$, $P_3$, $P_4$ of the point P on the lines $g_1$, $g_2$, $g_3$, $g_4$ lie on one circle.
[b]5.[/b] The reflections of the point P in the lines $g_1$, $g_2$, $g_3$, $g_4$ lie on one circle.
[b]6.[/b] If we consider the four triangles, each of them formed by three of the four lines $g_1$, $g_2$, $g_3$, $g_4$, then the point P has the same isogonal conjugate with respect to all of these triangles.
[b]7.[/b] The point P has the same isogonal conjugate with respect to two of these triangles. [This indeed is equivalent to [b]6.[/b]! In other words, if the point P has the same isogonal conjugate with respect to two of these four triangles, then this isogonal conjugate also coincides with the isogonal conjugates of the point P with respect to the other two triangles.]
[b]8.[/b] If we connect the point P with the point of intersection of any two lines $g_i$ and $g_j$, and reflect the connecting line in the angle bisector of the angle between the lines $g_i$ and $g_j$, then the six lines resulting in this way (by taking all six possible pairs of lines $g_i$ and $g_j$) concur at one point.
[b]9.[/b] The point P is the focus of a conic touching the lines $g_1$, $g_2$, $g_3$, $g_4$. [Note that the center of such a conic lies on the line joining the midpoints of the segments AA', BB', CC'. This gives the relation with your problem [b]a)[/b].]

Actually, all the points P satisfying the above equivalent conditions lie on a special cubic, the so-called [b]isoptic cubic[/b] of the complete quadrilateral formed by the four lines $g_1$, $g_2$, $g_3$, $g_4$. This cubic passes through the six points A, B, C, A', B', C' and through the Miquel point of the complete quadrilateral formed by the four lines $g_1$, $g_2$, $g_3$, $g_4$ (this point is the point of intersection of the circumcircles of the four triangles, each of them enclosed by three of the lines $g_1$, $g_2$, $g_3$, $g_4$).

For more details, see Hyacinthos messages #8242, #8280, #8283, #8287, #8289, #8290, #8303, #8304, #8305, #8370, #8380, which you can access through

  http://groups.yahoo.com/group/Hyacinthos/messages/8242?expand=1

(for the later messages, click on "Next").

  Darij

i actually proved myself that it is a cubic going through the six points. the proof is quite easy, however. by the way, looking at a special configuration, we can derive simson(i don't know how he's written) out of this: take one line to be the infinite line, the other three forming a triangle. then my conditions in b) are equivalent to P lying on the circumcircle of triangle ABC. but since the orthogonal projections of P lie on a circle in the finite case, but one points is infinite now, they must lie on a line, so the orthogonal projections of P to the three sides of the triangles lie on a line :)

Peter

Indeed I have never thought of this way to prove Simson's theorem! It's indeed nice to note that in the case of one of the lines $g_1$, $g_2$, $g_3$, $g_4$ being the line at infinity, the isoptic cubic degenerates to the circumcircle (together with the line at infinity).

Congratulations for having discovered the method of cubic curves in synthetic geometry on your own!

By the way, it is completely unimportant whether you spell the name Simson correctly or not: it is not his line, hence the naming is incorrect per definitionem...

  Greetings,
  Darij

Thank you very much darij for this usefull lesson of geometry!


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_{0}^{\pi}\sqrt{(\cos x\plus{}\cos 2x\plus{}\cos 3x)^{2}\plus{}(\sin x\plus{}\sin 2x\plus{}\sin 3x)^{2}}\ dx$.

[hide="Approach"]$ I \equal{}\int^{\pi}_{0}\sqrt{1\plus{}4\cos x\plus{}4\cos^{2}x}\,dx \equal{}\int^{\pi}_{0}\sqrt{(2\cos x\plus{}1)^{2}}\,dx$

[/hide]


The time on a digital clock reads 12:54. You must insert a symbol for addition, subtraction, multiplication or division between each pair of adjacent digits (ignoring the $ ^{\prime\prime}$:$ ^{\prime\prime}$). Using three distinct operations without any parentheses, what is the greatest possible value after the expression is simplified? Express your answer as a decimal to the nearest tenth.

Obviously, you're gonna want to multiply the 5 and the 4, then add it, so you have 1_2+5*4. Then dividing gives a higher value then subtracting, so you have 1/2+5*4=20.5


Let $ a,b,c>0$ s.t. $ a^2\plus{}b^2\plus{}c^2\equal{}3$.Prove that:
$ \frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{c^2\plus{}a^2}{c\plus{}a}\geq 3$.

[quote="Allnames"]I think it can be solve by SOS,the sum that i have received is $ \sum\ Sa(b \minus{} c)^{2}$ there $ Sa$ you can simpely find ,the last is easy,i will post my post in the next time (sorry because i have short time[/quote]

 What is SOS?

[quote="Marius Mainea"]What is SOS?[/quote]
Sum Of Squares.
The idea of the method is to move everything to one side of the inequality and represent it as a sum of positive expressions, normally squares.

[quote="Allnames"]I think it can be solved by SOS,the sum that i have received is $ \sum\ Sa(b \minus{} c)^{2}$ there $ Sa$ you can simpely find ,the last is easy,i will post my post in the next time (sorry because i have short time[/
I don't think so.Can you post your proof?
(I think you have much time-->he is my classmate :P )

Sorry,here is my proof
We can rewrite this ineq to this form
$ \sum\ Sc (a\minus{}b)^{2}$$ \geq$0 there  $ Sc$=$ \frac{1}{a\plus{}b} \minus{} \frac{2}{\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\plus{}a\plus{}b\plus{}c}$
And $ Sa$,$ Sb$ are simalar
Then,We can simplely prove $ Sb$,$ Sa\plus{}Sb$,$ Sc\plus{}Sb$\geq0
You can finish it,so use SOS theorem we have a nice proof

WLOG assume $ a\equal{}\min(a,b,c)$.  Then $ a^2\plus{}b^2 \leq b^2\plus{}c^2 \leq c^2\plus{}a^2$ and $ \frac{1}{a\plus{}b} \geq \frac{1}{b\plus{}c} \geq \frac{1}{c\plus{}a}$. Using Chebyshev`s inequality we get:
\[ \frac{1}{3}\cdot 2(a^2\plus{}b^2\plus{}c^2)\left( \frac{1}{a\plus{}b} \plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a} \right) \leq \sum \frac{a^2\plus{}b^2}{a\plus{}b}.\]

But \[ (a\plus{}b\plus{}c)^2 \leq 3(a^2\plus{}b^2\plus{}c^2)\equal{}9 \Longrightarrow a\plus{}b\plus{}c \leq 3.\]
According to the Cauchy-Schwarz inequality we get \[ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a} \geq \frac{9}{2(a\plus{}b\plus{}c)} \geq \frac{9}{3 \cdot 2}\equal{}\frac{3}{2}.\]
Finally
\[ \frac{1}{3}\cdot 2(a^2\plus{}b^2\plus{}c^2)\left( \frac{1}{a\plus{}b} \plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a} \right) \leq \sum \frac{a^2\plus{}b^2}{a\plus{}b} \Longleftrightarrow \sum \frac{a^2\plus{}b^2}{a\plus{}b} \geq \frac{1}{3} \cdot 2 \cdot 3 \cdot \frac{3}{2}\equal{}3\]

[quote="Allnames"]Sorry,here is my proof
We can rewrite this ineq to this form
$ \sum\ Sc (a \minus{} b)^{2}$$ \geq$0 there  $ Sc$=$ \frac {1}{a \plus{} b} \minus{} \frac {2}{\sqrt {3(a^{2} \plus{} b^{2} \plus{} c^{2})} \plus{} a \plus{} b \plus{} c}$
And $ Sa$,$ Sb$ are simalar
Then,We can simplely prove $ Sb$,$ Sa \plus{} Sb$,$ Sc \plus{} Sb$\geq0
You can finish it,so use SOS theorem we have a nice proof[/quote]
I don't think that is the best way.
Using Am-Gm, easily to see that $ Sa,Sb,Sc \geq\ 0$. So, it's obviously trues  :wink:

[quote="karis"]Mîndrilă Claudiu, i think you're wrong[/quote]
For what reason?

because he proved that :$ \frac{a^2\plus{}c^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{a^2\plus{}b^2}{c\plus{}a}\geq 3$ not:
$ \frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{c^2\plus{}a^2}{c\plus{}a}\geq 3$
you can see it when he used Chebychev.


$ [\sinh]+[\tgx]=[\cosh]+[%Error. "ctgx" is a bad command.
]$

_____________________
Azerbaijan my motherland  :D

hello, and what is your problem?
Sonnhard.

Look "under" the LATEX code he used! He means

$ [ \sin x ] \plus{} [ \tan x ] \equal{} [ \cos x ] \plus{} [ \cot x ]$

whatever that means to him (probably floor functions).


In how many positive four-digit integers that are not multiples of 1111 do the digits form an arithmetic sequence?

Actually, there are 21, you forgot 9630,6420,3210.

:P I got 210, because I thought the digits didn't have to be in order, so I counted numbers such as 4231.... :blush: thanks!


Which AMC 12 do you all think is the hardest?
For me, the AMC 12 2001 seems to be more difficult than the ones from 2002, 2003, 2004.

The 2001 AMC 12 was [i]very[/i] difficult. The cutoff score to qualify for AIME was only 84!!! (although then only 2 points were given for a skipped question)


I noticed that dvipng has a command to make the background image transparent.
However I haven't been  able to get it to work for me.

I have MiKTeX 2.7 installed and I am using the dvipng program from the MiKTeX 2.7\miktex\bin
directory it has these commands

[code]This is dvipng 1.9 Copyright 2002-2006 Jan-Ake Larsson

Usage: dvipng [OPTION]... FILENAME[.dvi]
Options are chosen to be similar to dvips' options where possible:
  -D #         Output resolution
  -l #         Last page to be output
  -o f         Output file, '%d' is pagenumber
  -O c         Image offset
  -p #         First page to be output
  -pp #,#..    Page list to be output
  -q*          Quiet operation
  -T c         Image size (also accepts '-T bbox' and '-T tight')
  -v*          Verbose operation
  -            Interactive query of options

These do not correspond to dvips options:
  -bd #        Transparent border width in dots
  -bd s        Transparent border fallback color (TeX-style color)
  -bg s        Background color (TeX-style color or 'Transparent')
  --depth*     Output the image depth on stdout
  --dvinum*    Use TeX page numbers in output filenames
  -fg s        Foreground color (TeX-style color)
  --follow*    Wait for data at end-of-file
  --freetype*  FreeType font rendering (default on)
  --gamma #    Control color interpolation
  --height*    Output the image height on stdout
  --noghostscript*  Don't use ghostscript for PostScript specials
  --nogssafer* Don't use -dSAFER in ghostscript calls
  --palette*   Force palette output
  --picky      When a warning occurs, don't output image
  --strict     When a warning occurs, exit
  --truecolor* Truecolor output
  -Q #         Quality (T1lib and PK subsampling)
  -z #         PNG compression level

   # = number   f = file   s = string  * = suffix, '0' to turn off
       c = comma-separated dimension pair (e.g., 3.2in,-32.1cm)[/code]

I tried
dvipng -T tight -bg Transparent -o testYes.png testmath.dvi

but the image is all screwed up after that.
dvipng was working fine without -bg Transparent but now I want the image to have a Transparent background so that it will look nice in different colored backgrounds of my html pages.

Does anybody know what I am doing wrong or an easy way to fix it.
I thought about just setting the \pagecolor in the latex document but I decided not to since
Then if I changed the background html I would have to change the .tex and recompile.
That would suck.

Also does anybody know what the artofproblemsolving site uses for it's font setting's of the latex because I am noticeing that my latex code looks a little different font wise then yours.
Do you guys set it thru your LatexRender program or do you just write the font stuff out
in the temp .tex file then compile and produce the image using  LatexRender?
Either way if I could get the font , font-size, font-type,...etc etc from you guys that would help
me alot.

Thanks again.

O and the latex for the image was

this
[code]\documentclass[12pt]{article}
\pagestyle{empty}

\begin{document}

 $\sqrt{x} = 5$ is $x=25$.
$\sum \int \sqrt{x} \sqrt[n]{x} \frac{a}{b} \oint \prod \coprod
\frac{\partial{f}}{\partial{x}} \lim_{a \rightarrow b}$
\end{document}
[/code]  :maybe:

You could easily examine the code for [url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url] yourself. Basically it writes a tex file and then compiles it:
[code]\documentclass[10pt]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\pagestyle{empty}
\begin{document}
$formula goes here$
\end{document}[/code]

Ok a few last questions.

First I used the same method LatexRender did.

i.e
[code]\documentclass[10pt]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\pagestyle{empty}
\begin{document}
$formula goes here$
\end{document}[/code]

But got---> see attachment

I am being very pick but it seems the font type is bolder
on [url]http://www.mayer.dial.pipex.com/latexrender/latexrender.htm[/url]
8th equation down corosponds to my $ formula goes here$.

Why is this occuring?

Second is their away to change the background color of the latex document?

Third
I know you can use \color{  } { } to change the text color for a specific part of the text.
But their must be a better way then wrapping the whole latex document in a color tag
to change all the color of the text on the page.

And is their away to specify the color by giving the rgb  0 thru 255 for valid values
Because I can only get the named colors to work like red , green but not something
like rgb =33,165,245

I have been unsuccessful fonts as well?
I would you make the text on a document larger then 12pt
Because I noticed you can only set the document to 8pt ,10pt, 12pt
But I know their must be away to set larger-size font , boldness , ...etc.

I have googled but I must not be typing it in correctly because it is not compiling
Some clarity would help.
Thanks

[quote="Mathematics!"]I am being very pick but it seems the font type is bolder
on [url]http://www.mayer.dial.pipex.com/latexrender/latexrender.htm[/url]
8th equation down corosponds to my $ formula goes here$.

Why is this occuring?[/quote]You don't say whether this is happening in the dvi file (so LaTeX fonts) or after using dvipng, so a dvipng problem. If the former you can use different LaTeX fonts but that is complicated and the attachment looks fine to me.

[quote="Mathematics!"]Second is their away to change the background color of the latex document?[/quote]\pagecolor. See page 6 of the [url=http://tug.ctan.org/tex-archive/macros/latex/required/graphics/grfguide.pdf]color package manual[/url].

[quote="Mathematics!"]I know you can use \color{  } { } to change the text color for a specific part of the text.
But their must be a better way then wrapping the whole latex document in a color tag
to change all the color of the text on the page.[/quote]Don't wrap the text in {} just use \color at the start and it will stay that colour.

[quote="Mathematics!"]And is their away to specify the color by giving the rgb  0 thru 255 for valid values
Because I can only get the named colors to work like red , green but not something
like rgb =33,165,245[/quote]See page 5 of the manual. Values are (decimals) between 0 and 1 not 0 and 255.

[quote="Mathematics!"]I have been unsuccessful fonts as well?
I would you make the text on a document larger then 12pt
Because I noticed you can only set the document to 8pt ,10pt, 12pt[/quote]Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=extsizes]extsizes package[/url] (included with MiKTeX which allows sizes up to 20pt.


rotation
geometry
geometric transformation

A rubber ball with a radius of 4.8 cm rolls along the horizontal surface of a table with a constant linear speed v. When the ball rolls off the edge of the table, it falls 0.89 m to the floor below. If the ball completes 0.71 revolutions during its fall, what was its linear speed, v?

The answer should be in cm/s

Thanks!

let me play with variables instead
let the the angular sped of the ball is $ \omega$ then we have $ v = \omega r$
also let the height below the table be $ h$ then we have time taken to fall down $ t = \sqrt {\frac {2h}{g}}$ so the angle rotated is $ \omega t$ so the no. of rotations is $ n = \frac {\omega t}{2\pi}$
so we have
$ n = \frac {\omega \sqrt {h}}{\sqrt {2g}\pi}$
so linear speed $ v = \frac {nr \sqrt {2g} \pi}{\sqrt {h}}$
plugging the data we get
$ v \approx 50.7 \text{cms}^{ - 1}$


Suppose the NCAA decided to have a single elimination tournament involving all Division 1A teams at the end of the basketball season. If there are 303 Division 1A teams, how many games will be played before a national champion is declared?

[hide]Since it's single elimination, for every match played, 1 team is eliminated.
In order to have a champion, 302 teams must be eliminated.
Thus, we need 302 matches.
>>302<<[/hide]

dwx314, i think your wrong, cuz they would have byes. and if they didn't have byes, there would be at least one, since there's an odd number of teams.

If there are 2 teams, they need 1 match.
3 teams, 2 matches.
4 teams, 3 matches.
5 teams, 4 matches.
6 teams, 5 matches.
$n$ teams, $n-1$ matches.
$303$ teams, $302$ matches.

assuming each game generates 1 loser and we need to declare 302 losers before we have a winner, there are 302 games that must be played.

[quote="yankeefan13"]302 losers 302 matches[/quote]
Ouch

my solution is based on the fact that all elimination games are being played concurrently.

[hide]

First off, you have 303 teams.

That means there would be [tex] \frac {302}{2} = 151[/tex] or 151 games being played the first round.

That would automatically eliminate 151 teams which would leave [tex]303 - 151 = 152[/tex] or 152 teams left.

For the second round, you would have [tex] \frac {152}{2} = 76[/tex] or 76 games being played.

Thus, that would eliminate another 76 teams

So far we have eliminated [tex]151 + 76 = 227[/tex] teams, leaving only [tex]303 - 227 = 76[/tex] or 76 teams left.

For the third round, you would have [tex] \frac {76}{2} = 38[/tex] or 38 games being played.

That would, therefore, eliminate another 38 teams.

After the third round, we have a total of [tex] 227 + 38 = 265[/tex] or 265 teams eliminated, leaving only 38 teams left.

For the fourth round, you would have [tex]\frac {38}{2} = 19[/tex] or 19 games being played, again eliminating another 19 teams.

After the fourth round, we have a total of [tex]265 + 19 = 284[/tex] or 284 teams eliminated, leaving 19 teams left.

For the fifth round, you would have [tex]\frac {18}{2} = 9[/tex] or 9 games being played, eliminating another 9 teams.

After the fifth round, we have a total of [tex]284 + 9 = 293[/tex] or 293 teams eliminated, leaving 10 teams left.

For the sixth round, you would have [tex] \frac {10}{2} = 5[/tex] or 5 games being played, eliminating 5 more teams.

After the sixth round, we have a total of [tex]293 + 5 = 298[/tex] or 298 teams eliminated, leaving 5 teams left.

For the seventh round, you would have [tex] \frac {4}{2} = 2[/tex] or 2 games being played, eliminating 2 more teams.

After the seventh round, we have a total of [tex] 298 + 2 = 300[/tex] or 300 teams eliminated, leaving only 3 teams left.

for the eighth round, you would have [tex]\frac {2}{2} = 1[/tex] or 1 game being played, eliminating 1 more team.

after the eighth round, we have a total of [tex] 300 + 1 = 301[/tex] or 301 teams
eliminated, leaving only 2 teams left.

The ninth round would obviously yield the champs.

Adding up the total number of games played we would have
[tex]151 + 76 + 38 + 19 + 9 + 5 + 2 + 1 + 1 = 302[/tex] or 302 games played. [/hide]

What about that last game for the two left?

hey 236factorial:

 I caught my mistake. Indeed, it came out at 302 games.
Actually, I like chess64's solution about the best.

its 303-1=302, you should have made it how many rounds must be played, that would have been a fun and tricky problem.


inequalities
function
limit
number theory
algebra unsolved
algebra

sorry if it is a silly question,but i had gotten this inequality when i wanted to solve a number theory problem:could u help me? :)
as we know if $ n > 3,n \in N$ then the function $ f(n) \equal{} n^{\frac {1}{n}}$ has this property: $ f(n \plus{} 1) < f(n)$
now i wanted to know that if for each constant positive integer like $ t$ there exists a positive integer $ n > 3$ for which we have $ n^{\frac {t}{n}} < 2$.
thx in advance for your answer.

For any fixed $ t$ and $ n > 3$, $ n^{\frac {t}{n}}$ is strictly decreasing and $ \lim_{n\to \infty}n^\frac {t}{n} \equal{} 1$, so there is at most one $ x\in \mathbb{R}\cap (3,\infty)$ satisfying $ x^{\frac {t}{x}} \equal{} 2$, and all $ n\geq x$ satisfy $ n^\frac {t}{n} < 2$.

thank u for your reply ,
how can we show that $ \lim_{n\to\infty}n^\frac{t}{n}\equal{} 1$?
does there exist a proof not using advanced things? :oops:
if there does, could u give a hint for that ?
thx again :)

It's very easy for fixed $ t$. Just
$ \lim_{n\to \infty} n^{\frac{t}{n}}\equal{} \left( \lim_{n\to \infty} n^\frac{1}{n} \right)^t\equal{}1^t\equal{}1$


LaTeX
logarithms
algebra
functional equation
real analysis
real analysis unsolved

Find all $f : (-1,1) \to \mathbb R$ that is differentiable and $f(\frac {x+y}{1+xy})=f(x)+f(y)$

If you want the solution I'll post it.

this is well known.I mean it is classical Cauchy's functional equation for $tg x$ ;)

Indeed, just replace $x,y$ with $\tanh(a),\tanh(b)$, and you get the classical Cauchy functional equation for $g=f\circ\tanh$ ($\tanh$ is the hyperbolic tangent function).

P.S.

Omid, try to avoid putting "(" right after ":" inside $\LaTeX$ dollar signs. I think this is what caused the errors in the initial message (which I see you have corrected now).


limit
logarithms
real analysis
real analysis unsolved

Please help me to solve this  :( :
- let $ a>0 $ be real number (fixed) and , $ n>=1 $ a natural number :
$\ c_{1}= 1$ ;
$\ c_{n+1} = a^{\frac{1}{ (n + 1)!}} * c_{n}^{\frac{1}{(n + 1)}} $
$b_{n}= \prod_{k=1}^{n} c_{k} $

 1 ) Find   $c_{n} $ and $ b_{n}$

 2 ) Find the limit $\lim_{ n->{\infty}} \ b_n $

Step 1: Take logarithms of everything, so you are working with sums rather than products.

$C_1=0$
$C_{n+1}=\frac A{(n+1)!}+\frac1{n+1}C_n$
$B_n=\sum_{k=1}^n C_k$

Now, just try some stuff. Calculate a few values, and guess a pattern so you can prove it inductively.

i found a pattern but i don't know how to compute the limit:
$ \lim_{ n->{\infty}} \ b_n $ = $ \lim_{ n->{\infty}} \prod_{k=1}^{n} c_{k} $ = $ \lim_{ n->{\infty}} a^{\sum_{k=1}^{n} \frac{k}^{(k+1)!}} $

$\sum{\frac{k}{(k+1)!}}$ is a classical telescopic sum.Use the identity $\frac{k}{(k+1)!}=\frac{k+1-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$


calculus
integration
geometry
calculus computations

How do you find the surface area of the surface revolution generated by revolving the graph $ y\equal{}x^3$, $ 0\leq x \leq 9$ around the $ x$-axis

That area is given by $ 2\pi\int_0^9x^3\sqrt{1\plus{}9x^4}\,dx,$ and the last integral it's quite doable.


trigonometry
integration
calculus
calculus computations

do you know $ \cos ^{\minus{}1}(x)\plus{}\sin ^{\minus{}1}(x)\equal{}\frac{\pi }{2}$?

How do you prove that? If its a long proof then no need to

From $ (\arcsin x)' \plus{} (\arccos x)' \equal{} 0$ we have $ \int_0^x(\arcsin x)'\,dx \plus{} \int_0^x(\arccos x)'\,dx \equal{} \int_0^x(0)\,dx$
$ \equal{} > \arcsin x \minus{} \arcsin(0) \plus{} \arccos x \minus{} \arccos(0) \equal{} 0$
$ \equal{} > \arcsin x \minus{} 0 \plus{} \arccos x \minus{} \frac {\pi}{2} \equal{} 0$
$ \equal{} > \arcsin x \plus{} \arccos x \equal{} \frac {\pi}{2}$.

[quote="bos1234"]Prove that $ \frac {d}{dx}(sin^{ \minus{} 1}x \plus{} cos^{ \minus{} 1}x) \equal{} 0$, and hence FIND THE VALUE of $ y \equal{} sin^{ \minus{} 1}x \plus{} cos^{ \minus{} 1}x$

(i) Considering the case x>0, determine the value of y geometrically from a diagram involving a right-angle triangle

$ \frac {d}{dx}(sin^{ \minus{} 1}x \plus{} cos^{ \minus{} 1}x)$ = $ {1}/{\sqrt {1 \minus{} x^2}}$ - $ {1}/{\sqrt {1 \minus{} x^2}}$ = 0

How to do the other two aprts?

if %\frac{d}{dx}=0% then the graph of y has no gradient. Is the fact that we have to use?[/quote]

Assume $ y \equal{} \sin x$ then $ x \equal{} \arcsin y \in \left[ { \minus{} \pi ,\pi } \right]$. Now
\[ x'_y  \equal{} \frac{1}
{{y'_x }} \equal{} \frac{1}
{{\cos x}} \equal{} \frac{1}
{{\sqrt {1 \minus{} \sin ^2 x} }} \equal{} \frac{1}
{{\sqrt {1 \minus{} y^2 } }}
\]
this means that
\[ \left( {\arcsin x} \right)^\prime   \equal{} \frac{1}
{{\sqrt {1 \minus{} x^2 } }}.
\]
Similarly we obtain
\[ \left( {\arccos x} \right)^\prime   \equal{}  \minus{} \frac{1}
{{\sqrt {1 \minus{} x^2 } }}.
\]
Thus
\[ \left( {\arcsin x \plus{} \arccos x} \right)^\prime  \equal{} 0.
\]
...

Oh right I get it now. Maybe the question should have come with limits?

thanks everybody


How many 3-digit numbers have a first digit which is triple the final digit?

Then it won't work. The valid pairs are 3-1 6-2 and 9-3. Thus we have $ 3(10) \equal{} 30$

Won't work? -- or you get a different answer? I don't see that the question rules out 0 as the 3rd digit, so it seems that the answer is incorrect.

Generally, a three digit number has three digits whose leading digit is not 0.

What crazypianist1116 means is, if the units digit is $ 0$, then the hundreds digit will be $ 3*0 \equal{} 0$
So any number ending in 0 will not be a 3 digit number anymore because $ 0 \_ \_ \equal{} \text{a 2 digit number}$

[hide]There are only $3$ possibilities for the first and the third digit:

$\begin{list*}3, 1\\ 6, 2\\ 9, 3\end{list*}$

There are $10$ numbers to choose from as the middle digit.  Thus, there are is $3\times 10=\boxed{30}$ possible numbers meeting the requirements.[/hide]


Do you study music theory? What do you think of it?

Yes, I study music theory (piano, to be exact). Personally, I think its not much of a thing to be studied and comprises mostly of memorizing tricks and facts. And the books aren't very good either. I always think about how AoPS books are so much better when I see a page telling you to memorize a bunch of stuff. Maybe to some music theory is interesting. But for me, I have no interest.

I have some interest in music theory, maybe because many of my friends are musically inclined :P  But I wasn't able to take it due to lack of space in my schedule... darn AP Chemistry -.-;

[quote="t0rajir0u"]Music theory is an interesting subject.  You need to study the classical rules so you can learn how to break them properly.  Innovation in music is all about defying expectation:  you can't do it properly if you don't understand those expectations.[/quote]

Hmm... Well, do you have any good music theory books you recommend? Note: It needs to cover Certificate of Merit Exam material, because I take it every year.

I like Kostka/Payne's tonal harmony.  It's pretty much the standard for music theory nowadays, at least at the collegiate level.  I think it's easy enough to understand that a clever high schooler could teach him/herself from it without too much of a problem, so I'd recommend it.  I'd also recommend that you really train your ears and aural skills correspondingly as you study music theory (it makes it much more interesting and relevant to what you're doing in practice).

As for whether or not I'm interested... I think at least a basic understanding of music theory is necessary to really successful performing.  I feel like many of the strides I've made as a musician have either been or have been accompanied by similar advances in my theoretical knowledge of the structure of music.  It is true that much of the initial study of music theory is a bit tedious because of new vocabulary and such, but dismissing the subject entirely because of that would be much like dismissing mathematics because you find arithmetic difficult (which I think quite a few people do).

In fact, much of the deeper music theory you encounter is actually very much like solving a mathematical problem.  Harmonizing a melody so that you obey all of the standard rules of harmony, as a very simple example, is very much a problem-solving experience.  I would encourage you to look a little deeper into the application of music theory, rather than the vocabulary, if you want to enjoy it more.

Well, so far it's been pretty uninteresting, probably because I've not gotten to a very high level. Use [url]http://www.dolmetsch.com/theoryintro.htm[/url] as reference.

When I first found out that musical tuning was messed up (IMO), and how we use equal temperament, and how after listening to just intonation for a while equal temperament sounds bland, etc., music was less beautiful for me.  :|

I play guitar and there's tons of good books out there. You can check out Chord Chemisty by Ted Green for chord progressions. Mick Goodrick's Advancing guitarist is fun. Mel Bay's Complete book of harmony is good too.

I play both violin and piano. sometimes it gets a bit annoying, since the same terms may tell u to do different things for the two instuments. i particularly am not fond of shifting on either of the instruments.  :lol:

I love music theory, but then I did start college as a music composition major. I think it is extremely important to understand what the composer was doing if you want a good interpretation when you listen or play a piece.

@dreambold22: shifting is a pain. I don't bring my cello to school with me (usually three months at a time), so when I come home and try to play my positions are really imprecise. I am always amazed at how hard you have to work every day to try to get better shifts.

[quote="dreambold22"]i particularly am not fond of shifting on either of the instruments.  :lol:[/quote]

Ahem... in order to be able to REALLY play piano and not just be a beginner, you need to do... a LOT of shifting. Take La Campanella, for example, or any other "simple" piece. If one does not shift, then the maximum range of notes one can play is... F, F#, G, G#, A, A#, B, C, and maybe C#. 9 notes. La Campanella has... many notes. :lol:

Piano... shifting??!!  :huh:  :rotfl:  :rotfl: piano shifting!

That's funny.

Piano shifting.

Sorry.

@ yongyi781: well i'm sorry that even tho i've been playing piano for 3 years i'm going half the rate of improvement i'm supposed to go... i'm just not good at piano! i dont like piano much. i like violin much mor. i've only been playing violin for not even a year now and i'm already on the 4th suzuki book, going on the 5th soon. yes, i am MUCH better at violin.

Hmm, okay. Now that I've studied music theory a bit more, it's starting to make a bit more sense, if you get my drift. Although I still cannot say that it's very fun.

@ dreambold22 FYI there is no such thing as shifting on piano, you can play about 0 "grade 1" pieces (I don't know how the grade thing works, but you get the point) with fixed hand position. Also there's no such thing as a rate of progress that you "should be [making]". Possibly a vague one, but you can't exactly measure it.

I do musical theory off and on with my piano teacher, who's a professional musician, so the theory can go over my head sometimes.  I think it's useful for understanding pieces, and really helps me memorize them.

Oh true. I never even thought of that, but the theory really makes memorizing pieces so much easier.

Yes, I have been studying music theory for quite some time. Over the years, I have been dumbfounded at how closely mathematics and music are related. In fact, I once saw an interesting book about how some scientists took two tuning forks at different intervals, struck them, hung them, and recorded the path they take in the swing. Amazingly, they all had 1-, 2-, 3-, 5-, and 8-fold symmetry -- good ol' Fibonacci never fails us.

Personally I like Music Theory because it's really about math and logic. Most everything that you should "memorize" is really just logic that they figure everyone's too lazy to think through.

Out of curiosity, who here has perfect pitch? I self-taught out of Kostka/Payne's Tonal Harmony for the AP Music Theory this year, but it wasn't a lot of work because I didn't have to study at all for most of the free response or any of the listening ones. They say that you have to be born with perfect pitch, but I'm like 98% sure I developed it later on, because I remember when I was little I played a C once and I kept telling my mom the piano was out of tune because I swore it sounded like an E.

[quote="cyu2012"]I remember when I was little I played a C once and I kept telling my mom the piano was out of tune because I swore it sounded like an E.[/quote]
Whoa. When was the last time you tuned your piano?? :D

I myself have perfect pitch. I never trained for it, but trough my years of playing my instruments (piano & cello), I have developed it. Right now, I need to work on figuring out two notes at a time. (Right now, I always think everything a third up, for example C and E sounds like E and G to me...)

[quote="Dojo"]Right now, I always think everything a third up, for example C and E sounds like E and G to me...[/quote]

Are you sure you don't mean E and G#?

[quote="dreambold22"]i can do 1st, 3rd, and 5th position on violin, but not 2nd or 4th.  :([/quote]
I'm guessing that "half position" would fall into the same category for you as 2nd or 4th. What's that? Play G#-A-B-C# on your G string using fingers 1-2-3-4. That's half position.

There's an interesting role played by even-odd parity for violin and viola players that affects how fast and how accurately you can read music. It's this: if you are in an odd position (the ones you like, 1st, 3rd, 5th), then any note written on a line will be played with an odd finger (1 or 3) and any note written on a space will be played with an even finger (2 or 4). The problem with even positions - 2nd, 4th, 6th, half - is that they are parity-reversing. Every violinist or violist learns this, whether they realize it or not. Part of the problem with the even positions is that they tend to impose an extra step in the cognitive processing between eyes and fingers - unless one trains oneself to completely overcome this.

Transposing something up or down an octave is also parity-reversing and causes some of the same problems. That's not too much of a problem with something that written as 8va, because that's probably in who-knows-what position on your E string anyway - but it might become a problem if you transpose something down an octave to make it easier to play.

Presumably this whole discussion of even-odd parity makes no sense to a cellist.

[quote]Presumably this whole discussion of even-odd parity makes no sense to a cellist.[/quote]

Or to this guitarist.  Could you explain further?  I'm curious.

Basically, violins and violas are tuned in fifths, a fifth apart from each other. They're also sized so that a whole step between adjacent fingers is easily reached anywhere on the fingerboard. Without frets, you don't want beginners moving their fingers up and down the string too much; it's also common to mark some standard finger positions with tape on the fingerboard.
Put this all together, and a student learning violin or viola learns everything in "first position" to start. A two-octave C major scale on the viola is fingered $ 012^*3012^*3012_*3012_*$ on the viola ($ 2^*$ is the "high 2" and $ 2_*$ the "low 2". All of those zeros and twos are written on the spaces. All of those ones and threes are written on the lines. When you start using fourth fingers, they're usually the same pitch as the next open string up. This correspondence is part of how you learn to read music fast enough to matter.
The rule then continues a little. The next position a student learns to use is third position, with the first finger a fourth above the open string. It's easy to find by sound, and then by sheer repetition. If it's a choice between shifting between first and third several times or staying in second for a while, I'll usually take the shifting.

Incidentally, if you show me something in bass clef, I don't read it that way. I can barely read it at all, but what I do have is as a singer. That's all relative pitch for me.

One brain-breaker I've seen a few times as a violist (the one I remember right now is in the Siegfried Idyll): the B# down two ledger lines below alto clef. If I'm sightreading that, I probably don't play anything; it looks like an impossible note. On actually thinking, it's the open C string.

[quote="jmerry"]Incidentally, if you show me something in bass clef, I don't read it that way. I can barely read it at all, but what I do have is as a singer. That's all relative pitch for me.[/quote]
I have faint familiarity with bass clef from some long ago piano lessons.

I've been going to a community college class this summer that's a reading orchestra, with some of its meetings for chamber groups. A few nights ago, one of our members wanted to try out some chamber arrangements he'd made of some movie music. He's a cellist, and he'd written parts for violins, flute, clarinet, cello, and piano, leaving out viola. But he wanted to play the piano himself, and there was no other cello. I volunteered to play the cello part on viola. Sometimes I played the notes as written (bass clef may be strange to a violist, but even-odd parity still holds) and I transposed some notes up an octave (which is parity-reversing). Fortunately, the pieces were slow and the notes were long - I did survive, mostly. But it made for an interesting mental exercise. I said something above about an extra step in cognitive processing between eye and finger - this was more like two or three extra steps of cognitive processing.

The violinist's version of jmerry's B# would be an F double sharp. At least the rarity of the double sharp symbol would be a little extra warning that something was up.

This - enharmonics - is a place where wind players and keyboard players think differently than string players. A wind player has to know that G# and A flat are the same note, played the same way, and a keyboard player has to know that as well. A violinist/violist/cellist does [i]not[/i] have to know that. Oh, we know it intellectually, but at the instinctive/instant reaction level, we don't act like we know it. In first position on the D string, we'll play G# with a third finger and A flat with a fourth finger - in the same spot. We reach up for sharps and down for flats. jmerry's problem with the B# (and I've had the exact same problem - I know what he's talking about) is that it violates our usual thinking of not acknowledging enharmonics.

[quote]Basically, violins and violas are tuned in fifths, a fifth apart from each other. They're also sized so that a whole step between adjacent fingers is easily reached anywhere on the fingerboard.[/quote]
Cellos are also tuned in fifths, but the second sentence of this quote does not apply on the cello - a whole step of pitch is generally larger than the comfortable space between adjacent fingers. Guitars are very different because the tuning interval between strings is not constant - mostly fourths, but with one major third. So you can't generalize across different strings.

[quote]A two-octave C major scale on the viola is fingered $ 012^*3012^*3012_*3012_*$[/quote]
I think that for a cellist, that would be 013401340124012. Only to play some other scales, you'd have to shift - so cellists learn to shift sooner than violinists or violists.

[quote="cyu2012"]
Out of curiosity, who here has perfect pitch? [/quote]

I'm not sure if you've seen this, a discussion of perfect pitch in this forum awhile back:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=336841008&t=34600[/url]

[quote]Guitars are very different because the tuning interval between strings is not constant - mostly fourths, but with one major third. So you can't generalize across different strings.[/quote]

Ah.  Probably more to the point, a whole step is generally an uncomfortable reach with adjacent fingers.  I would never have tried it, so I would have never noticed a pattern even on the strings tuned fourths apart.

I think that music theory is quite important. It helps me the most in music composition and sight-reading. In music composition, you don't want to be estimating rhythms and notes; you want to know how it sounds in your head first. For sight-reading, all this theory comes in as notes, accidentals, rhythms, dynamics, and other important things.

[quote="Osud"]Ah.  Probably more to the point, a whole step is generally an uncomfortable reach with adjacent fingers. [/quote]
That's the same as the situation with a cello. I don't actually know how the string length (bridge to bridge) of a guitar compares to that of a cello, but both are significantly longer than a viola, which is in turn longer than a violin. And a string bass (or bass guitar) is longer yet.

Getting back to the subject of music theory: I was just thinking about a particular passage in a particular piece, and thought of something theoretical that helps explain to me what is going on and maybe why it works so well.

Most Western music from about 1600 to 1900, along with some Western music that's older or younger than that, is strongly based on either the major key or mode or the minor key or mode. If you've been trained in that music, you've been trained to play scales, both major scales and minor scales.

Chromaticism tends to contrast with major/minor tonality. And a complete chromatic scale - playing every half step - would seem to exist in denial of major/minor tonality. But can the same be said of a fraction of a chromatic scale? Major scales don't allow much room, but minor scales are different just because there are three different versions of a minor scale.

Suppose we think about the key of D minor. The lowest five notes of the D minor scale - D, E, F, G, A - are fixed in any minor scale. But the upper notes vary. There's the natural minor, played according to the one-flat key signature, in which the upper notes are B flat, C, D. There's the harmonic minor, which incorporates the notion of a leading tone and has as its upper notes B flat, C sharp, D - with that step-and-a-half interval in there. And then there's the melodic minor: B, C sharp, D going up, but D, C, B flat going down. The point I'm trying to make is that all of the chromatic notes in there - B flat, B, C, C sharp - all exist in at least some version of the D minor scale. And using those chromatic notes might not be a denial of the minor tonality at all but a reinforcement of it.

Now, here's the passage. We're in 2/4 time, not too fast, and at first we have steady 8th notes, making two-measure phrases of 8 notes each.

First: chromatic down, melodic up: D, C sharp, C, B, B flat, A, B, C sharp. Repeat that pattern several times.

Second: the harmonic minor pattern: D, A, B flat, A, G, B flat, C sharp. Repeat that twice.

Third: a new rhythmic patten, with more than one note on each pitch, stressing each pitch for about a measure. The order uses several of these pitches in an order somewhat like the melodic minor: C, C, B, C sharp. Do that twice.

And finally - inevitably, because we're overwhelmingly and unmistakeably in D minor - a D minor descending arpeggio. And back up to the finish.

The chromatic notes do not deny D minor, they emphasize it. (And perhaps you recognize the piece and who gave us that pattern?)

[quote]I don't actually know how the string length (bridge to bridge) of a guitar compares to that of a cello, but both are significantly longer than a viola, which is in turn longer than a violin. And a string bass (or bass guitar) is longer yet.
[/quote]

A string length of 650 mm is more or less standard for modern classical guitars, but somewhat longer or shorter string lengths are also possible.

[quote](And perhaps you recognize the piece and who gave us that pattern?)[/quote]

Actually, I don't, but then again, I'm not familiar with most repertory outside of the guitar's.

[quote="Osud"]A string length of 650 mm is more or less standard for modern classical guitars, but somewhat longer or shorter string lengths are also possible.[/quote]
I made some measurements for a presentation I did on the one dimensional wave equation and string instruments. I measured my viola's bridge-to-bridge length as 371 mm and my violin as 324 mm. So you're saying that the strings of a guitar are almost exactly twice as long as the strings of a full-sized violin.

As part of the same presentation, I also destroyed some used-up strings so I could weigh measured lengths of them (on a chemistry department's balances) to get an estimate of the mass per unit length, which I could then use to determine the tension on the strings. The tension worked out to around 10-15 pounds (45-65 N). If I had to guess, I'd guess that guitar strings are under a tension several times as great as that. Of course the best way to be sure would be to weigh some guitar strings to determine the mass per unit length.

I'll just wait to see if anyone else recognizes the music I was talking about.

[quote]The tension worked out to around 10-15 pounds (45-65 N). If I had to guess, I'd guess that guitar strings are under a tension several times as great as that. Of course the best way to be sure would be to weigh some guitar strings to determine the mass per unit length. [/quote]

That's surprising.  I checked the specs for D'Addario, and their "normal" tension strings range from 11.6-15.6 pounds, while their "extra hard" tension strings (what I use) range from 12.5-16.4 pounds.  (Source:  [url]http://store.daddario.com/category/146041/EJ45_Pro-Arte_Normal_Tension[/url] and [url]http://store.daddario.com/category/146033/EJ44_Pro-Arte_Extra-Hard_Tension[/url])

For comparison, my 13 course lute has a string length of 700 mm.  I don't know the tension of its top course, but one source gives 37.5 N as a reasonable figure.

Interesting. I think I was being fooled by the different-looking pegs.

So the tension on guitar strings is about the same as the tension on violin strings. The highest-pitched string on a guitar - the upper E - is one octave below a violin E string in pitch. And the guitar string is twice as long. That means that the mass per unit length of those two strings has to be essentially the same. A violin E string is very thin and light; it can be a simple wire. I guess that means that your guitar (upper) E is also very thin and light. The pitch difference between your top and bottom strings is two octaves, which is a third more than the difference between the violin's top and bottom strings. I would then expect your bottom E string to be just a little thicker and heavier than a violin G string - but not as thick and heavy as my viola C string.

Actually, they go for dramatic differences in the materials used. The bottom three strings are at about the same thicknesses as the top three strings, but are much heavier. For the linked sets, it's bare nylon for upper strings and silver-wound nylon for lower strings. Nylon doesn't quite float, but it's close. Silver is ten times as dense.

i have perfect pitch and i play violin. that helps when i shift. i have gone to like 12th position! and did F sharp major in 3 octaves.

speaking of out of tune, this music teacher at my old school tunes her A string (and the whole orchestra's) to an A sharp. OUCH

I take music Theory tests every year. I have completed level six for the Virgina Music Teachers Association tests. I hope to be able to take the AP music theory test.

[quote="Yongyi781"]By the way, the AP Music Theory test was today, which I took.[/quote]
Death.
Let's say that test was probably the hardest test I have taken short of USAMO.

Outside of dictation, the AP Music Theory is pretty easy.

But then again, dictation is pretty killer and self-esteem lowering.

Funny how dictation was the part that I actually didn't completely bomb.
I honestly had no clue what half the terms were, and I actually sat through 2 semesters of AP music theory.

haha, I haven't gotten to AP testing level yet, I'm only in seventh grade. But is sounds so fun! err, I mean horrible!