limit
function
real analysis
real analysis theorems

this is a proof that i sorta came up with for l'hopitals rule.  i saw a proof one time but it didnt give any reguard to the indeterminat forms, this is a slight modification that i've done.  i've seen other proofs that are far more complex so i now have doubts that mine's correct.

i start out with the alternative form for derivitives:

$\lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$
let x + h = u,
h = u - x
as h approaches zero, u approachs x
$\lim_{u \to x} \frac{f(u) - f(x)}{u-x}$

l'hopitals rule:

given:
$\lim_{x\to a} f(x) = 0$
$\lim_{x\to a} g(x) = 0$

$\frac{f(x)}{g(x)} = \frac{\frac{f(x)-0}{x-a}}{\frac{g(x)-0}{x-a}} = \frac{\lim{x\to a}\frac{f(x)-f(a)}{x-a}}{\lim{x\to a}\frac{g(x)-g(a)}{x-a}} = \frac{f'(x)}{g'(x)}$, at x = a

i've stemmed the rest of the indeterminant from this proof: $\pm\frac{\infty}{\infty}$, just say $\lim_{x\to a} \frac{u}{v} = \frac{\infty}{\infty}$ where u and v are functions of x, $u = 1 / g(x)$ and $v = 1 / f(x)$.

$0 * \infty$ is equavelent to $\frac{0}{1 / \infty}$ or $\frac{0}{0}$  $1^{\infty}$ is $ln(1) * \infty$ or $0 * \infty$ and so on for the rest of the indeterminant forms

You *really* need the mean value theorem, because that is what connects the values of $f(x)$ and $f'(x)$, not so much taking a million limits at once, which may correspond to different $\epsilon$s and $\delta$s.

is that for $\frac{f^{(d)}(x)}{g^{(d)}(x)}$ isnt indeterminant?  if no, could you explain the idea behind using the mean value theorem?  the complex proofs i've seen all use that, but the one in my book just has for a given z, there exists... but doesnt show that it is z.


limit
topology
geometry
incenter
geometry proposed

Given triangle $ABC$ inscribed $(O)$ three bisectors of $ABC$ are concurrent at $I$ and intersect $(O)$ formed a triangle $A_{1}B_{1}C_{1}$, three bisectors of $A_{1}B_{1}C_{1}$ are concurrent at $I_{1}$ and intersect $(O)$ formed a triangle $A_{2}B_{2}C_{2}$ similarly we have sequence $A_{n}B_{n}C_{n}$ and $I_{n}$. Now find the limit
$\lim_{n\rightarrow\infty}I_{n}$

You understand quite highly, here is only elementary box of course the topology is  induced by Euclide's metric of $R^{2}$ is $d(x,y)=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$
That's all
[quote="kunny"]Let $A_{n}B_{n}C_{n}$ be the triangle with side lengths $B_{n}C_{n}=a_{n},\ C_{n}A_{n}=b_{n},\ A_{n}B_{n}=c_{n}.$

We have $\overrightarrow{OI_{n}}=\frac{a_{n}\overrightarrow{OA_{n}}+b_{n}\overrightarrow{OB_{n}}+c_{n}\overrightarrow{OC_{n}}}{a_{n}+b_{n}+c_{n}}.$ We are to make the recursive formula for $\overrightarrow{OI_{n}}.$[/quote]
I don't understand your ideal, your fomular is well know, everybody can know it, post all solution please!

kunny the best you should post all solution instead of the definition of incenter
 :D


$ 42$
would be my guess, but i can't think of a reasonable way of solving this in 45 seconds.

The quotient  numbers have to be even because you cant use the digit 5.
even multiples are....
270
540 (has a 5 and thus cant be used)
810 (has a 1 and thus cant be used)
...then you cant use 1080 and 1350 because they have 4 digits " * * * * "
so the only digit that can go in the quotient is 2
ANS: 22


I set my blog as available only to users in order to prevent its contents from being publicly available.  However, I can now find cached versions of my blog using Google, even when I'm not on.  I believe that this is due to bots such as googlebot, etc.  Therefore, would it possible for the bots to be banned from the users-only blogs (although I think that there might be one, Yahoo! slurp, who is in fact a human)?  Or can we do this ourselves with the "enemies" function?

Thank you very much.

OK, we'll take a look.

I'm reluctant to make global changes (i.e. make all blogs invisible to Google) because I assume that most people *do* want their blogs to be visible to search engines.

Thanks.  Would it be feasible to make only the blogs which are user-view only invisible to Google?  (Presumably those people are those who don't want their blogs published.)

That may be possible.  I'll look into it (or, rather, I'll ask the people who actually know about this stuff to look into it).

[quote="Phelpedo"]On this subject, has the ability to make new blogs been taken away? I can't figure out how to make one   :([/quote]
Just go to your profile, and one of the tab options on the left should be "Create a New Blog" (or something like that). Click on it, choose the settings for your blog, and submit them to create the blog. (However, note that once you create a blog you can't delete it, which is what happened to me.  :roll:  :P )


Let $ a \geq b \geq c \geq 0 $
  Prove that :
 $ \frac{a^2-b^2}{c} + \frac{b^2-c^2}{a} + \frac{c^2-a^2}{b} \geq 3a - 4b +c $

Hmmm,.... :D  so I guess this inequalities will do the hard work here:

            $\frac{a+b}{c}\geq2, 0<\frac{b+c}{a}\leq 2, \frac{a+c}{b}\geq 1$  ;)

:roll:  This is not true, for counter-example,  a=4, b=1, c=3/4.


modular arithmetic
number theory unsolved
number theory

I havethe following question:

Show that if a composite number n passes the Fermat test to a base a, but fails the Miller-Rabin test for the same base then we can factor the number n.

Please offer help, thanks...

If $ n$ passes Fermat test but fails Miller-Rabin test for a base $ a$ then we have such integer $ k$ that
\[ \begin{cases} a^k\not\equiv \pm 1\pmod{n} \\
a^{2k}\equiv 1\pmod{n}\end{cases}\]
In other words, $ n$ divides $ a^{2k} \minus{} 1 \equal{} (a^k \minus{} 1)(a^k \plus{} 1)$ but neither of $ a^k \minus{} 1$ and $ a^k \plus{} 1$. That is, $ \gcd(n,a^k \minus{} 1)$ and $ \gcd(n,a^k \plus{} 1)$ give nontrivial factors of $ n$.


$ A\equal{} \left[\begin{array}{ccccc} 1 & 1& 1& \cdots & 1\\1& 2& 2& \cdots & 2\\1& 2& 1& \cdots & 1\\\cdots\\1& 2& 1& \cdots \end{array}\right]$
find $ A^{\minus{}1}$

Are the 3rd through $ n$th rows of this matrix identical? That's what it looks like, although you could be more explicit with the notation. If so, then the determinant is zero and $ A^{\minus{}1}$ doesn't exist.

Except for the $ 3\times 3$ case where the usual Gaussian elimination computation produces the inverse, which happens to have all its entries integers.

[hide="Answer in that case"]$ A^{\minus{}1}\equal{}\begin{bmatrix}2&\minus{}1&0\\\minus{}1&0&1\\0&1&\minus{}1\end{bmatrix}$[/hide]

I believe Sunlight means something like (exemplified for the 5x5 case)

$ \left[\begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 & 2 \\ 1 & 2 & 1 & 1 & 1 \\ 1 & 2 & 1 & 2 & 2 \\ 1 & 2 & 1 & 2 & 1 \end{array}\right]$

The inverse is of the form a tridiagonal matrix like

2  -1  0.......  0
-1  0  1 0..... .0
0  -1  0 -1 0...0
.....................
0........... ......W

where W is two by two matrix if the size of A is even

W=
(0  1)
(1  -1)

if the size of A is odd take -W

yes, and it is one problem in IMC 1994 :)


Olga purchases a rectangular mirror
(the shaded region) that fits exactly
inside a frame. The outer perimeter of
the frame measures 60 cm by 80 cm.
The width of each side of the frame
is 10 cm. What is the area of the
mirror?

since the width of the frame is 10 cm, that means the actual mirror is 20 cm less in length and width
so we have.,...

$ 40 \cdot 60 \equal{} 2400$


trigonometry
ratio
complex numbers
geometric series
algebra

Prove that the average of the numbers $n \sin n^{\circ} \; (n = 2,4,6,\ldots,180)$ is $\cot 1^{\circ}$.

[hide="Alternate Solution"]Instead of proving that
\[\frac{2\sin 2^\circ+4\sin 4^\circ+6\sin 6^\circ+\cdots+180\sin 180^\circ}{90}=\cot1^\circ,\]we'll prove that
\[2\sin 1^\circ\sin 2^\circ+4\sin 1^\circ\sin 4^\circ+6\sin 1^\circ\sin 6^\circ+\cdots+180\sin 1^\circ \sin 180^\circ=90\cos 1^\circ.\]
By the Product to Sum formulas, our sum simplifies to
\begin{eqnarray*}\sum_{n=1}^{90}2n \sin 1^\circ \sin 2n^\circ &=& \sum_{n=1}^{90}n[\cos(2n-1)^\circ-\cos(2n+1)^\circ] \\ &=&-90\cos 181^\circ+\sum_{n=1}^{90}\cos(2n-1)^\circ .\\ &=& 90\cos 1^\circ+\sum_{n=1}^{90}\cos(2n-1)^\circ . \end{eqnarray*} It now suffices to show that $ \sum_{n=1}^{90}\cos(2n-1)^\circ =0.$ Using the identity $\cos n^\circ =-\cos (180-n)^\circ,$
\begin{eqnarray*}\sum_{n=1}^{90}\cos(2n-1)^\circ &=& \cos 1^\circ+\cos 3^\circ++\cdots+\cos 89^\circ+\cos 91^\circ+\cdots+\cos 177^\circ+179^\circ \\ &=& \cos 1^\circ+\cos 3^\circ+\cdots+\cos 89^\circ-(\cos 89^\circ+\cdots+\cos 3^\circ+\cos 1^\circ)\\ &=& 0. \end{eqnarray*}$\mathbb{QED}$
[/hide]

[hide="complex numbers"]

For now, disregard $n=180$, and consider the sum $\sum_{k=1}^{89}2k\sin{2k^{\circ}}$. By the fact that $\sin{\theta}=\sin{(180^{\circ}-\theta)}$, we have that the first sum is equal to $\sum_{k=1}^{89}(180-2k)\sin{2k^{\circ}}$.

Now, we add these two sums together and divide by two, to find that they are equal to $90\sum_{k=1}^{89}\sin{2k^{\circ}}$.

$\sin{180^{\circ}}=0$, when we add that to the sum, the sum stays the same. Since we are taking the average of 90 terms, all that is left to show is that

${\sum_{k=1}^{90}\sin{2k^\circ}}=\cot{1^{\circ}}$.

Note that this is equal to $\Im({\sum_{k=1}^{90}\text{cis }2k^{\circ})}$, which we can express, by Euler's formula, as $\Im(\sum_{k=1}^{90}e^{2k\pi i/180})$. This is the imaginary part of a finite geometric series with first term $e^{2k\pi i/180}$, common ratio $e^{2k\pi i/180}$, and 90 terms. By the formula for the sum of a geometric series, we have

$\Im(\sum_{k=1}^{90}e^{2k\pi i/180})$
$=\Im(\frac{e^{2\pi i/180}(e^{\pi i}-1)}{e^{2\pi i/180}})$
$=\Im(\frac{-2e^{\pi i/90}}{e^{\pi i/90}-1})$
$=\Im(\frac{-2e^{\pi i/90}(e^{-\pi i/90-1})}{(e^{\pi i/90}-1)(e^{\pi i/90}-1)})$
$=\Im(\frac{-2+2e^{\pi i/90}}{2-e^{-\pi i/90}-e^{\pi i/90}})$
$=\Im(\frac{-2+2\cos{2^{\circ}}+2i\sin{2^{\circ}}}{2-\cos{(-2^{\circ})}-i\sin{(-2^{\circ})}-\cos{2^{\circ}}-%Error. "isin" is a bad command.
{(-2^{\circ})}})$
$=\Im(\frac{-2+2\cos{2^{\circ}}+2i\sin{2^{\circ}}}{2-2\cos^{2^{\circ}}})$
$=\frac{\sin{2^{\circ}}}{1-\cos{2^{\circ}}}$

The cotangent half angle formula tells us that $\cot{\frac{\theta}{2}}=\frac{\sin{\theta}}{1-\cos{\theta}}$. Applying this, we find that

$\cot{\frac{2^{\circ}}{2}}$
$=\cot{1^{\circ}}$, as desired. $\Box$

$\mathbb{QED}$
[/hide]

This type of problem is trivial once you know about complex numbers; I believe that they should not be considered mathematical Olympiad problems anymore.

Well, this problem is flattened by the following formula:

Let $a_1,a_2,\cdots ,a_n$ be an A.P. with common difference $d$. Then
$\sin a_1+\sin a_2+\cdots +\sin a_n= \frac{\sin(\frac{a_1+a_n}{2})\sin (\frac{nd}{2})}{\sin(\frac{d}{2})}$.

As a sidenote, we also have
$\cos a_1+\cos a_2+\cdots +\cos a_n= \frac{\cos(\frac{a_1+a_n}{2})\sin (\frac{nd}{2})}{\sin(\frac{d}{2})}$.

My proof is nice and simple:

$2\sin2\sin1+2(2\sin4\sin1)+3(2\sin6\sin1)+....+90(2\sin180\sin1)$

$=\cos1-\cos3+2(\cos3-\cos5)+3(\cos5-\cos7)+....+90(\cos179-\cos181)$

$=\cos1+\cos3+\cos5+...+\cos179-90\cos181$

$=(\cos1+\cos179)+(\cos3+\cos177)+...+(\cos89+\cos91)+90\cos1$

$90\cos1$.

Dividing the first and the last line by $90\sin1$ the result follows....

[u][b]Note[/b][/u]:All the angles are in degrees.

Sorry to revive, but I don't think I'm just restating anything here.

[hide]
We apply [b]vincenthuang bashing[/b]

Recall that if $\omega = cis(\theta)$, $|\omega| = 1$.
Now observe that

\[ \frac{\omega-\omega^{-1}}{2i} = \frac{\omega - \frac{|\omega|^2}{\omega}}{2i} = \frac{2i\sin{\theta}}{2i} = \sin{\theta} \]

Now observe that the desired sum is

\[ 2\sin{2} + 2\sin{4} \cdots +2\sin{88}+1\]

Let $\omega = cis(2)$ and observe that $\cot{1} = \frac{i(\omega^{\frac{1}{2}} + \omega^{\frac{-1}{2}})}{\omega^{\frac{1}{2}} - \omega^{\frac{-1}{2}}} = \frac{i(\omega+1)}{\omega-1}$

The desired sum is

\[ 2\sin{2} + 2\sin{4} \cdots +2\sin{88}+1\]

\[= \frac{\omega-\omega^{-1}}{i} + \frac{\omega^2-\omega^{-2}}{i} \cdots +\frac{\omega^{44}-\omega^{-44}}{i}+1\]

\[= \frac{(\omega^{44}+\omega^{43}\cdots + \omega)-(\omega^{-44} + \omega^{-43} \cdots + \omega^{-1})}{i} +1\]

\[= \frac{(\omega^{45}-1)\cdot(\frac{1}{\omega} + \frac{1}{\omega^2} \cdots +\frac{1}{\omega^{44}})}{i} +1\]

\[=\frac{(i-1)(\frac{\omega}{\frac{\omega^{44}-1}{\omega-1}})}{i} + 1 \]

\[=\frac{(i-1)(i-\omega)}{(\omega-1)(i)(\omega^{45})}+1\]

\[=(1-i)(\frac{i-\omega}{\omega-1}) + 1 \]

\[ = \frac{(1-i)(i-\omega)+\omega-1}{\omega-1} \]

\[ = \frac{i+1-\omega+i\omega+\omega-1}{\omega-1} \]

\[ = \frac{i\omega+i}{\omega-1} \]

\[ = \frac{i(\omega+1)}{\omega-1} \]

\[ = \cot{1} \], as desired.
[/hide]

im pretty bad at solving problems and my solutions are like super long. also im bad at proof writing too

[hide=Solution]Let $S$ be the sum of the numbers $n \sin n^{\circ} \; (n = 2,4,6,\ldots,180).$ In summation notation, $\boxed{S = \sum_{n=1}^{90} 2n \sin (2n)^\circ} .$
Note: Although $\sin 180^\circ$ is $0$, the problem asks for the [b]average[/b] so it is important to keep in mind that $S$ has 90 terms.
Let $T$ be the sum of the numbers $n \sin (180 - n)^{\circ} \; (n = 2,4,6,\ldots,178).$ In summation form, $T = \boxed{ \sum_{n=1}^{89} 2n \sin (180-2n)^\circ} .$ We want to show that the [b]values[/b] of $S$ and $T$ are equal. Because $\sin \theta = \sin (180 - \theta)$ for any $\theta ,$ we can change $T$ from $\sum_{n=1}^{89} 2n \sin (180-2n)^\circ$ to $\sum_{n=1}^{89} 2n \sin (2n)^\circ .$ Without the last term, the value of the two sums would be equal, but $180\sin 180^\circ = 0$, so $S = T.$
Now notice what happens when we add $S$ to $T.$
We know $S+T$ is just $2S$; this will be the LHS. However, notice that for every term $2n \sin (2n)^\circ$ in $S$, there is a corresponding term $(180 - 2n)\sin (2n)^\circ $ in $T$, and these terms sum to $180\sin (2n)^\circ .$ So, $S + T = 2S = \sum_{n=1}^{90} 180 \sin (2n)^\circ $ (don’t forget the $180\sin 180^\circ$!)
The original average is $\dfrac{S}{90} = \dfrac{2S}{180} = \sum_{n=1}^{90} \sin (2n)^\circ .$ We want to show that this is equal to $\cot 1^\circ .$
Since I am not smart enough, I had to prove it with complex numbers (used radians instead because that is the proper notation when using exponential form).
Because $e^{i \theta} = \cos \theta + i\sin \theta ,$ we can change $\dfrac{S}{90}$ into $\sum_{n=1}^{90} \text{Im} \left( e^{2n\pi i / 180} \right) = \text{Im} \left( \sum_{n=1}^{90} e^{2n\pi i / 180} \right) .$
The sum we are taking the imaginary part of is a 90-term geometric series with first term $e^{2\pi i / 180} = e^{\pi i / 90}$ and common ratio $e^{\pi i / 90} .$ So, using the geometric sum formula, we have \begin{align*} \text{Im} \left( \dfrac{e^{\pi i /90} (e^{90\pi i /90} - 1)}{e^{\pi i /90} - 1} \right) \\ \\ =\text{Im} \left( \dfrac{-2 e^{\pi i /90}}{e^{\pi i /90} - 1} \right) \\ \\ =\text{Im} \left( \dfrac{-2 \cos \tfrac{\pi}{90} - 2i\sin \tfrac{\pi}{90}}{\cos \tfrac{\pi}{90} - 1 + i\sin \tfrac{\pi}{90}} \cdot \dfrac{\cos \tfrac{\pi}{90} - 1 - i\sin \tfrac{\pi}{90}}{\cos \tfrac{\pi}{90} - 1 - i\sin \tfrac{\pi}{90}} \right) \\ \\ =\text{Im} \left( \dfrac{-2\cos^2 \tfrac{\pi}{90} +2\cos \tfrac{\pi}{90} + 2i\cos \tfrac{\pi}{90}\sin \tfrac{\pi}{90} - 2i\cos \tfrac{\pi}{90}\sin \tfrac{\pi}{90} + 2i\sin \tfrac{\pi}{90} - 2\sin^2 \tfrac{\pi}{90} }{(\cos \tfrac{\pi}{90} - 1)^2 + \sin^2 \tfrac{\pi}{90}} \right) \\ \\ =\dfrac{2\sin \tfrac{\pi}{90}}{\cos^2 \tfrac{\pi}{90} - 2\cos \tfrac{\pi}{90} + 1 + \cos^2 \tfrac{\pi}{90}} = \dfrac{2\sin \tfrac{\pi}{90}}{2 - 2\cos \tfrac{\pi}{90}} = \dfrac{2\sin \tfrac{\pi}{90}}{2 - 2\cos \tfrac{\pi}{90}} = \dfrac{1\sin \tfrac{\pi}{90}}{1 - \cos \tfrac{\pi}{90}} \\ \\ =\dfrac{1}{\dfrac{ 1 - \cos \tfrac{\pi}{90} }{1\sin \tfrac{\pi}{90}}} = \dfrac{1}{\tan \dfrac{ \tfrac{\pi}{90}}{2}} = \cot \tfrac{\pi}{180} = \cot 1^\circ . \blacksquare . \end{align*}[/hide]

$\sum_{n = 1}^{90} (2n) \sin (2n) = \left( \sum_{n = 1}^{44} 180 \sin (2n) \right) + 90 \sin 90$, so the average of the numbers is $\left( \sum_{n = 1}^{44} 2 \sin (2n) \right) + 1$. Note that $\sum_{n = 1}^{44} 2 \sin 1 \sin (2n) = \sum_{n = 1}^{44} \cos (2n-1) - \cos (2n + 1)$, which clearly telescopes to $-\cos (89) + \cos (1)$. Thus, $$\sum_{n = 1}^{44} \sin (2n) = \frac{-\cos (89) + \cos (1)}{\sin (1)} = \frac{-\sin 1 + \cos 1}{\sin 1} = -1 + \cot 1$$
The problem statement follows.

Nothing too new here. We use "Vincent Huang bashing."
[hide=Solution]
Let $z=e^{\frac{i\pi}{180}}$. Then
\begin{align*}
&\sum_{k=1}^{90} 2k(\sin{2k^{\circ}})\\
=&\sum_{k=1}^{90} 2k\cdot \frac{1}{2i}\left(z^{2k}-\frac{1}{z^{2k}}\right)\\
=&\frac{1}{i}\sum_{k=1}^{90}k\left(z^{2k}-\frac{1}{z^{2k}}\right)\\
=&\frac{1}{i}\left[\left(z^2+2z^4+3z^6+\dots+90z^{180}\right)-\left(\frac{1}{z^2}+\frac{2}{z^4}+\dots+\frac{90}{z^{180}}\right)\right]
\end{align*}
The key observation here is that $z^{180}=-1$. So we multiply everything on the right-hand section by $-z^{180}$. Then it becomes
\begin{align*}
&\frac{1}{i}\left[\left(z^2+2z^4+3z^6+\dots+90z^{180}\right)-\left(\frac{1}{z^2}+\frac{2}{z^4}+\dots+\frac{90}{z^{180}}\right)\right]\\
=&\frac{1}{i}\left[\left(z^2+2z^4+3z^6+\dots+90z^{180}\right)+\left(z^{178}+2z^{176}+\dots+90\right)\right]\\
=&\frac{1}{i}\left(90z^2+90z^4+\dots+90z^{178}+90z^{180}+90\right)
\end{align*}
But again $z^{180}=-1$, so $90$ and $90z^{180}$ cancel. This is a geometric series, and equals
\begin{align*}
\frac{90}{i}\left(\frac{z^{180}-1}{z^2-1}-1\right)=\frac{90}{i}\left(\frac{z^{180}-z^2}{z^2-1}\right)=90i\left(\frac{z^2+1}{z^2-1}\right)=90\cdot \frac{\cos 1^{\circ}}{\sin 1^{\circ}},
\end{align*}
as desired. $\blacksquare$
[/hide]

[hide=Storage]
(Omitting the degree symbol for ease).
$$\frac{1}{90}\sum_{i=1}^{90}2i\sin 2i=\frac{1}{90}(2\sin 2+178\sin 178+4\sin 4+176\sin 176+\cdots+90\sin 90)$$$$=\frac{1}{90}\left(180(\sin 2+\sin 4+\cdots+\sin 88)+90\right)=1+2\sum_{i=1}^{44}\sin 2i$$$$=1+2\sum_{i=1}^{22}2\sin 45\cos(2i-1)=1+2\sqrt2\sum_{i=1}^{22}\cos(2i-1)\quad\because\text{Sum to Product Identities}.$$Now, since $\cot 1=\frac{\cos 1}{\sin 1}$, we can multiply through by $\sin 1$ and show that the resulting expression is equal to $\cos 1$.
$$\sin 1\left(1+2\sqrt2\sum_{i=1}^{22}\cos(2i-1)\right)=\sin 1+2\sqrt2(\cos1\sin1+\cos3\sin1+\cdots+\cos43\sin1)$$$$=\sin1+\sqrt2(\sin2+\sin0+\sin4-\sin2+\cdots+\sin44-\sin42)\quad\because\text{Product to Sum Identities}$$$$=\sin 1+\sqrt2\sin 44=\sin 1+2\sin 44\cos 45$$$$=\sin 1+\sin 89-\sin 1=\sin 89=\cos 1\quad\because\text{Product to Sum Identities}.$$
[/hide]

$\dagger \ \color{green}{\textrm{\textbf{A different solution using Abel's Summation by Parts:}}}$
As we know that $\color{blue}{\sin (a)+\sin (a+d)+\sin (a+2d)+\ldots +\sin (a+(n-1)d)=\frac{\sin \left(a+\frac{n-1}{2}d\right)\cdot \sin \left(\frac{n}{2}d\right)}{\sin \left(\frac{d}{2}\right)}}$ and $\color{magenta}{\cos (a)+\cos (a+d)+\cos (a+2d)+\ldots +\cos (a+(n-1)d)=\frac{\cos \left(a+\frac{n-1}{2}d\right)\cdot \sin \left(\frac{n}{2}d\right)}{\sin \left(\frac{d}{2}\right)}}$, we apply Abel's Summation by Parts Formula to get:
\begin{align*}
&\frac{1}{90}\sum_{n=1}^{90}2n\sin (2n) \\ =&\frac{1}{90}\left(180\cdot \frac{\sin (91)}{\sin (1)}-\frac{2}{\sin (1)}\sum_{n=1}^{89} (\sin(n+1)\cdot \sin (n))\right) \\ =&2\cot (1)-\frac{1}{90\sin (1)} \sum_{n=1}^{89} (\cos (1) - \cos (2n+1)) \\=& 2\cot (1)-\cot (1)+\frac{2}{\sin (1)}\sum_{n=0}^{89} \cos (2n+1) \\ =& \cot (1)+\frac{2}{\sin (1)} \cdot \frac{\cos (1+89)\cdot \sin (90)}{\sin (1)}\\ =&\cot (1). \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \blacksquare
\end{align*}

Solution from [i]Twitch Solves ISL[/i]:

Because \[ n \sin n^{\circ} + (180-n) \sin(180^{\circ}-n^{\circ}) = 180 \sin n^{\circ} \] So enough to show that \[ \sum_{n=0}^{89} \sin (2n)^{\circ} = \cot1^{\circ} \] Let $\zeta = \cos2^{\circ} + i \sin 2 ^{\circ}$ be a primitive root. Then \begin{align*} 	\sum_{n=0}^{89} \frac{\zeta^n - \zeta^{-n}}{2i} 	&= \frac{1}{2i} 	\left[ \frac{\zeta^{90}-1}{\zeta-1} - \frac{\zeta^{-90} - 1}{\zeta^{-1}-1} \right] \\ 	&= \frac{1}{2i} 	\left[ \frac{-2}{\zeta-1} - \frac{-2}{\zeta^{-1}-1} \right] \\ 	&= \frac{1}{-i} \frac{\zeta^{-1}-\zeta}{(\zeta-1)(\zeta^{-1}-1)} 	= i \cdot \frac{\zeta+1}{\zeta-1}. \end{align*} Also, \begin{align*} 	\cot 1^{\circ} &= \frac{\cos 1^{\circ}}{\sin 1^{\circ}} 	= \frac{(\cos1^{\circ})^2}{\cos1^{\circ}\sin1^{\circ}} \\ 	&= \frac{\frac{\cos2^{\circ}+1}{2}}{\frac{\sin2^{\circ}}{2}} 	= \frac{\frac{1}{2}(\zeta+\zeta^{-1})+1}{\frac{1}{2i}(\zeta-\zeta^{-1})} \\ 	&= i \cdot \frac{(\zeta+1)^2}{\zeta^2-1} = i \cdot \frac{\zeta+1}{\zeta-1}. \end{align*} So we're done.

not a desirable method oops

Work in radians.

Let $S$ be the set of even numbers between $2$ and $180$ inclusive. Then, it suffices to show that $\sum_{n\in S} n\sin \frac{n\pi}{180} = 90 \cot \frac{\pi}{180}$. However, since $n \sin n = \Im(ne^{n\pi i / 180})$, it suffices to show that $\sum_{n\in S} \Im(ne^{n\pi i / 180}) = \Im(\sum_{n\in S}ne^{n\pi i / 180}) = 90 \cot \frac{\pi}{180}$.

Now, let $A = \sum_{n\in S}ne^{n\pi i / 180}$. We can write out $A$ and $e^{2\pi i / 180}A$ out and subtract to get $$(1 - e^{2\pi i / 180})A = 2(e^{2i\pi/180} + e^{4\pi i / 180} + \cdots + e^{180\pi i / 180}) + 180e^{2\pi i / 180}.$$

Now, we compute $e^{2i\pi/180} + e^{4\pi i / 180} + \cdots + e^{180\pi i / 180}$. Let $z = e^{2\pi i / 180}$; then this sum becomes $$z + z^2 + \cdots + z^{90} = z(1 + z +\cdots + z^{89}) = z\left(\frac{1 - z^{90}}{1-z}\right) = z\left(\frac{2}{1-z}\right) = \frac{2z}{1-z} = \frac{2z\left(1 - \frac{1}{z}\right)}{(1-z)\left(1 - \frac{1}{z}\right)} = \frac{2z - 2}{2 - 2\Re(z)} = -\frac{1-z}{1 - \Re(z)}.$$ Thus, we can write $$(1-z)A = 2\left(-\frac{1-z}{1-\Re(z)}\right) + 180z \iff A = \frac{-\frac{2-2z}{1-\Re(z)} + 180z}{1-z} = -\frac{2}{1-\Re(z)} + \frac{180z}{1-z}.$$ We want to find $\Im(A)$. However, since $-\frac{2}{1-\Re(z)}$ is real, it suffices to find the imaginary part of $$\frac{180z}{1-z} = \frac{180z\left(1 - \frac{1}{z}\right)}{(1-z)\left(1 - \frac{1}{z}\right)} = \frac{180z - 180}{2 - 2\Re(z)}.$$ Since $-\frac{180}{2-2\Re(z)}$ is real, it thus suffices to find $$\Im\left(\frac{180z}{2-2\Re(z)}\right) = \Im\left(\frac{90z}{1 - \Re(z)}\right) = \frac{90\Im(z)}{1 - \Re(z)} = \frac{90 \sin \frac{2\pi}{180}}{1 - \cos \frac{2\pi}{180}}.$$ Now, using double angle, this is equal to $$\frac{180\sin \frac{\pi}{180}\cos \frac{\pi}{180}}{1 - (1 - 2\sin^2 \frac{pi}{180})} = \frac{180\sin \frac{\pi}{180}\cos \frac{\pi}{180}}{2\sin^2 \frac{\pi}{180}} = \frac{90\cos \frac{\pi}{180}}{\sin \frac{\pi}{180}} = 90 \cot \frac{\pi}{180},$$ as desired.

We rewrite what we want to prove as $$\sum_{n=1}^{90} n \sin 2n = 45\cot 1.$$ Note that $$n\sin(2n)+(90-n)\sin(2(90-n))=90\sin(2n),$$ so this becomes $$45+90\sum_{n=1}^{44}\sin(2n)=45\cot(1),$$ where the 45 comes from the isolated term $45\sin(90).$ We can divide out the 45 to get $$1+2\sum_{n=1}^{44}\sin(2n)=\cot(1).$$ Multiplying this by $\sin(1),$ this becomes $$\sum_{n=1}^{44}2\sin(2n)\sin(1)=\cos(1)-\sin(1).$$ Then, by product-to-sum, this becomes $$\sum_{n=1}^{44}\cos(2n-1)-\cos(2n+1)=\cos(1)-\sin(1).$$ This is true since we now have a happy tale of telescoping: $$LHS=\cos(1)-\cos(3)+\cos(3)-\cos(5)+\cos(5)\cdots -\cos(89)=\cos(1)-\cos(89)=\cos(1)-\sin(1),$$ so we are done.

First, notice that$$n \sin n^\circ + (180-n) \sin(180-n)^\circ = 180 \sin n^\circ,$$so the condition is equivalent to$$\sum_{n=0}^{89} \sin(2n)^\circ = \cot 1^\circ.$$This is just a complex bash: note that$$\sum_{n=0}^{89} \frac{\omega^n-\omega^{-n}}{2i} = \frac{\frac 1{1-\omega} - \frac 1{1-\omega^{-1}}}{i} = \frac{\omega-\omega^{-1}}{i(1-\omega^{-1})(1-\omega)}$$where $\omega = \exp\left(\frac{\pi}{90}\right)$. However, for $\zeta = \exp\left(\frac{\pi}{180}\right)$, we have$$\cot 1^\circ = \frac{\zeta+\zeta^{-1}}{-i(\zeta-\zeta^{-1})} = \frac{\zeta^2- \zeta^{-2}}{i(1-\zeta^{-2})(1-\zeta^2)},$$as required.


i can heres my time [hide=".   "]$ \infty$[/hide]

This should go to Other Problem Solving Topics.

I can solve it but I'm pretty sure there's a more efficient method than mine. It takes me forever. Also the faces are pretty hard to turn and the stickers peel off really easily.

[quote="i_like_pie"]I can solve it, but my best time is just under 10 minutes. Considering the world record is around 1.5 minutes, that's really slow. Considering I know nobody else who can solve one, that's really fast.[/quote]

Someone likes the concept of relativity.

BTW, there are youtube videos of 100 x 100 x 100 cubes being solved and also 1000 x 1000 x 1000 cubes (although this was completed using a computer program). There are also videos of a 3 year old completing a rubik's cube, and lots of other ridiculously incredible things as well.

~20 minutes
I can do a 4x4 in 10 min. I'm slow :( but I made up my own methods

[quote=Kevin K.]Can anybody here solve a 5x5 Rubik's Cube?

To try to solve a 5x5, I tried some algorithms that work for the 3x3, but so far, I only have one layer and 2 edges (as in the 3 pieces that form an edge).

Can anyone here solve a 5x5? What's your fastest time?[/quote]

yes


quadratics
calculus
calculus computations

Im having some trouble with complex numbers. Can someone find the roots of this equation?

z^4-(sqrt 2)*(-1+i)*z^2-2i=0

$z^{4}-\sqrt{2}(-1+i)z^{2}-2i=0$

I can't think of an easier way than solving for $z^{2}$ using the quadratic formula... then solving for $z$...

Cant get the right answer. Can this be solved just like as if there were real coefissients? Sorry, but complex numers is new for me. Can someone please post a complete solution? I really need help here. Its urgent


Two distinct integers, $ x$ and $ y$, are randomly chosen from the set $ \{1,2,3,4,5,6,7,8,9,10\}$.  What is the probability that $ xy\minus{}x\minus{}y$ is even?

Read the problem.  The two numbers must be [b]distinct[/b], which means they must be different.

Thus, the probability that the first number will be even is $ \frac{5}{10}\equal{}\frac{1}{2}$, while the probability that the second number will be even is $ \frac{4}{9}$.  Multiplying these two yields $ \frac{1}{2} \times \frac{4}{9}\equal{}\frac{4}{18}\equal{}\boxed{\frac{2}{9}}$.

Different solution:

$ xy\minus{}x\minus{}y$ being even is the same as $ xy\minus{}x\minus{}y\plus{}1$ being odd. This is the same as $ (x\minus{}1)(y\minus{}1)$. This is only odd if both $ x\minus{}1$ and $ y\minus{}1$ are odd, which means that $ x$ and $ y$ must be even. The number of ways to choose even $ x$ and $ y$ is $ \binom{5}{2}$ and the number of total ways to choose $ x$ and $ y$ is $ \binom{10}{2}$, so the answer is their ratio, or $ \frac{2}{9}$.

We use SFFT to find $ xy-x-y+1=(x-1)(y-1)$ must be odd.  Thus, $x$ and $y$ must both be even.  Our probability is $\dfrac{\binom52}{\binom{10}{2}}=\dfrac{2}{9}$.

We note that $xy-x-y$ is very close to the expansion of $(x-1)(y-1)$. (This is basically a use of Simon's Favorite Factoring Trick.)

If $xy-x-y$ is even, then $xy-x-y+1 = (x-1)(y-1)$ is odd. This only occurs when $x-1$ and $y-1$ are both odd, so $x$ and $y$ must be even. There are $\binom{5}{2}$ distinct pairs of even integers, and $\binom{10}{2}$ distinct pairs of integers, so the probability is $\dfrac{\binom{5}{2}}{\binom{10}{2}} = \boxed{\frac{2}{9}}$.

$Make$ $sure$ $that$ $you$ $understand$ $the$ $problem$ $!$

[hide=easier solution]

so basically we note that this is equal to $xy-(x+y)$ because of algebra stuff

then we note that both $xy$ and $x, y$ must be even for their subtraction to be even.

Therefore, both $x$ and $y$ are even

we can choose this in $5C2$ ways which is equal to $10$.

total is $10C2$

then we get answer

[quote=MyRascal]Shouldn't it be 1/4?
xy-(x+y) is even---> xy is even and x+y is even----> x is even and y is even.
the probability of both x and y being even should be 1/2*1/2 which is 1/4 right?[/quote]

dude I put that as my first answer toooooo(it's wrong bc all the numbers have to be distinct)
also another sfft problem in adv probability with combinations lol

[quote=Arsenal_FC]Does this have ties to Chicken McNugget Theorem?[/quote]

no


Okay, here's my problem.

I go onto the usamts site and I try to log in using my username and password, however, when I try that the screen says that I should register for the usamts to log in.

If I type in an incorrect password, I can't get into the main screen, however, as I just said, typing the correct password doesn't help either.

Any ideas?

It happens for me to; just click the login link on that page and log in again.

I can't recreate this bug - what OS and browser are you using?

OK, I'm using Firefox(Deer Park Alpha 2) and Windows XP. Here is what I see:
[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=298[/img][/hide]
I click Submit:
[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=299[/img][/hide]
I click "Click here to login.":
[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=300[/img][/hide]
And I click Submit again:
[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=301[/img][/hide]

Finally, after inputting my username and password 2 times, I get to my scores.

Ditto everything solafidefarms said--same browser, same OS, same workaround.

Hurm, very peculiar.  Works fine for me on IE. .  .  I'll see if I can figure out what's up, but this may be something specific to firefox.

Works on my Firefox 1.0 . . .  I'll try upgrading and see if it fails then.

Ah - the Deer Park Alpha 2 is only in alpha release - I suspect the problem is in the browser.  You may want to report it to the developers.

[quote="solafidefarms"]I'll check if it is a problem with my other Firefox versions - maybe it'll work there.[/quote]Most certainly it's a browser issue. I [b]strongly[/b] recommend Firefox 1.0.7 for regular browsing activities, and all other versions (1.5 beta 1 etc.) for testing / fun purposes only.

OK, out-of-the-box 1.0 - no bugfixes - has that problem...

EDIT: as does 1.0.1...

EDIT2: I'll try more later...class is more important then testing. (As you can tell, I collect old Firefoxes! :lol:)

Do you think you could put those pictures in hide tags?  They're really big and it makes this topic harder to read.  Thanks :)

[quote="The Original Pi Guy"]The thing is, I have the latest version of Firefox, but it still isn't working.

*is frustrated*[/quote]

a lot of ppl use firefox. it should work, it doesnt seem to be troubling anyone i know. maybe the site is on ur forbidden list, check ur firewalls and automatic virus blockers.

EDIT: oh yea, and ur immunity list. tho i doubt its cuz of that.

[quote="The Original Pi Guy"]Hm... tried that, but nothing seems to work there either. I also tried clearing the cache.
Is there anyway to check scores outside of the site?[/quote]

If it becomes too big of an issue you can come over and use my computer, I have no problems logging in.  ;)

Now it works with only 1 login. Hmmmm....


Prove that for avery positiv $a_1,a_2...,a_n$ this ineq. is true.
$\frac{a_1+\sqrt{a_1a_2}+...+\sqrt[n]{a_1a_2...a_n}}{n}\leq \sqrt[n]{a_1\frac{a_1+a_2}{2}...\frac{a_1+...+a_n}{n}}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18574]www.mathlinks.ro/Forum/viewtopic.php?t=18574[/url]
Darij has uploaded two files in post#10,and there is two solutions :)

i have an idea but i'm not sure if it works, is it possible to reorder the elements appropriately so that we can directly apply holder?

For example, for n = 3, we can use
$(a+a+a+a+a+a)(a+a+a+a+a+a)(a+a+a+b+b+b)(a+a+b+a+b+b)(a+a+b+b+c+c)(a+a+b+b+c+c)$
and then apply holder.


trigonometry
floor function
modular arithmetic
algebra
function
domain
arithmetic sequence

For each integer $ n > 1,$ let $ F(n)$ be the number of solutions of the equation $ \sin x \equal{} \sin nx$ on the interval $ [0,\pi].$ What is $ \sum_{n \equal{} 2}^{2007}F(n)?$

$ \textbf{(A)}\ 2,014,524 \qquad \textbf{(B)}\ 2,015,028 \qquad \textbf{(C)}\ 2,015,033 \qquad \textbf{(D)}\ 2,016,532 \qquad \textbf{(E)}\ 2,017,033$

[quote="adalton"][hide]Note that $F(2)=3$ $F(3)=4$ $F(4)=5$ $F(5)=5$ $F(6)=7$ so in general $F(n)=n+1$ unless $n\equiv1(mod4)$.

Since the sequence ends at $F(2007)=2008$ we can use the formula $\sum_{i=1}^{n}{i}=\frac{n(n+1)}{2}=\frac{(2008)(2009)}{2}$

Subtracting $\lfloor\frac{2007}{4}\rfloor$ (since every fourth number is one off of the arithmetic sequence) and $1+2$ (since the first term is 3, we over counted with the summation formula).

This yields $2017036-501-3=2016532$ $\boxed{D}$[/hide][/quote]
that doesn't show that the pattern actually works for all $n$...

[hide="the horrible solution i came up with"]
\[\sin nx-\sin x=2\left( \cos \frac{n+1}{2}x\right) \left( \sin \frac{n-1}{2}x\right)\]
So $\sin nx=\sin x$ if and only if $\cos \frac{n+1}{2}x=0$ or $\sin \frac{n-1}{2}x=0$.

The first occurs whenever $\frac{n+1}{2}x=(j+1/2)\pi$, or $x=\frac{(2j+1)\pi}{n+1}$ for some nonnegative integer $j$. Since $x\leq \pi$, $j\leq n/2$. So there are $1+\lfloor n/2 \rfloor$ solutions in this case.

The second occurs whenever $\frac{n-1}{2}x=k\pi$, or $x=\frac{2k\pi}{n-1}$ for some nonnegative integer $k$. Here $k\leq \frac{n-1}{2}$ so that there are $\left\lfloor \frac{n+1}{2}\right\rfloor$ solutions here.

However, we overcount intersections. These occur whenever
\[\frac{2j+1}{n+1}=\frac{2k}{n-1}\]
\[k=\frac{(2j+1)(n-1)}{2(n+1)}\]
which is equivalent to $2(n+1)$ dividing $(2j+1)(n-1)$. If $n$ is even, then $(2j+1)(n-1)$ is odd, so this never happens. If $n\equiv 3\pmod{4}$, then there won't be intersections either, since a multiple of 8 can't divide a number which is not even a multiple of 4.

This leaves $n\equiv 1\pmod{4}$. In this case, the divisibility becomes $\frac{n+1}{2}$ dividing $(2j+1)\frac{n-1}{4}$. Since $\frac{n+1}{2}$ and $\frac{n-1}{4}$ are relatively prime (subtracting twice the second number from the first gives 1), $\frac{n+1}{2}$ must divide $2j+1$. Since $j\leq \frac{n-1}{2}$, $2j+1\leq n<2\cdot \frac{n+1}{2}$. Then there is only one intersection, namely when $j=\frac{n-1}{4}$.

Therefore we find $F(n)$ is equal to $1+\lfloor n/2 \rfloor+\left \lfloor \frac{n+1}{2}\right\rfloor =n+1$, unless $n\equiv 1\pmod{4}$, in which case it is one less, or $n$. Then finish as in the above "solution".

Amazingly, i pulled this off in <10 minutes... :rotfl:  A rather bad use of time, though; i must have been focused on just getting through every problem.
[/hide]

Wow, you did all that in ten minutes?
Yikes.
Anyway, it's easy to see that if $\sin{x}= \sin{nx}= 1$, $x = \frac{\pi}{2}$, and since $\sin$ has a period of $2\pi$, all $nx-x = 2\pi k \Rightarrow n = 4k+1$.
Too bad I didn't even glance at this during the test.

Having just read a topic on that on AOPS the day before, I tried the same thing! ... And wasted liked 10 minutes...

[quote="t0rajir0u"]Hmm.  I tried to solve this with a complex number substitution and failed horribly.  Anyone have better luck?[/quote]

That was my initial thought as well, though I wasn't taking the test officially.

I think the problem with that approach is it's hard to say anything definitively about either whether the solutions are in the right range or anything about multiplicity.

I didn't take the 12A officially but that same day in my math class I "took" it (so I guess both AMC's in one day) and I thought I got a factorization...
I ended up with something like $(x^{n+1}+1)(x^{n-1}+1)$ but I can't completely remember... a couple signs might be flipped.
But I ended up abandoning that...

I just graphed $\sin x=\sin nx$ for $n=2,3,4,5,6,7,8,9,10,11,12,13$ on my calculator and counted the solutions. Then I found a pattern, and solved the problem. Took me about 10 minutes, seemed easier than going through all the steps to find that $F(n)=n$ when $n\equiv 1\pmod{4}$, and $F(n)=n+1$ otherwise.

meh...finding f(n)=<n+1 is pretty easy...but you are like WAIT A SECOND THATS THE LARGEST ANSWER....so obviously there is some overcount...a bit of f2 solving gives you the overcount for n=1mod 4

[quote="fizix"][hide]sin x = sin nx
n = 1
QED

 :rotfl: [/hide][/quote]

Um... did you read the question?

I rushed the last 5 question in 15 minutes.. and I had no time to pick out the errors i made in #22 and #24.. end up with a 138...

I forgot about the tangents when n=4k+1  :(

[quote="scorpius119"][quote="adalton"][hide]Note that $ F(2) \equal{} 3$ $ F(3) \equal{} 4$ $ F(4) \equal{} 5$ $ F(5) \equal{} 5$ $ F(6) \equal{} 7$ so in general $ F(n) \equal{} n \plus{} 1$ unless $ n\equiv1(mod4)$.

Since the sequence ends at $ F(2007) \equal{} 2008$ we can use the formula $ \sum_{i \equal{} 1}^{n}{i} \equal{} \frac {n(n \plus{} 1)}{2} \equal{} \frac {(2008)(2009)}{2}$

Subtracting $ \lfloor\frac {2007}{4}\rfloor$ (since every fourth number is one off of the arithmetic sequence) and $ 1 \plus{} 2$ (since the first term is 3, we over counted with the summation formula).

This yields $ 2017036 \minus{} 501 \minus{} 3 \equal{} 2016532$ $ \boxed{D}$[/hide][/quote]
that doesn't show that the pattern actually works for all $ n$...

[hide="the horrible solution i came up with"]
\[ \sin nx \minus{} \sin x \equal{} 2\left( \cos \frac {n \plus{} 1}{2}x\right) \left( \sin \frac {n \minus{} 1}{2}x\right)
\]
So $ \sin nx \equal{} \sin x$ if and only if $ \cos \frac {n \plus{} 1}{2}x \equal{} 0$ or $ \sin \frac {n \minus{} 1}{2}x \equal{} 0$.

The first occurs whenever $ \frac {n \plus{} 1}{2}x \equal{} (j \plus{} 1/2)\pi$, or $ x \equal{} \frac {(2j \plus{} 1)\pi}{n \plus{} 1}$ for some nonnegative integer $ j$. Since $ x\leq \pi$, $ j\leq n/2$. So there are $ 1 \plus{} \lfloor n/2 \rfloor$ solutions in this case.

The second occurs whenever $ \frac {n \minus{} 1}{2}x \equal{} k\pi$, or $ x \equal{} \frac {2k\pi}{n \minus{} 1}$ for some nonnegative integer $ k$. Here $ k\leq \frac {n \minus{} 1}{2}$ so that there are $ \left\lfloor \frac {n \plus{} 1}{2}\right\rfloor$ solutions here.

However, we overcount intersections. These occur whenever
\[ \frac {2j \plus{} 1}{n \plus{} 1} \equal{} \frac {2k}{n \minus{} 1}
\]

\[ k \equal{} \frac {(2j \plus{} 1)(n \minus{} 1)}{2(n \plus{} 1)}
\]
which is equivalent to $ 2(n \plus{} 1)$ dividing $ (2j \plus{} 1)(n \minus{} 1)$. If $ n$ is even, then $ (2j \plus{} 1)(n \minus{} 1)$ is odd, so this never happens. If $ n\equiv 3\pmod{4}$, then there won't be intersections either, since a multiple of 8 can't divide a number which is not even a multiple of 4.

This leaves $ n\equiv 1\pmod{4}$. In this case, the divisibility becomes $ \frac {n \plus{} 1}{2}$ dividing $ (2j \plus{} 1)\frac {n \minus{} 1}{4}$. Since $ \frac {n \plus{} 1}{2}$ and $ \frac {n \minus{} 1}{4}$ are relatively prime (subtracting twice the second number from the first gives 1), $ \frac {n \plus{} 1}{2}$ must divide $ 2j \plus{} 1$. Since $ j\leq \frac {n \minus{} 1}{2}$, $ 2j \plus{} 1\leq n < 2\cdot \frac {n \plus{} 1}{2}$. Then there is only one intersection, namely when $ j \equal{} \frac {n \minus{} 1}{4}$.

Therefore we find $ F(n)$ is equal to $ 1 \plus{} \lfloor n/2 \rfloor \plus{} \left \lfloor \frac {n \plus{} 1}{2}\right\rfloor \equal{} n \plus{} 1$, unless $ n\equiv 1\pmod{4}$, in which case it is one less, or $ n$. Then finish as in the above "solution".

Amazingly, i pulled this off in <10 minutes... :rotfl:  A rather bad use of time, though; i must have been focused on just getting through every problem.
[/hide][/quote]

I was wondering...
Can you explain how you get the first line of your solution? Is it some common knowledge?

The [i]sum-to-product[/i] formulas allow us to write a sum or difference of sines or cosines as a product of sines or cosines.  For example, because the sine angle summation formula says

$ \sin (a \plus{} b) \equal{} \sin a \cos b \plus{} \cos a \sin b$
$ \sin (a \minus{} b) \equal{} \sin a \cos b \minus{} \cos a \sin b$

we conclude that

$ \sin (a \plus{} b) \minus{} \sin (a \minus{} b) \equal{} 2 \cos a \sin b$.

Letting $ a \plus{} b \equal{} x, a \minus{} b \equal{} y$, we have the formula

$ \sin x \minus{} \sin y \equal{} 2 \cos \left( \frac {x \plus{} y}{2} \right) \sin \left( \frac {x \minus{} y}{2} \right)$.

As you can see, one use of the sum-to-product formulas is to discern the behavior and distribution of roots.  :)

[quote="adalton"][hide]in general $F(n)=n+1$ unless $n\equiv1(mod4)$.

[/hide][/quote]

Can someone explain this part?

Well, that was just a pattern he noticed.
However, after looking at the equations, you get....
$sin(x) = sin(nx)$
For all $n>1$, one of two things will solve the equation:
$nx=x+2k\pi$ or $x=\pi-nx$
For the first relation, $x=\frac{2k\pi}{n-1}$
Thus, x is only in the domain if $n-1\geq 2k$ for $k\geq 0 $
The number of solutions to the first relation is equal to the number odd numbers less than or equal to n. However, in this case we still have missed one solution, namely, the k that gives $\frac{\pi}{2}$.
For example, in $sin(x) = sin(2x)$, $x=\frac{\pi}{2}$ is a solution. We add this to the number of solutions, +1.
For the other case, $x=\pi - nx$ will always yield one solution, and we add this to the total, which is the number of odd numbers up to n +2. From here you can add up everything relatively easily.

OK, so clearly my solution is wrong. However, why is it wrong?

Sorry to bump an extremely long thread, but in scorpius119, how does he get

[quote] $j \leq \frac{n-1}{2}$ [/quote]

in the part where he overcounts intersections?

[quote=OlympusHero]Hi! Here's what I did so far: This can occur if $x \equiv nx \pmod {360}$, obviously, or if $180-x \equiv nx \pmod {360}$. The first condition happens when $x(n-1)$ divides $360$, and the second happens when $x(n+1)$ divides $180$ but not $360$ ($180$ multiplied by odd number). But I don't know how to count the amount of working values for these... maybe I'm being stupid because it's late :P Hint?[/quote]

$x$ isn't necessarily an integer amount of degrees, so modular arithmetic is probably not the right approach.


My friend and I are going to be teaching part of a 7th grade math class tomorrow. We are teaching probability. Our subject matter is going to include very basic theoretical probability, basically getting their minds around
$ P(A)\equal{}\frac{Ways A can happen}{Ways anything can happen}$ Do you know of any good exercise problems we could give them that aren't from the Intro To Counting & Probability Book?

Well, the Intro to C and P book covers pretty much everything of basic probability. If it is really basic, you could always do those "classics". i.e. What is the probability of rolling a 6 on a 6 sided die? or What is the probability of flipping two heads in a row after you have flipped two coins? (I think drawing a tree diagram is what most people do to explain the basics of the second question, even if you could just multiply 1/2 by 1/2.)

Its a really very nice comment in this forum site about the speaks of problem Middle school classroom in maths.
I know that its a very hard but you can teach if you can problem in maths site you teach in class.
Thanks.

[url=http://ezinearticles.com/?Heel-Tastic-Review---Does-it-Work?&id=2852393]Heel Tastic[/url]


algebra
polynomial
calculus
integration
algebra proposed

The polynomial $ P(x)\equal{}a_{0}x^{k}\plus{}a_{1}x^{k\minus{}1}\plus{}...\plus{}a_{k}$, where $ a_{0}, a_{1}, ..., a_{k}$ are integers, is said to be divisible by an integer $ m$ if $ P(x)$ is a multiple of $ m$ for every integral value of $ x$. Prove that if $ a, k, m$ are positive integers such that $ ak!$ is a multiple of $ m$, then a polynomial $ P(x)$ with leading term $ ax^{k}$ can be found that is divisible by $ m$.

[quote="t0rajir0u"]$ P(x) \equal{} ak!{x\choose k}$[/quote]
Solution?

$ {x\choose k}$ is a polynomial with leading term $ \frac{1}{k!}$ which is integer for integer $ x$... I think the rest follows pretty naturally.  To emphasize that the coefficients are integer I suppose I should write out the polynomial as

$ P(x) \equal{} a(x\minus{}1)(x\minus{}2)...(x\minus{}k)$

This is divisible by $ ak!$ and hence by $ m$ by a well-known lemma.


trigonometry
function
algorithm
number theory
relatively prime

Find the period of sin(ax/b)+cos(cx/d), where a,b,c, and d are integers and a/b and c/d are in lowest terms.

[hide="As far as I get..."]

Period of $ sin(\frac{ax}{b})$:

$ \frac{ax}{b}=2\pi$

$ x=\frac{2\pi b}{a}$

Period of $ cos(\frac{cx}{d})$:

$ \frac{cx}{d}=2\pi$

$ x=\frac{2\pi d}{c}$

So the period of the sum is:

$ LCD(\frac{2\pi b}{a},\frac{2\pi d}{c})$

$ =2\pi * LCD(\frac{b}{a},\frac{d}{c})$

I don't know if LCD applies to rational numbers properly, so I'll put it another way:

When I say $ LCD(\frac{b}{a},\frac{d}{c})$, I actually mean:

What is $ e\frac{b}{a}$, where $ e$ and $ f$ are the lowest possible positive integers which satisfy:

$ e\frac{b}{a}= f\frac{d}{c}$

[/hide]

Not sure if this can be taken any further... unless someone wants to cite an algorithm for LCD or something.

Maybe rewriting as a single sine would work.
$ f(x) = \sin(\frac{ax}{b})+\sin(\pi-\frac{cx}{d})$
Replace $ \frac{a}{b}$ with A and $ \pi-\frac{c}{d}$ with B.

Then we have $ f(x) = \sin(Ax)+\sin(Bx)$

by a sum to product rule, this gives

$ f(x) = 2\sin(\frac{Ax+Bx}{2})\cdot\cos(\frac{Ax-Bx}{2})$
$ f(x) = 2\sin(\frac{x(A+B)}{2})\cdot\cos(\frac{x(A-B)}{2})$

$ A+B = \frac{a}{b}+\pi-\frac{c}{d}$
$ A-B = \frac{a}{b}-\pi+\frac{c}{d}$

$ f(x) = 2\sin(\frac{x(\frac{a}{b}+\pi-\frac{c}{d})}{2})\cdot\cos(\frac{x(\frac{a}{b}-\pi+\frac{c}{d})}{2})$

Does this help?

Now we have a multiplicative function which looks much uglier than the first one..

I'm lost.

[quote="DiscreetFourierTransform"]Does this help?

Now we have a multiplicative function which looks much uglier than the first one..

I'm lost.[/quote]

bahahahahaha... it was a good idea though... who knows, maybe it will help someone else...

and you made a mistake DiscreetFourierTransform

when you substitute $ B = \pi-\frac{c}{d}$, you are not left with $ \sin(Bx)$

sorry

None of you have the right answer yet.

I found this in AoPS Volume II, so I'll post that solution, with more detail (even though AoPS is pretty detailed).

[hide]Let $ z$ be the desired period.

Since $ \sin (ax/b)$ repeats every $ 2b\pi/a$ and $ \cos (cx/d)$ repeats every $ 2c\pi/d$ times, $ z$ needs to be "divisible" by both $ 2b\pi/a$ and $ 2c\pi/d$.

This is what I mean when I say that:
$ z/(2b\pi/a)=az/2b\pi$ and $ z/(c\pi/d)=dz/2c\pi$
must be both integers.

With this observation, the problem breaks down.
Since $ a$ and $ b$ are relatively prime, but $ az/2b\pi$ is still an integer, the numerator of $ z$ must be divisible by $ 2$, $ b$, and $ \pi$. From $ cd/2d\pi$, the numerator of $ z$ must be divisible by $ 2$, $ d$, and $ \pi$.

Hence, the numerator of $ z$ is divisible by $ 2$, $ \pi$, $ b$, and $ d$. Since we want the numerator of $ z$ to be minimized, the numerator of $ z$ is $ [\pi, 2, b, d]$ (the brackets mean LCM). Since $ \pi$ is irrational, we can move it outside:
the minimum numerator of $ z$ is $ \pi\cdot [2,b,d]$.

Let's do a similar analysis on the denominator of $ z$. We want this quantity to be maximized, not minimized, since a fraction gets smaller when the denominator gets bigger. Since $ az/2b\pi$ and $ dz/2c\pi$ are both integers, the denominator of $ z$ can be no greater than $ (a,c)$ (the parentheses mean GCF). If it were, $ az$ and $ cz$ wouldn't be integers anymore, so the fractions cannot possibly be integers. So
the maximum denominator of $ z$ is $ (a,c)$

So,

$ z = \frac{\pi\cdot [2,b,d]}{(a,c)}$[/hide]


function
complex numbers
real analysis
real analysis unsolved

Let $ a_{n}$ be the sequence of complex numbers and let $ b_{n}$ be the sequence of real numbers. Prove that the set of these
complex numbers $ z$, for which series:
\[ \sum_{n>0} a_{n}e^{b_{n}z}\]
is absolutely convergent, is convex set

suppose $ A$ is the set when our serie converge abs.
or:
$ \sum |a_{n}\parallel{}e^{b_{n}z}| = \sum a_{n}e^{b_{n}Re (z)} < \infty$
note that $ e^{b_{n}x}$ for all real $ b_{n}$ is convex function . then :
$ e^{b_{n}(tx_{1} + (1 - t)x_{2})} \leq te^{b_{n}x_{1}} + (1 - t)e^{b_{n}x_{2}}$ for all $ t\in [0,1]$
Hence if $ z_{1},z_{2} \in A$ then
$ \sum |a_{n}|e^{b_{n}(tRe(z_{1}) + (1 - t)Re(z_{2}))} \leq \sum |a_{n}|(te^{b_{n}Re(z_{1})} + (1 - t)e^{b_{n}Re(z_{2})}) < \infty$
thus
$ tz_{1} + (1 - t)z_{2} \in A$
Hence $ A$ is convex .


Let $x,y,z>0$ with $xyz=1$.  Prove that $x+y+z \le x^2+y^2+z^2$.

Let $x=a^3,y=b^3,z=c^3$. The desired inequality can be rehomogenised (?) as
$(x+y+z)(xyz)^{\frac 1 3} \leq x^2+y^2+z^2$
that is :
$a^6+b^6+c^6 \geq abc(a^3+b^3+c^3)$
which is a direct consequence of Muirhead's inequality since $(6,0,0)$ majorizes $(4,1,1)$.

Pierre.

Here's another approach.  We will use the inequality $a^2 - a \geq a - 1$. We get
\begin{eqnarray*}
(x^2 - x) + (y^2 - y) + (z^2 - z)
  & \geq & x + y + z - 3 \\
  & \geq & 3 \sqrt[3]{xyz} - 3 \\
  & =  & 3 - 3 \\
  & = & 0.
\end{eqnarray*}

Yet another approach.  By the QM-AM inequality and then the AM-GM inequality, we have
\begin{eqnarray*}
x^2 + y^2 + z^2
  & \geq & \frac{(x + y + z)^2}{3} \\
  & = & \frac{x + y + z}{3} \cdot (x + y + z) \\
  & \geq & \sqrt[3]{xyz} \cdot (x + y + z) \\
  & = & x + y + z.
\end{eqnarray*}


Solve the following system of equation :
$ \frac{3x}{x\plus{}1}\plus{}\frac{4y}{y\plus{}1}\plus{}\frac{2z}{z\plus{}1}\equal{}1$
$ 8^9.x^3.y^4.z^2\equal{}1$

[quote="N.T.TUAN"]Are $ x,y,z$  positive numbers?[/quote]
Yes, $ x,y,z$ positive numbers.
I'm sorry.  :(

The following is true:

Let $ x_1, x_2, ..., x_9$ be positive real numbers such that $ \prod x_i \equal{} 8^9$. Then $ \sum \frac {1}{x_i \plus{} 1} \ge 1$, and the equality occurs if and only if $ x_i \equal{} 8 \text{ }\forall i \in \{1, 2, ..., 9\}$.

Using the particular case in which $ x_1 \equal{} x_2 \equal{} x_3 \equal{} \frac {1}{x}$, $ x_4 \equal{} x_5 \equal{} x_6 \equal{} x_7 \equal{} \frac {1}{y}$ and $ x_8 \equal{} x_9 \equal{} \frac {1}{z}$ (observe that $ \frac {1}{1 \plus{} \frac {1}{t}} \equal{} \frac {t}{t \plus{} 1}$, so $ \sum \frac {1}{x_i \plus{} 1} \equal{} \frac {3x}{x \plus{} 1} \plus{} \frac {4y}{y \plus{} 1} \plus{} \frac {2z}{z \plus{} 1}$), we get $ x \equal{} y \equal{} z \equal{} \frac {1}{8}$.

Now, observe that

$ \sum \frac {1}{x_i \plus{} 1} \ge 1$
$ \Longleftrightarrow \sum (1 \plus{} x_1)(1 \plus{} x_2)...(1 \plus{} x_9)/(1 \plus{} x_i) \ge \prod (1 \plus{} x_1)(1 \plus{} x_2)...(1 \plus{} x_9)$
$ \Longleftrightarrow \sum_{sym} (1 \plus{} x_1)(1 \plus{} x_2)(1 \plus{} x_3)...(1 \plus{} x_8) \ge \frac {1}{9} \sum_{sym} (1 \plus{} x_1)(1 \plus{} x_2)...(1 \plus{} x_9)$
$ \Longleftrightarrow \sum_{sym} \binom{8}{0} \plus{} \binom{8}{1}x_1 \plus{} \binom{8}{2}x_1x_2 \plus{} ... \ge \sum_{sym} \frac {1}{9}\binom{9}{0} \plus{} \frac {1}{9}\binom{9}{1}x_1 \plus{} \frac {1}{9}\binom{9}{2}x_1x_2 \plus{} ...$

It is not difficult to prove that $ \binom {8}{k} \ge \frac {1}{9}\binom{9}{k}$ (for 0 <= k <= 8). So the above inequality is equivalent to some inequality like

$ a\sum x_1 \plus{} b\sum x_1x_2 \plus{} c\sum x_1x_2x_3 \plus{} ... \plus{} h\sum x_1...x_8 \ge x_1x_2...x_9 \equal{} 8^9$,

where $ a, b, c, ..., h \ge 0$ and the sums are symmetric. Using AM-GM, we see that each sum attains its minimum precisely when all variables are equal. Also, if $ x_i \equal{} 8 \text{ } \forall i$, the equality occurs (because the equality occurs in the initial inequality). So this is the only case of equality.


induction
function
graph theory
combinatorics proposed
combinatorics

$ 300$ students participates in an international contest. Each student speaks exactly $ 2$ of $ 6$ official contest languages (and no other languages). Moreover, each official language is spoken by exactly $ 100$ students. Prove that the organizers can arrange these students in a circle such that no two consecutive students speak a common language.

[quote="ehsan2004"][quote="Erken"][quote="ehsan2004"]no one ???? :maybe:[/quote]
Do you need a hint?[/quote]

I need a complete solution. :)[/quote]
I'm sorry,because its solution is too long,and my english is not enough good,to translate it :(
But if you want i could scan the solution and post it here and maybe someone will translate it, :wink:
Or maybe freemind will post solution.
Thank you.

Ok, here is the solution. It is long but has few main steps. But I advise you to think more about it. A hint is to use induction on $ n$, where $ 3n$ is the number of contestants.
[hide="Solution"]
We prove the general statement when we have $ 3n$ students. Each students speaks exactly $ 2$ of the official languages, and each official language is spoken by exactly $ n$ students.
Use induction on $ n$. Base cases are $ n \equal{} 1$ and $ n \equal{} 2$, which are trivial.
Assume the statement true for all $ k$, $ 1\le k < n$. We'll prove the problem statement for $ 3n$ students, $ n > 2$. To use induction we need to have some students, whose removal leaves a graph where each language is still spoken by an equal number of students. This means that the removed students (their number must clearly be a multiple of $ 3$) form a graph which also obeys the problem statement. Moreover, for the induction step to be small, the number of students which should be 'removed' must be small. For $ 3$ contestants there is only one possibility: First students speaks languages $ A,B$. Second - $ C,D$ and third -$ E,F$. Call such a graph $ M_3$. For $ 6$ contestants there are 2 possibilities: the graph consists of two $ M_3$, or the bipartite graph ($ 6$ contestants - $ 6$ languages) is made up of two $ K_{3,3}$.

As you have probably guessed, we should prove that the graph determined by the $ 3n$ contestants and $ 6$ languages contains either a $ M_3$ or two $ K_{3,3}$. This is going to be long, but not so difficult. Let the set of languages be $ L \equal{} \{A,B,C,D,E,F\}$. For shortness of notations, let $ f: L^2\rightarrow\{\text{True}, \text{False}\}$ a function for which $ f(XY)$ gives the truth of the statement: [i]There is a contestant speaking languages $ X$ and $ Y$[/i]. When I use "[i]if[/i] $ f(XY)$" I'll usually mean "[i]if[/i] $ f(X,Y) \equal{} \text{True}$".

Assume the graph has no $ M_3$ or $ K_{3,3}$.
Now WLOG, assume $ f(AB)$. If all contestants speak either $ A$ or $ B$ then $ A$ or $ B$ will be spoken by at least $ \dfrac{3n}2 > n$ students, impossible. So, WLOG assume $ f(CD)$. If $ f(EF)$ then we get a $ M_3$. Hence $ f(EF) \equal{} \text{False}$. Note that students who speak $ E$ speak at least one of $ A,B,C,D$. Because $ f(AB)$ and $ f(CD)$, there exist two students who both speak $ E$ but for whom the second language is different (otherwise if all $ n$ students who speak $ E$ speak wlog also just $ A$, then $ A$ is spoken by at least $ n \plus{} 1$ students). WLOG we have 2 cases: $ f(EA) \equal{} f(EB) \equal{} \text{True}$ and $ f(EA) \equal{} f(EC) \equal{} \text{True}$.

[i]Case 1.[/i]
$ f(EA) \equal{} f(EB) \equal{} \text{True}$.
If $ f(AF) \equal{} \text{True}$ then we have $ M_3$: $ f(AF), f(CD), F(BE)$. If we have $ f(BF)$ then we have $ M_3$: $ f(BF)$, $ f(AE)$, $ f(CD)$. Let now $ f(AF) \equal{} f(BF) \equal{} F(EF) \equal{} \text{False}$. Then students who speak $ F$ speak also $ C$ or $ D$. Moreover $ f(BF) \equal{} f(CF) \equal{} \text{True}$. If one of them is false, then we get that all $ n$ contestants who speak $ F$ speak also, wlog, $ C$, hence $ C$ is spoken by at least $ n \plus{} 1$ contestants ($ f(CD) \equal{} \text{True}$). Then we get 2 $ K_{3,3}$: $ f(AB),f(BE),f(EA)$ and $ f(CD),f(DF),f(CF)$.

[i]Case 2.[/i]
$ f(EA) \equal{} f(EC) \equal{} \text{True}$.
If $ f(FD) \equal{} \text{True}$ then we get the $ M_3$: $ f(AB),f(EC),f(FD)$. If $ f(FB) \equal{} \text{True}$ then we have $ M_3$: $ f(BF),f(AE),f(CD)$.
So assume $ f(FE) \equal{} f(FD) \equal{} f(FB) \equal{} \text{False}$. Now as in Case 1, both $ f(FA)$ and $ f(FC)$ must be true. Note that $ f(ED) \equal{} f(EB) \equal{} \text{False}$, otherwise we get $ M_3$'s (for example, if $ f(ED) \equal{} \text{True}$ then we have $ f(AB)$, $ f(ED)$ and $ f(CF)$).
So all students who speak $ E$ or $ F$ speak $ A$ or $ C$. This means $ A$ and $ C$ are spoken by at least $ 2n$ students. But except these students, we have $ f(AB) \equal{} f(CD) \equal{} \text{True}$, so we have a contraditcion ($ A$ or $ C$ are spoken by $ > n$ students).

Now let's finish the solution.
Assume we have a $ M_3$.
Let it be $ f(AB), f(CD), f(EF)$. The remaining $ 3n \minus{} 3$ contestants can be arranged, according to the induction hypothesis around a circle so that no two adjacent students speak the same language. If in this arrangement there is some contestant also speaking $ AB$ (or $ CD$, $ EF$) then to his right insert $ EF, CD, BA$ and we are done. If there is no such contestant, then take a contestant $ I$ such that the contestants next to him have at most one language in common (that is - they speak different sets of languages). Assume $ I$ speaks $ AC$ and that the contestant to his right (or left) $ J$ doesn't speak both $ B$ and $ D$. Then we can put $ I, f(EF),f(CD),f(AB),J$ or $ I,f(EF),f(AB),f(CD),J$. This is possible because $ J$ doesn't speak $ A,C$ and at least one of $ B$ and $ D$.

Assume we have $ 2$ $ K_{3,3}$. Let them be $ f(AB),f(BC),f(CA)$ and $ f(DE),f(EF),f(FD)$. Arrange the remaining $ 3n \minus{} 6$ contestants in a good way (according to the induction hypothesis this is possible). Now if among them there is a student speaking one of the set of languages found in the two $ K_{3,3}$'s, say $ (AB)$, then to his right insert $ DE,BC,EF,CA,DF,AB$. It is easy to see that we get a good arrangement. Assume no contestant speaks a set of languages from the two $ K_{3,3}$'s. Then take any two contestants. Suppose they speak $ AC$ and $ BD$. Then between them insert $ DE,AB,DF,BC,EF,CA$ and we get a good arrangement.

I hope I didn't misplace some letters:).
[/hide]
If you like Saint-Petersburg Problems, you may try http://www.mathlinks.ro/viewtopic.php?t=182443

Thanks very much dear Freemind and dear Erken . :lol:


Show that a function $ f(x)\equal{}2^x$ for $ x$ from real can be expressen like a sum of odd and even function

Am I missing something here?

$ 2^x\equal{}\frac{2^x\plus{}2^{\minus{}x}}{2}\plus{}\frac{2^x\minus{}2^{\minus{}x}}{2}$

and those fractions are even and odd functions, respectively.

Generalization: $ f(x)\equal{}\frac{f(x)\plus{}f(\minus{}x)}{2}\plus{}\frac{f(x)\minus{}f(\minus{}x)}{2}$


Edit: wrong answer: not considered distinct digits.

Certainly you have a solution with that answer, right? :)

[quote="Andreas"][hide="one-liner"]Note that with $c = 1$ $\Longrightarrow$ $2$, with $c = 2$ $\Longrightarrow$ $2 \cdot 3$, ..., with $c = 9$ $\Longrightarrow$ $9 \cdot 10$.[/hide][/quote]

Huh? How can you have any ringed numbers for $c=1, 2, 3, 4$ if all the digits must be distinct?

[hide]
If c=5 or 6, there are 2 distinct pairs which equal c
If c=7 or 8, there are 3 distinct pairs which equal c
If c=9 there are 4 distinct pairs

Any number [b]abcde[/b] can also be arranged as [b]ebcda[/b], [b]edcba[/b], and [b]adcbe[/b] = 4 arrangements

So the answer is 4(2P2 + 2P2 + 3P2 + 3P2 + 4P2) = [b]112[/b][/hide]

[hide="case work"]
Note that for every $w+x=y+z$, there are $8$ distinct "ring" numbers that can be derived (switching $y$ and $z$, $w$ and $x$, and then putting them on the opposite ends of the number).

For $c \leq 4$ there are no "ring numbers" possible.
$c=5 \implies 5=1+4=2+3 \Longrightarrow 8 \text{ numbers}$
$c=6 \implies 6=1+5=2+4 \Longrightarrow 8 \text{ numbers}$
$c=7 \implies 7=1+6=2+5=3+4 \Longrightarrow 3 \cdot 8=24 \text{ numbers}$
$c=8 \implies 8=1+7=2+6=3+5 \Longrightarrow 3 \cdot 8=24 \text{ numbers}$
$c=9 \implies 9=1+8=2+7=3+6=4+5 \Longrightarrow 6 \cdot 8=48 \text{ numbers}$

This gives a total of $14 \cdot 8 = \boxed {112 \text{ \"ring\" numbers}}$[/hide]

Can you come up with the number of "ring" numbers of length $n$?

(edit: [i]2[/i] minor change[i]s[/i])

[quote="Altheman"][quote="E^(pi*i)=-1"]Call a five-digit positive integer "ringed" if all its digits are distinct and the sum of the outermost two digits is equal to both the middle digit and the sum of the other two digits, i.e. in the number abcde, a+e=b+d=c.  How many ringed numbers are there?[/quote]

this is almost the same as my first aime problem  :) , except the digits are distinct[/quote]

Sorry Altheman, I didn't realize you had used this problem before.

I think the result is $\sum\limits_{c=1}^9(\sum\limits_{k=0}^c\cdot\sum\limits_{k=1}^c)$


MATHCOUNTS
geometry
calculus
integration
absolute value

I am beginning a new game for any one that would like to participate. You begin the game with 1,000,000 points, or the jackpot. I am going to think of a number, and I am going to post three hints to help you guess what my number is. You will private message me your guess. When I recieve your guess, I will find the absolute value of the difference of your guess and the actual number. When I find this number, I will subtract it from your total number of remaining points. If you guess the right number, then I will give you 10,000 points. If you run out of points, then you will lose and be eliminated from the competition. In order to join, all you have to do is private message me your answers. There is also one more condition; I will also be posting a mathcounts question below the hints and if you get it correct, then you will recieve 1,000 points. Good luck and you have until June 10th to join. Here are the hints;

1. It is a positive integer.
2. It is divisible by 100.
3. It is less than 1000.
4. It has an even number of factors.

MATHCOUNTS 1988 School Target #5.

Nine squares whose sides are one inch long are placed together to form a three by three square. The figure also contains squares with other dimensions. What is the sum of the areas of all squares with integral dimensions contained in the figure?

me 2

srry 4 the last post...

i keep messing up!

do u no how 2 delete an accidentally created post?

I think you have to ask a mod to delete it.

put into games and fun-factory. Mods, plz move.

So far, only three people have submitted their guesses and their answer to the problem, and these are their scores.

1. tinytim                           999,900
2. Havingfitz                       1,010,000
3. BOGTRO                         1,000,900

  So far the rankings are shown below;

1st Place: Havingfitz
2nd Place: BOGTRO
3rd Place: tinytim

If you think that you have a good shot at winning, then you should submit your answer.

i thought i joined

cuz i pm-ed u the answers :D

:icecream:

Can you please submit your answers to me again ernie, because I must not have gotten them

I joined and have submitted my answer twice, yet I am not on the list of entrants. Why?

Oh, sorry 12markkram34, I think that I did get your answer, but I forgot to post it, but I do have it. Now these are the scores.

1. tinytim - 999,900
2. Havingfitz - 1,010,000
3. BOGTRO - 1,000,900
4. 12markkram34 - 1,000,900

1st Place is Havingfitz
2nd Place is BOGTRO
2nd Place is 12markkram34
4th Place is tinytim

Games belong in G&FF, even if they're slightly mathematical by answering a mathcounts question. Sorry.

I will join.
I will pm later. (Hopefully I will remember)

Here are the scores so far as of this morning;

 1. tinytim - 999,900
 2. Havingfitz - 1,010,000
 3. BOGTRO - 1,000,900
 4. 12markkram34 - 1,000,900
 5. erabel - 1,000,800

Here are the scores now that another person has submitted their guess.

1. tinytim - 999,900
2. Havingfitz - 1,010,000
3. BOGTRO - 1,000,900
4. 12markkram34 - 1,000,900
5. erabel - 1,000,800
6. budi713 - 1,000,900

There are now seven people that have submitted their answers. Here are the results.

1. tinytim - 999,900
2. Havingfitz - 1,010,000
3. BOGTRO - 1,000,900
4. 12markkram34 - 1,000,900
5. erabel - 1,000,800
6. budi713 - 1,000,900
7. chenhsi - 999,900

  I am now going to wait either until this weekend is over, or until there are up to 10 contestants. Which ever happens [b]last[/b].

There are now eight people that have submitted their guesses and answers.

tinytim - 999,900
Havingfitz - 1,010,000
BOGTRO - 1,000,900
12markkram34 - 1,000,900
erabel - 1,000,800
budi713 - 1,000,900
chenhsi - 999,900
abacadaea - 1,000,900

I think that I will only except answers until June 10th, but if I don't have ten competitors by then, then I will wait until there are ten. Good job to the 8 of you.

Um...

dingzhou just sent me a PM titled "Jackpot game"

with a number of like 99999999999999999999999999999...

For this week's round of jackpot, I have decided that I am actually going to change the question, and the number. sorry for those people that already sent me the answers.

These are the current results of the jackpot game. There is a star beside the names of the people that didn't send me their answers yet;

*12markkram34 - 900,900
abacadaea - 792,900
*BOGTRO - 900,900
*budi713 - 900,900
*chenhsi - 899,900
*erabel - 900,800
*Havingfitz - 910,000
mathforlife - 822,000
*tinytim - 899,900

Okay the scores are completely different now, but there have been two people that have reached zero, and they are both eliminated. Those two people are;

chenhsi
erabel

There are only seven people left.

Okay, nobody is answering the question and guessing numbers any more, so we are going to have to stop.

This was the top three

1. abacadaea
2. mathforlife
3. Havingfitz

Congratulations abacadaea!!! You've won the game.


geometry
circumcircle
trigonometry
geometry unsolved

In triangle $ ABC$ we have $ BC > CA > AB$. On side $ BC$  and on extension of $ BA$ lie points $ D$ and $ E$ (respectively) and $ BD\equal{}BE\equal{}AC$. Circumcircle of $ BED$ intersect side $ AC$ in point $ P$ and line $ BP$ intersect  circumcircle of $ ABC$ in point $ Q$. Prove that $ AQ \plus{} QC \equal{} BP$

See [i]Mathematical Olympiads 1999-2000: Olympiad Problems from Around the World, published by the American Mathematics Competitions[/i]-Page 81

I have two solution for it

solution1)We produced AQ from A toT such that QT=QC
then it is going to prove that AT=BP ,We cliam that triangle BDP and ACT are congruent
  We easily to see that $ BD \equal{} AC$ and$ \angle PBD \equal{} \angle TAC$,to prove our cliam it is only to see that $ \angle BPD \equal{} \angle CTA$
Because $ QC \equal{} QT$ then we get $ \angle AQC \equal{} 2\angle QTC$
By cyclic $ A,Q,C,B$ we get that$ \angle ABC \equal{} 180 \minus{} \angle AQC$
since $ BED$ is a  isoceles triangle with $ BE \equal{} BD$
then we get $ \angle BED \equal{} \dfrac{180 \minus{} (180 \minus{} \angle AQC)}{2} \equal{} \angle QTC$
but by cyclic $ B,E,P,D$ we  conclude that $ \angle BPD \equal{} \angle BED$
Then It is easy to see that our cliam is true

solution2)We will prove this problem byTrigonometry
(Note that we can easily to find every angle which is use in this solution)
Let $ \angle EBP \equal{} x$ By Law of sine on the circumcircle of triangle BED we get
$ BP : sin(90 \minus{} \frac {B}{2} \plus{} x) \equal{} BE : sin(90 \minus{} \frac {B}{2})$
or equivalent to
$ BP \equal{} \dfrac{BE*cos(\frac {B}{2} \minus{} x)}{cos\frac {B}{2}}$
and then by law of sine again with circumcircle of triangle ABC,we conclude that
$ AQ : sin x \equal{} QC : sin(B \minus{} x) \equal{} AC : sinB$
Hence,$ AQ \equal{} \dfrac{BE*sinx}{sinB} and QC \equal{} \dfrac{BE*sin(B \minus{} x)}{sinB}$
then it is easy to conclude that
$ \dfrac{sinx}{sinB} \plus{} \dfrac{sin(B \minus{} x)}{sinB} \equal{} \dfrac{cos(\frac {B}{2} \minus{} x)}{cos\frac {B}{2}}$
as desired


linear algebra
matrix
linear algebra unsolved

Hi friends,

I got stuck up with the following problem.

Let us assume that I have a matrix $A$, of size $n\times n$ and  $rank(A)=p$, whose entries are complex. If I decompose it into real and imaginary matrices $C$ and $D$ i.e, $A=C+jD$, then what would be the rank of the matrices $C$ and $D$. Will they be same as $p$ or they adjust in a way that $C+jD$ has a $rank\;\;p$ ? and what about the converse like

If I have two matrices $C$ and $D$, of different ranks say $p$ and $q$ respectively, then what would be the rank of matrix $A=C+jD$?

thank you and regards

bhupala

Consider two examples: $A=\begin{pmatrix}1&0\\0&i\end{pmatrix}$ and $A=\begin{pmatrix}1+i&0\\0&0\end{pmatrix}$. There is also the issue of real vs. complex rank: do you count the rows that are linearly independent over $\mathbb R$ or over $\mathbb C$?


superior algebra
superior algebra theorems

problem 1: In an ordered field 0<a<b implies b^-1<a^-1

problem 2:In an ordered field a^2+b^2=0 iff a = b = 0

help with the proof please. thanks.  :blush:

1: Multiply both sides by $ ab$.

2: Squares of nonzero numbers are positive.


Έστω $ n$ στρατηγοί που θέλουνε να τοποθετήσουν απόρρητα σχέδια σε ένα χρηματοκιβώτιο.Για τη μεγαλύτερη δυνατή ασφάλεια τοποθετούν σε αυτό $ a_n$ κλειδαριές και ο κάθε στρατηγός εφοδιάζεται με $ b_n$ κλειδιά έτσι ώστε το χρηματοκιβώτιο να ανοίγει αν και μόνο αν η πλειοψηφία των στρατηγών είναι παρούσα.Να βρεθεί το ελάχιστο $ a_n$ και στη συνέχεια να προσδιοριστεί το $ b_n$.
Το πρόβλημα το έχω λύσει ως ένα σημείο(αν θέλει κάποιος μου λέει να ποστάρω τη λύση) και βρήκα $ a_n\equal{}\binom{n}{k}$ και $ b_n\equal{}\binom{n\minus{}1}{k}$.Για να είναι πλήρης η λύση όμως δεν πρέπει να δείξουμε ότι όντως υπάρχει μια τέτοια διανομή κλειδιών με αυτά τα $ a_n,b_n$?Πώς βρίσκουμε μια τέτοια διανομή; :(
Ευχαριστώ εκ των προτέρων για τις απαντήσεις :)

Τι είναι το κ; Νομίζω η απάντηση σου λύνει την ερώτηση "Να βρεθεί το μικρότερο $ a_n$ ώστε το χρηματοκιβώτιο να ανοίγει αν και μόνο αν τουλάχιστον κ+1 στρατηγοί είναι παρόντες". (Άρα όπως έθεσες την άσκηση, έχουμε  $ k \equal{} \lfloor n/2 \rfloor$) Εκτός και αν χάνω κάτι.

Αν κατάλαβα καλά την άσκηση, νομίζω πως έλυσες το δύσκολο μέρος της και δεν κατάφερες να λύσεις το εύκολο κομμάτι.  :P

Ναι $ k \equal{} \left\lfloor\frac {n}{2}\right\rfloor$ ξέχασα να το βάλω sorry
[quote="Demetres"]Νομίζω η απάντηση σου λύνει την ερώτηση "Να βρεθεί το μικρότερο $ a_n$ ώστε το χρηματοκιβώτιο να ανοίγει αν και μόνο αν τουλάχιστον κ+1 στρατηγοί είναι παρόντες". (Άρα όπως έθεσες την άσκηση, έχουμε  $ k \equal{} \lfloor n/2 \rfloor$) Εκτός και αν χάνω κάτι.
[/quote]
Ναι ακριβώς αυτό είναι η άσκηση. :)

Νομίζω πως από την λύση του πρώτου μέρους φαίνεται (σχεδόν) αμέσως πως μπορείς να κάνεις την διανομή των κλειδιών. (Εκτός και αν έχουμε βρει τελείως διαφορετικές λύσεις.)

Μπορείς να γράψεις την δική σου αναλυτικά;

Aπό τα δεδομένα συνάγουμε ότι σε κάθε ομάδα $ k\equal{}\left \lfloor \frac{n}{2} \right\rfloor$ στρατηγών υπάρχει μία κλειδαριά της οποίας κανείς δεν έχει το κλειδί(θα τη λέμε απαγορευμένη).Επίσης δύο ομάδες $ k$ στρατηγών δεν μπορεί να έχουν την ίδα απαγορευμένη κλειδαριά γιατί τότε το χρηματοκιβώτιο δε θα άνοιγε με την πλειοψηφία παρούσα.Επομένως υπάρχουν τουλάχιστον $ \binom{n}{k}$ κλειδαριές δηλαδή $ a_n \geq \binom{n}{k}$.
Αν υποθέσουμε ότι γίνεται $ a_n \equal{} \binom{n}{k}$ τότε θα δείξουμε ότι $ b_n\equal{}\binom{n\minus{}1}{k}$.Πραγματικά ένας τυχαίος στρατηγός ανήκει σε $ \binom{n\minus{}1}{k\minus{}1}$ ομάδες $ k$ στρατηγών και καθεμία έχει διαφορετική απαγορευμένη κλειδαριά άρα ο $ b_n$ δεν έχει τουλάχιστον $ \binom{n\minus{}1}{k\minus{}1}$ κλειδιά δηλαδή $ b_n \leq \binom{n}{k}\minus{}\binom{n\minus{}1}{k\minus{}1}\equal{}\binom{n\minus{}1}{k}$.Επιπλέον ο τυχαίος στρατηγός ανήκει σε $ \binom{n\minus{}1}{k}$ ομάδες $ k\plus{}1$ στρατηγών από τους οποίους οι $ k$ (πλην του σταθερού που επιλέξαμε) έχουν μια απαγορευμένη κλειδαριά που υποχρεωτικά (αφού οι $ k\plus{}1$ έχουν όλα τα κλειδιά) την έχει ο σταθερός στρατηγός.Άρα $ b_n \geq \binom{n\minus{}1}{k\minus{}1}$

Λίγο διαφορετική από την δική μου που όντως δεν σε βοηθούσε άμεσα να βρεις την διανομή των κλειδιών.

Πάρε κ στρατηγούς και μια κλειδαριά που δεν μπορούν να την ανοίξουν. Αλλά κάθε κ+1 στρατηγοί μπορούν να την ανοίξουν άρα...

Sorry αλλά πάλι δεν βλέπω πώς προκύπτει η διανομή :blush:

Πάρε κ στρατηγούς και μια κλειδαριά που δεν μπορούν να την ανοίξουν. Αλλά κάθε κ+1 στρατηγοί μπορούν να την ανοίξουν άρα... οι υπόλοιποι ν-κ στρατηγοί μπορούν να ανοίξουν την συγκεκριμμένη κλειδαριά.

Για την διανομή, για κάθε ν-κ άτομα έχουμε μια κλειδαριά την οποία αυτοί (και μόνο αυτοί) μπορούν να την ανοίξουν. Άρα για κάθε κ άτομα υπάρχει κλειδαριά που δεν μπορούν να την ανοίξουν, αλλά κάθε κ+1 άτομα ανοίγουν όλες τις κλειδαριές (αφού στους κ+1 στρατηγούς, τουλάχιστον ένας θα έχει το κλειδί της).

[quote="giannis18"]Μάλλον δεν βλέπω κάτι προφανές εδώ αλλά ποια είναι η διανομή των κλειδιών;Δηλαδή αν αριθμήσουμε τα κλειδιά και τους στρατηγούς ποια κλειδιά ακριβώς θα πάρει ο κάθε στρατηγός;[/quote]

Το λάθος σου είναι πως ντε και καλά προσπαθείς να τους αριθμήσεις.  :P

Σε κάθε μία από τις $ \binom{n}{k} \equal{} \binom{n}{n\minus{}k}$ κλειδαριές, αντιστοιχούμε ένα από τα $ \binom{n}{n\minus{}k}$ υποσύνολα του $ \{1,\ldots,n\}$ με $ n\minus{}k$ στοιχεία και άρα με μια ομάδα $ n\minus{}k$ στρατηγών. Ο κάθε στρατηγός αυτής της ομάδας θα πάρει από ένα κλειδί για αυτήν την κλειδιαριά.

Π.χ. για ν=5, κ=2 έχουμε

Στρατηγοι: 1,2,3,4,5

Κλειδαριες: 123,124,125,134,135,145,234,235,245,345 (Γράφω 123 για την αντιστοιχεία της συγκεκριμμένης κλειδαριας με το σύνολο $ \{1,2,3\}$.)

Ο στρατηγός 1 παίρνει κλειδιά για τις 123,124,125,134,135,145
Ο στρατηγός 2 παίρνει κλειδιά για τις 123,124,125,234,235,245
 κτλ

Οι 1 και 2 π.χ. δεν μπορούν να ανοίξουν το χρηματοκιβώτιο γιατί δεν μπορούν να ανοίξουν την κλειδαρία 345. Αντίθετα οι 1,2 και 3 μπορούν να ανοίξουν όλες τις κλειδαριές.

Αν έγραφα 1,2,...,10 για τις κλειδαριές θα ήταν όντως ζόρικο να βρω ποιοι στρατηγοί παίρνουν πια κλειδιά.

Ok ευχαριστώ πολύ :)


Find all functions $f: \mathbb R \rightarrow \mathbb R$, such that:

$f(f(x)+y)=f(x^{2}-y)+4f(x)y$, for all $x,y \in \mathbb R$.

[quote="hucht"]Hallas el intervalo donde $f(x)<x^{2}$ luego el intervalo donde $f(x)>x^{2}$, como ambos intervalos son vacios se sigue que $f(x)=x^{2}$ ;)[/quote]
I'm not a very good translator, but I'll try so that you can understand:
If you try to find the interval where $f(x)<x^{2}$, then the interval where $f(x)>x^{2}$, both intervals are empty, therefore it must be $f(x)=x^{2}$.


geometry
3D geometry
tetrahedron
analytic geometry

In an $ ABCD$ tetrahedron $ G$ is the centroid and $ I$ is the incenter. $ GI$ is parallel to $ ABC$ face. Prove that area of the $ ABC$ face is the arithmetic mean of the area of the other faces.

[hide]Place the tetrahedron, which we will call $ \mathcal{T}$, on the coordinate plane such that face $ ABC$ is on the xy-plane.  Let $ h$ be the z-coordinate of point $ D$, noting that this is also the height.  Notice that $ I$ has the same z-coordinate as $ G$, wihose z-coordinate is the average of the z-coordinates of $ A$, $ B$, $ C$, and $ D$, or $ \frac{h}{4}$.  This means that $ I$ is a distance $ \frac{h}{4}$ from each face of $ \mathcal{T}$.

Now consider the four tetrahedra formed with $ I$ as the vertex and faces of $ \mathcal{T}$ as bases.  We will denote these tetrahedra as $ \mathcal{A}$, $ \mathcal{B}$, $ \mathcal{C}$, and $ \mathcal{D}$, where $ \mathcal{A}$ denotes the one whose base is $ \triangle ABC$.  Letting brackets denote area:

\begin{eqnarray*}
[\mathcal{T}] &=& [\mathcal{A}]+[\mathcal{B}]+[\mathcal{C}]+[\mathcal{D}] \\
\frac{1}{3}h[ABC] &=& \frac{1}{3}\cdot\frac{h}{4}[ABC]+\frac{1}{3}\cdot\frac{h}{4}[ABD]+\frac{1}{3}\cdot\frac{h}{4}[ACD]+\frac{1}{3}\cdot\frac{h}{4}[BCD] \\
\text{}[ABC] &=& \frac{1}{4}\left([ABC]+[ABD]+[ACD]+[BCD]\right) \\
4[ABC] &=& [ABC]+[ABD]+[ACD]+[BCD] \\
3[ABC] &=& [ABD]+[ACD]+[BCD] \\
\text{}[ABC] &=& \frac{[ABD]+[ACD]+[BCD]}{3}
\end{eqnarray*}
Q.E.D.[/hide]


Let a convex quadrilateral ABCD satisfying $ \angle BCD\equal{}\angle CDA$. The bisector of angle ABC cuts CD at E. Prove that $ \angle AB\equal{}90^o$ if and only if AB=AD+BC.

[quote="mr.danh"]Let a convex quadrilateral ABCD satisfying $ \angle BCD = \angle CDA$. The bisector of angle ABC cuts CD at E. Prove that $ \angle AB = 90^o$ if and only if AB=AD+BC.[/quote]
If $ \angle AEB=\frac{\pi}{2}$ then ${ \angle CEB}<\frac{\pi}{2}$
So there is a point $ F$ on side $ AB$ such that $ \angle BEF=\angle BEC$
Then triangle $ BEC$ and $ BEF$ are congruent,so $ BC=BF$ and $ \angle BFE=\angle BCE=\angle EDA$
Thus,quadrilateral $ ADEF$ is cyclic.Because $ \angle AEB =\frac{\pi}{2}$ and $ \angle CEB=\angle BEF$,we have ${ \angle FEA}=\angle AED$
So $ \angle FDA=\angle FEA=\angle AED=\angle AFD$
Hence $ AF=AD$ and $ AB=AF+BF=AD+BC$


What's so great about skipping school? I mean, especially Mon. That's when I have MC and band (yea! no choir). And, I get logic probs for Algebra :) .

alright, so some days are ok. the rest of the days...phbbt...

Mondays, Wednesdays, and Fridays are the best cuz that's when I have band (except for every other Friday) and I have MC on those days, even when I have choir on Fridays, it's not that bad 'cuz we always do Countdown in the mornings...

whoa  you have MC 3 times a week? lucky you. our school had it once a week. and we still got 1st in state. shows you what the [i]rest[/i] of the state is like

jk we just had some good peeps on the team

yea...from 7:15 til 8:15 and for like the 2-3 weeks b4 a competition we meet every day after school fo an hour and a half...last year we got fifth cuz Matt bombed the sprint round. also cuz I wasn't aon the team, stupid peter stupid peter, he just had to take my place just cuz he was in 8th grade and it was his last year :evil:

Woah, nice! I only started (well, resurrected) MATHCOUNTS so we're lucky to have our short weekly meeting, nonetheless [b]3[/b]...

We're doomed...

not necessarily...lessee... at our school (which had practice once a week), we had one guy place 10th in state, one guy place 6th in state, and one guy place 1st in state. The NC team had like 3 practices before nats but they still got 5th place and one guy (not me) got 13th place or someat.  So... you're not necessarily doomed (yet).

alright, thanks for the encouragement MysticTerminator.

Let's end this thread here since this is [b]way[/b] off-topic, although this is still on forum :)
consider this thread locked.

[quote="Syntax Error"]we meet once a week for like an hour i think. it starts this week. and it was the same last year and the team still got 1st. yup yup

i believe that mathcounts practices are pretty much useless, you can accomplish a whole lot more by yourself. but its still fun nonetheless[/quote]
agreed. still, it helps to know the peeps a bit.

Mmm, also I'd agree. Although we haven't officially started, I can tell we aren't going to learn much from a former theater major...I mean, I know a lot more than our coach and our coach is like "yeah, I know. but I'm still a better comedian than you, so ha!" :) </no offence to anyone who recongnized who I was talking about, and the person who I'm talking about>

another reason why we're doomed...

my stupid middle school got second in the state 2 yrs in a row when i was on the team. for some strange reason, we failed to beat a rival middle school even though our team was quite well prepared. we met 3 times a week too. now that im in high school, i assist in coaching the mc team. this year well meet at least 4 a week.

yeah, you do get to know the peeps better. but all we do is grade the warmups and workouts there. and eat and drink.

Doesn't the last part make it worthwhile? :) </jk>

notice where i said "its fun nonetheless" in one of my previous posts


integration
real analysis
real analysis unsolved

Suppose $x_n$ is a enumeration of all rational numbers in $(0,1)$. For any $x$ we define $f(x)=\sum_{x_n<x}{\frac{1}{2^n}}$. What is $\int_{0}^{1}{f(x)dx}$?

$\int_0^1 f(x)\,dx$ is strictly between 0 and 1. It can be arbitrarily close to zero, and it can be arbitrarily close to 1. I suspect it can be any real number in that range.


On the AMC, if a problem involves trig and does not specify whether an angle ie in deg or rad, what should be assumed?

Well, usually you can tell if its going to be radians or not. But if there is no degree sign, then you must take it to be radians.

Yeah, there was this weird question about finding the number of roots of cos(log(x)) over 0<x<1 I saw the other day, but it turns out that the answer was infinity so it didn't matter whether x was in deg or rad.

That is such an awesome problem. The point is that if you are using a graphing calculator, you'll see two roots regardless of which window size you use. But the answer isn't two, and it's easy to do it with thinking and hard to do it with a calculator.

you should try both just in case, thats what i do

[quote="tetrahedr0n"]i missed an mmpc problem this year by assuming the answer was in degrees not radians. i chose the "none of the above" option. lol[/quote]

oy veh, i hate ones with "none of the above" grrrr :-x

"none of the above" just makes you want to choose it more, that's usually why they put it.

why does calculus use radians? when i change to degree mode on my ti-89 the derivatives give weird stuff.

The derivative of sin(x) in radians is cos(x). sin(x) in degrees is the same as sin(pi/180 x) in radians, so the derivative is pi/180*cos(pi/180 x) radians or pi/180*cos(x) in degrees. Radians are definitely nicer for calculus.


integration
calculus
MATHCOUNTS
rate problems

Joe has a laser gun that fires lasers once he presses the trigger. For the first second, it charges up, and then for the next 1.72 seconds, it fires laser beams at the rate of $\frac{2}{x}$ laser beams per second, where x is the amount of seconds since the trigger has been pressed. After the 2.72 seconds (1 second of charging, 1.7 seconds of firing) have elapsed, the gun sprouts legs and walks away or something.

Figure out how many laser beams Joe's gun fires. I know you people can do this in some way. Think of any way you can do this - estimating, educated guessing, whatever. And don't forget to post your solution - I want to know how many different ways people can solve this.

Don't you need more information? Like the total amount of time or how long he pressed the trigger or something?

[quote="lotrgreengrapes7926"]Sounds like this needs calculus. Obviously, there's an easier way but whatever. [hide="Probably wrong solution"] $\displaystyle\int^{2.7}_{1} \frac{5}{x} \,dx=4.9662589$ according to my graphing calculator. So $\boxed{4}$, maybe?? [/hide][/quote]

It is correct (I changed the numbers to make it a little easier), but it is possible without calculus.

[hide] I guess the best way is to first calculate the number of times per second the laser will fire at 0.1 second intervals. Then find the number of beams it will fire in the 1.72 second period and average them.
$\displaystyle{\frac{1.72\displaystyle\sum_{i=0}^{16} {\frac{2}{1+0.1i}}}{17}}=2.15$. So the answer is $\boxed{2}$? [/hide]
[hide="Checking with Calculus"] $\displaystyle\int^{2.72}_{1} \frac{2}{x} \,dx=2.00126$. $\boxed{2}$. :D [/hide]

[quote="jclarkemathL314159"]Well since this is the mathcounts forum, you probably don't have to use summations or calculus.

I guess you could try graphing, but that would just lead to calculus[/quote]
 :roll: I guess I should explain the summation. [hide] $\displaystyle{\frac{1.72\displaystyle\sum_{i=0}^{16} {\frac{2}{1+0.1i}}}{17}}=2.15$. Ok, the sum from 0 to 16 of $\frac{2}{1+0.1i}$ is basically $\frac{2}{1}$, $\frac{2}{2.1}$, etc., the rate of laser beams at intervals of 0.1 seconds: at 1 second, 1.1 second, 1.2 seconds, etc. Adding these together, we divide this by 17 to get the average rate of the laser beams per second. Then we multiply this by 1.72 because there are 1.72 seconds in which the laser beam is on and firing. :D [/hide]

[quote="Yooper"]Nice explanation.  Though I get 2.075 for your sum

Here's a nice use of sum(seq( keys on TI to to provide a non-calculus estimate for the solution to a calculus problem.  I used:
1.72sum(seq(2/(1+0.1N),N,0,16,1))/17=2.075

Find [i]sum([/i] on 2nd LIST MATH
   and [i]seq([/i]  on 2nd LIST OPS  menus

Since the choice of using 10ths was not mentioned in problem, 100ths would give a better estimate:

 [hide]1.72sum(seq(2/(1+0.01N),N,0,169,1))/170=2.016[/hide][/quote]
I think your second one should be [hide] 1.72sum(seq(2/(1+0.01N),N,0,172,1))/173=2.0033 :D Since it's firing for 1.72 seconds. [/hide]

I am sorry, but I am slow.   I still don't understand how you were able to do it non-calculusy, though it was written perfectly :huh:

BTW, we are 'sort of' of doing calculus in the sense that this could be from the section in a calculus textbook 1 or 2 sections ahead of learning the offical rules of differentiation.  On the other hand, it is 'just' rate*time, so it is algebra.  The algebra approach should make sense, but is very tedious (but TI does the dirty work this time) and only gives estimates.  Calculus is quick and exact.

I'll try to elaborate on greengrapes post #7, a very nice explanation.  Since the rate 2/x (shots per second) from x=2 to x=3.72 seconds is constantly changing, we can't do rate*time for EVERY instant.  So we break the 1.72 second interval into a convenient interval, say 0.1 sec intervals - so there will be 17 intervals.

From t=2 to t=2.1 sec, we use a rate of 2/2=1 shot per second.  From t=2.1 to t=2.2 sec, we use a rate of 2/2.1 shots per second. Etc for a total of 17 rates.  The last one is:From t=3.6 to t=3.7 sec, we use a rate of 2/3.6 shots per second.  Then greengrapes summed these 17 rates and divided by 17 to find the average rate of shots per second = 1.206.  Then to get the total number of shots, he used rate*time:  1.206 shots/sec * 1.72 sec = 2.075 shots

That's the math, hope it helps :)   Then I suggested using an interval of 0.01 sec to get a more accurate answers.  The rest is to understand summation notation and the TI key punching.

Yooper: this is the kind of answer I was looking for - an answer that explains a complicated concept in a way involving much simpler ideas.

You get a cookie and a helicopter attached to your head. :heli:

This question really confused me. I first thought of my calculator, but I couldn't think of anything to put in it. I let common sense take over.
[hide]
Obviously the firing is going to slow down. At the beginning one second has passed, so the rate is 2/1 or 2 beams per second. at the end almost 3 seconds have passed so the rate is 2/3 beam per second. now the real average rate is going to be somewhere in between there, but since it's inverse and because of the shape of inverse graphs it'll be closer to the lower bound. the shooting goes on for 2 seconds, making the number of beams shot at most 4 and at least 4/3.  The real number would be between there, and closer to the lower bound. Now My mental-mathing self had a problem with partial laser beams. she either had a good justification or she was too lazy. so she concluded it had to be 2 or 3 beams. closer to the lower bound...2. [/hide]
Thank-you for posting this problem... I now understand a little of what calculus is...


The product of three consecutive odd integers is 1287. What is
the sum of the three integers?

Notice that $ \sqrt[3]{1287}$ is almost $ 11$, hinting that $ 11$ would be your middle number. The answer is $ 9\plus{}11\plus{}13\equal{}(9\plus{}2)\plus{}11\plus{}(13\minus{}2)\equal{}11(3)\equal{}\fbox{33}$.


For real numbers $a,b,c$, prove that \[(a^{2}+b^{2}+c^{2}-ab-bc-ca)^{2}\leq \frac{3}{2}\sqrt{4(a-b)^{4}(b-c)^{4}+(b-c)^{4}(c-a)^{4}+(c-a)^{4}(a-b)^{4}}.\] When does the equality hold? (note that RHS is NOT cyclic.)

Let $x=a-b,y=b-c,z=c-a$, so that $x+y+z = 0$
the ineq becomes $\frac{1}{4}(x^{2}+y^{2}+z^{2})^{2}\leq \frac{3}{2}\sqrt{4x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4}}$
Notice two identities
$x^{2}+y^{2}+z^{2}=-2(xy+yz+zx)$
$x^{2}y^{2}+z^{2}x^{2}+y^{2}x^{2}= (xy+yz+zx)^{2}$
Using the first and squaring both sides, we have to prove
$(xy+yz+zx)^{4}\leq \frac{9}{4}(4x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4})$
Writing $\frac{9}{4}= \frac{1}{4}+1+1$, and use Cauchy-Scharz + second identity, we obtain the desired result.


geometry
3D geometry
number theory
prime factorization

A solid rectangular block is formed by gluing together $N$ congruent $1$-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly $231$ of the $1$-cm cubes cannot be seen. Find the smallest possible value of $N$.

Okay, i'm not the best at stuff like this but here's a shot
[hide]
$x = width$ $y = height$ $z = depth$
okay, if you can full see 3 faces of a cube then there is
$(x-1)(y-1)(z-1)$ cubes not visible.
There fore
$(x-1)(y-1)(z-1)=231$
Then if you take the prime factorization of $231$ you get $3*7*11$
so then just let
$x=4$ $y = 8$ $z = 12$
$x*y*z=384$
There is my answer $\boxed{384}$
[/hide]
Sorry if i'm in the completely wrong direction.

Edit:  Misread the question.  The above solution seems right.

[quote="t0rajir0u"]If $a, b, c$ are the side lengths of the block, then $231 = ab + bc + ac$ and we want to minimize $N = abc = \sqrt{ab \times bc \times ac}$.  With a constant sum, minimizing the product occurs when $a, b, c$ are as different as possible, so $a = 115, b = c = 1$ gives us $N = \boxed{115}$.

(A short logical check for the other extreme, when $a = b = c = \sqrt{231/3} > 8$, gives us a value of $N > 8^3 = 512$)[/quote]

I think you misread the question. It says 231 blocks CANNOT be seen.


[quote="vinny919"]I think the USAMTS Solutions Submissions Page says that they don't want any "half-baked ideas." What if, however, I do not have a complete solution, but I have two or three possible avenues which might lead to a solution, except that I am missing some insight and was thus unable to get the answer.  Should I include all two or three of these partial solutions and go as far as I can in hopes that one of them is the way you're "supposed" to do it and thereby get partial credit? Thanks[/quote]

This is the kind of question we can't answer while a round is in progress.

[quote="vinny919"]I think the USAMTS Solutions Submissions Page says that they don't want any "half-baked ideas." What if, however, I do not have a complete solution, but I have two or three possible avenues which might lead to a solution, except that I am missing some insight and was thus unable to get the answer.  Should I include all two or three of these partial solutions and go as far as I can in hopes that one of them is the way you're "supposed" to do it and thereby get partial credit? Thanks[/quote]

Half-baked ideas are very hard for us graders to judge. If an unfinished idea matches the beginning of one of the common submissions, either a correct solution or a fallacious one, we can grade it by assuming it would follow the common pattern and give it partial credit by how far it went down that path. However, if the idea is uncommon, we have no good way of judging it.

Pierre Fermat once scribbled in the margin of his copy of "Arithmetica" the half-baked idea, "It is impossible for a cube to be the sum of two cubes, a fourth power to be the sum of two fourth powers, or in general for any number that is a power greater than the second to be the sum of two like powers. I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain." (except that he wrote it in Latin). As you can see from that example, some half-baked ideas are very hard to evaluate. And if one of you writes a half-baked idea as grand as Fermat's Last Theorem, we graders will probably miss the grandeur and never think to mention it to anyone.

And writing down three avenues of possible solutions looks like you want the grader to choose which method you should use. When a participant submits two correct and complete solutions by distinct methods, I assume that that person is just showing off and I grade the best-written one. But with uncompleted solutions, the participant can't be showing off. Instead he is being deliberately indecisive. That is rude.

What would be easier on us graders would be if you simply list the significant facts about the problem that you discovered already and then confess that you appear to lack the mathematical tools to solve the problem with those facts. Some solutions have key facts that we graders use as milestones for assigning partial credit. If you found and verified such a fact, you would earn the partial credit. However, the key fact that is most important is the answer itself. The answer (or a wrong answer easily corrected to the right one) is almost always required to score more than 2 points.

Erin Schram
USAMTS grader


let x,y are non-gative numbers such as : 2xy+$ {x}^{2}$ $ \ge$ 0

and let a,b,c, are base of triagle ABC ,S is the area of ABC

prove that:$ {x}^{2}$ $ \times$ ($ {a}^{2}$ +$ {c}^{2}$ )+$ {y}^{2}$ $ \ge$ 4S$ \sqrt{2}${x}^{2}${ +xy }$

[quote="trannguyenquyND_VN"]let x,y are non-gative numbers such as : 2xy+$ {x}^{2}$ $ \ge$ 0

and let a,b,c, are base of triagle ABC ,S is the area of ABC

prove that:$ {x}^{2}$ $ \times$ ($ {a}^{2}$ +$ {c}^{2}$ )+$ {y}^{2}$ $ \ge$ 4S$ \sqrt {2}${x}^{2}${ + xy }$[/quote]

What is with the attached picture? :huh:

By the way did you mean:
$ x, y$ are nonnegative numbers such that $ 2xy + x^{2}\ge 0$,
and let $ a, b, c$ be the sides of triangle ABC with area $ S$, prove that:
$ x^2\cdot (a^2+b^2+c^2) + y^2\ge 4S\sqrt{2x^2 + xy}$

If not please rectify. I did not understand your question. :(
[hide="PS"]Are you nervous when using LaTeX ??  :wink:  [/hide]

[quote="trannguyenquyND_VN"]let x,y are non-gative numbers such as : 2xy+$ {x}^{2}$ $ \ge$ 0

and let a,b,c, are base of triagle ABC ,S is the area of ABC

prove that:$ {x}^{2}$ $ \times$ ($ {a}^{2}$ +$ {c}^{2}$ )+$ {y}^{2}$ $ \ge$ 4S$ \sqrt {2}${x}^{2}${ + xy }$[/quote]
 :lol:  :lol:  :)  LHS=x(a^2+c^2)+y.b^2
it very easy (look 5).