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Does there exist a function $s\colon \mathbb{Q} \rightarrow \{-1,1\}$ such that if $x$ and $y$ are distinct rational numbers satisfying ${xy=1}$ or ${x+y\in \{0,1\}}$, then ${s(x)s(y)=-1}$? Justify your answer.

[i]Proposed by Dan Brown, Canada[/i]

Where did you find this problem? It is realy interesting.

I believe the answer to be 'yes', and I think I have a way of constructing the function, but it's only verified empirically, and I haven't fixed my ideas well enough to provide a full proof. Here's what I thought we could do:

It suffices to determine the function on the positive rationals, of course, so let's restrict our attention to those. Construct the [url=http://www.cut-the-knot.org/blue/Stern.shtml]Stern-Brocot tree[/url] step by step. At each step, to each leaf you attach a leaf to the left and a leaf to the right. Label the leaf to the left with $-1$ and the leaf to the right with $1$. It seems to work.

We have $s(x)=s(1+x)=-s(1/x)$. I see a direct way to use continuous fractions here, since having these rules we can obtain $s(x)$ from $s(1)$. :?
Did I say something stupid?

Solution: Let x=a/b be a positive rational, where a,b are coprime positive integers. Consider the squence of consecutive remainders given by euclidean algorithm for the ordered pairs (a,b). if  u mod v denotes the least non negative remainder of u modulo v, this sequence can be written as
r_0=a,   r_1=b,  r_2=a mod b, ... , r_(i+1)=r_(i-1) mod r_i, ... , r_n=1, r_(n+1)=0    (1)

the index n=n(x) of the least nonzero remainder r_n=1 is uniquily determined by x, so
t(x)=(-1)^n(x) is a well-defined function from the positive rationals into {-1,1}. Now define s: Q-->{-1,1} by

       s(x)=t(x) for x in Q, x>0 or s(x)=-t(-x) for x in Q, x<0 or s(0)=1

We prove that s has the desired properties. Let x and y be distinct rational numbers.
*  If x+y=0, let x>0 and y<0 (x,y are nonzero). Then , by definition, s(x)=t(x),
    s(y)=-t(-y)=-t(x), hence s(x)s(y)=-1.
**If xy=1 then x and y are of the same sign and neither equals 1 or -1. Suppose first
   that x=a/b >0, y=b/a >0, with a,b coprime positive integers, one may also assume
   that a>b. Euclidean algorithm for (a,b) starts a,b, (a mod b), ... .On the other hand,
   the algorithm for pair (b,a) gives b,a,b,(a mod b)m,... .
   Because a>b implies r_2=(b mod a)=b. Each term in (1) depends only on the previous
   two, so the sequence  for x has length by 1 less than the sequence for y, that is,
   n(y)=n(x)+1. Hence t(y)=-t(x), and so s(y)=-s(x), as needed. Now the case of x,y
   follows from the definition...
**If x+y=1 then ....

by the same logic it'll be true...

We see that $t(x)$ is a number of subfractions in the continuous fraction representing $x$.

I knew it's imoral relive topic but I have a question: we can prove that $ s(1) \equal{} 0$ if we use $ s(x) \equal{} \minus{} s(\frac {1}{x})\Leftrightarrow s(1) \equal{} \minus{} s(1)\Leftrightarrow s(1) \equal{} 0$ contradiction.

I dont really understand :maybe:

x and y are distinct rational numbers if you use s(x) =-s(1/x) for x=1 then we have x=1=1/1=1/x=y.

Every positive rational number has a unique continued fraction of the form $ a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{\ddots\frac{1}{a_{m - 1} + \frac{1}{a_m}}}}} $ for some $ m \in \mathbb{N} $ where $ a_i \in \mathbb{N} $ for all $ i $.  From now on this will be denoted as $ [a_0; a_1, a_2, \dots, a_m] $.

For every $ a \in \mathbb{Q}^{+} $, let $ s(a) = (-1)^m $.  Then let $ s(0) = -1 $ and for all $ b \in \mathbb{Q} \setminus \{0\} $ let $ s(-b) = -s(b) $.  I claim that this $ s $ satisfies the desired properties.  To prove this, consider two rational numbers $ x < y $.

$ 1) $ If $ x + y = 0 $ then by definition $ s(x)s(y) = -1 $ as desired.

$ 2) $ If $ x + y = 1 $ and $ x = 0 $ then by definition  $ s(x)s(y) = -1 $ as desired since $ s(1) = 1 $.

$ 3) $ If $ x + y = 1 $ and $ x < 0 $ then to prove that $ s(x)s(y) = -1 $ it suffices to show that $ s(-x) = s(y) $.  Let $ -x = [a_0; a_1, a_2, \dots, a_m] $.  Then $ y = [a_0 + 1; a_1, a_2, \dots, a_m] $ and so $ s(-x) = s(y) = (-1)^m $ as desired.

$ 4) $ If $ x + y = 1 $ and $ x > 0 $ we can let $ y = [0; 1, a_2, \dots, a_m] $ so that $ x = [0; a_2 + 1, a_3, \dots, a_m] $ so $ s(y) = (-1)^m $ and $ s(x) = (-1)^{m - 1} $ which implies that $ s(x)s(y) = -1 $ as desired.

$ 5) $ If $ xy = 1 $, we can assume WLOG that $ 0 < x < 1 $.  Then letting $ x = [0; a_1, a_2, \dots, a_m] $ we have that $ y = [a_1; a_2, a_3, \dots, a_m] $ so $ s(x) = (-1)^m $ and $ s(y) = (-1)^{m - 1} $ which implies that $ s(x)s(y) = -1 $ as desired.

This implies that my claim is true and  so we are done.  Note that this also implies that Grobber's idea about the Stern-Brocot tree works as well.

Note also that this function $s$ is unique up to multiplication by $-1$. Indeed let $x\in\mathbb{Q}^{+}$ and $x=[a_0;a_1,\ldots,a_m]$. Using $s(x+1)=s(x)$ and $s(x)=-s(1/x)$ it's easily obtained $s(x)=(-1)^m s(1)$.

Fake alg, pure NT :furious:. Anyway, here's my solution: For two integers $0<p<q$, let $f(p,q)$ denote the number of steps taken in the Euclidean Algorithm to find $\gcd(p,q)$. In other words, if we let $b_0=q,$ $b_1=p$ and write the system of equations $$b_r=a_{r+1}b_{r+1}+b_{r+2} \text{ such that } 0<b_{r+1}<b_r \quad \forall r \in \{0,1, \dots ,n\} \text{ and } b_{n+2}=0$$
then $f(p,q)=n,$ where $n \geq 0$. Then we claim that the following function works-
[quote=The ProGawd Function]
[list=1]
[*] $s(n)=1$ for $n \in \mathbb{N},$ and $s(-m)=-1$ for $m \in \mathbb{N} \cup \{0\}.$
$\text{ }$
[*] $s \left(\frac{-1}{2} \right)=1.$
$\text{ }$
[*] $s(1+x)=s(x)$ for $x \neq \frac{-1}{2},0.$
$\text{ }$
[*] For integers $p,q$ with $0<p<q$, $$s \left (\frac{p}{q} \right)=(-1)^{f(p,q)+1}$$
[/list]
[/quote]
It's easy to see that this uniquely defines the image of all rationals. Now we show that this indeed satisfies all the given properties.
[list]
[*] First we prove $s(x)s(-x)=-1$ when $x \neq 0$. For $x \in \mathbb{N}$ and $x=\pm \frac{1}{2}$, this easily follows from the definition, so we can ignore these cases. WLOG assume $x>0$ and suppose $x \in (n-1,n)$ for some $n \in \mathbb{N}$ with $2x \neq 1$. Then from property 3 of our function, we have $s(x)=s(x-(n-1))$ and $s(-x)=s((n-1)-x)$. Since $0<x-(n-1)<1,$ so it suffices to show that $s(x) \cdot s(-x)=-1$ for $x \in (0,1)$.

Take $x=\frac{p}{q}$ for $0<p<q$, and first suppose $p< \frac{q}{2}$. We have $$s \left( \frac{-p}{q} \right)=s \left(1-\frac{p}{q} \right)=(-1)^{f(q-p,q)+1}$$
But, as $q-p>\frac{q}{2}>p$, so the first step of our Euclidean Algorithm while finding $\gcd(q-p,q)$ will be $q=(q-p)+p$. So $f(q-p,q)=1+f(p,q-p)$. Also, using $q>2p$, we get $f(p,q-p)=f(p,q)$. Thus, we have $$s \left( \frac{-p}{q} \right)=(-1)^{f(q-p,q)+1}=(-1)^{1+f(p,q)+1}=-s \left( \frac{p}{q} \right)$$
Thus, for $x \in \left(0, \frac{1}{2} \right),$ we have $s(x)s(-x)=-1$. Finally, if $x \in \left(\frac{1}{2},1 \right),$ then $s(x)=s(x-1)$ and $s(-x)=s(1-x)$. Since $0<1-x<\frac{1}{2}$, then the previous result gives $s(1-x)s(-(1-x))=-1$. Combined with the above equalities, we get $s(x)s(-x)=-1$. Thus, the result is true for all $x \in \mathbb{Q}$.
$\text{ }$
[*] Now we show that $s(x)s(1-x)=-1$ for $x \neq \frac{1}{2}$. For $x=1$, the result is true by definition. And for $x \neq 1,\frac{1}{2}$, by the above result, we have $s(1-x)=-s(x-1)=-s((x-1)+1),$ which directly gives the desired property.
$\text{ }$
[*] Finally we show that $s(x)s \left(\frac{1}{x} \right)=-1$ when $x \neq 0,\pm 1$. Since $s(-x)=-s(x)$ (proved before), so it suffices to show the result for positive $x$. Also, WLOG we can take $0<x<1$, and write $x=\frac{p}{q}$ with $p<q$ and $p,q \in \mathbb{N}$. Then $$s \left(\frac{1}{x} \right)=s \left( \frac{q}{p} \right)=s \left( \frac{q}{p} -1\right)= \dots =s \left( \frac{q \pmod{p}}{p} \right)=(-1)^{f(q \pmod{p},p)+1}$$
But, while finding $\gcd(q\pmod{p},p)$ we simply start from the second step in the Euclidean Algorithm to find $\gcd(p,q)$. So we have $f(q \pmod{p},p)+1=f(p,q)$. This gives us $$s \left(\frac{1}{x} \right)=(-1)^{f(q \pmod{p},p)+1}=(-1)^{f(p,q)}=-s \left(\frac{p}{q} \right)=-s(x)$$
Thus, we have $s(x)s(-x)=-1$ as desired. $\blacksquare$

[hide=Solution] The answer is yes. For any positive rational number $r$, if $r$ $\not \in$ $\mathbb{Z}$, then we can write $r = \frac ab$, where $a$, $b$ $\in$ $\mathbb{Z}^+$ and $gcd(a, b) = 1$, and we define $f(r) = \frac{b}{a\ mod\ b}$, where $0 \le a\ mod\ b < b$. For positive rational $r$, if $r$ $\not \in$ $\mathbb{Z}$, we define $g(r)$ to be the minimum repeated applications of $f(r)$ until an integer is obtained, and $g(r) = 0$ if $r$ $\in$ $\mathbb{Z}^+$. We claim that the following function satisfies the conditions in the problem: \[\{s(0) = -1,\ s(x) = (-1)^{g(x)}|\ x > 0,\ s(x) = (-1)^{g(-x) + 1}|\ x < 0\}\ \forall\ x\ \in\ \mathbb{Q}\]

First, this function clearly satisfies $s(x) = s(-x)$ $\forall$ $x \ne 0$ $\in$ $\mathbb{Q}$. Also, for any positive rational $x \ne 1$, $min(\frac 1x, x) < 1$, so assume WLOG $x < 1$. Then $x$ $\not \in$ $\mathbb{Z}$, and we can write $x = \frac ab$, where $a$, $b$ $\in$ $\mathbb{Z}^+$ and $gcd(a, b) = 1$, and $\frac 1x = \frac ba$. Since $x < 1$, $a < b$, therefore $f(x) = \frac ba = \frac 1x$, so $g(x) = g(\frac 1x) + 1$, hence $s(x) \ne s(\frac 1x)$ $\forall x > 0$ $\in$ $\mathbb{Q}$, and since $s(-x) = -s(x)$, we obtain the same result for all negative rationals  $\ne -1$ as well. Finally, we will show that for any $x \ne \frac 12$ $\in$ $\mathbb{Q}$, $s(x) \ne s(1-x)$. We consider several cases:

[b]Case I: $x$ $\in$ $\mathbb{Z}$[/b]

If $x$ $\in$ $\{0, 1\}$, then exactly one of $x$, $1 - x$ is equal to $0$ and the other is $1$, and $s(0) \ne s(1)$, so we are done. Otherwise, exactly of one $x$, $1 - x$ is positive and the other is negative, but since $s(x) = 1$ $\forall$ $x$ $\in$ $\mathbb{Z}^+$ and $s(x) = -1$ $\forall$ $x$ $\in$ $\mathbb{Z-}$, we must have $s(x) \ne s(1 - x)$ $\forall$ $x$ $\in$ $\mathbb{Z}$.

[b]Case II: $x$ $\not \in$ $\mathbb{Z}$, $x$ $\not \in$ $[0, 1]$ [/b]

Since $x$ $\not \in$ $[0, 1]$, exactly one of $x$, $1 - x$ is positive and the other is negative, so WLOG assume $x > 0$. Furthermore, we can express $x$ as $\frac ab$, where $a$, $b$ $\in$ $\mathbb{Z}^+$, $gcd(a, b) = 1$, $a > b$. Then, $1 - x = \frac{b - a}{b}$, so $s(1-x) = -s(x - 1) = -s(\frac{a - b}{b})$, however since $g(\frac ab) = g(\frac{a - b}{b})$, $s(\frac{a - b}{b}) = s(\frac ab)$, so $s(1 - x) \ne s(x)$ $\forall$ $x$ $\in$ $\mathbb{Q}$, $\not \in$ $\mathbb{Z}$, $\not \in$ $[0, 1]$

[b]Case III: $x$ $\not \in$ $\mathbb{Z}$, $x$ $\in$ $[0, 1]$ [/b]

Since $x$ $\in$ $[0, 1]$, $1 - x$ $\in$ $[0, 1]$ as well. Moreover, exactly one of $x$, $1 - x$ is less than $\frac 12$ and the other is greater than $\frac 12$, so assume WLOG $x > \frac 12$ Therefore, we can express $x$ as $\frac ab$, where $a$, $b$ $\in$ $\mathbb{Z}^+$, $gcd(a, b) = 1$, $a < b < 2a$. Then, $1 - x = \frac{b - a}{b}$, and $f(1 - x) = \frac{b}{b - a}$. Also, since $x > \frac 12$, $\frac ba$ $\not \in$ $\mathbb{Z}$, so we have $f(f(x)) = f(\frac ba) = \frac{a}{b - a}$. However, $g(\frac{b}{b - a}) = g(\frac{a}{b - a})$, so $g(x) = g(1 - x) + 1$, hence $s(x) \ne s(1 - x)$ $\forall$ $x$ $\in$ $\mathbb{Q}$, $\not \in$ $\mathbb{Z}$, $\in$ $[0, 1]$

Therefore, $s(x) \ne s(-x)$ $\forall$ $x \ne 0$ $\in$ $\mathbb{Q}$, $s(x) \ne s(\frac 1x)$ $\forall$ $x$ $\in$ $\mathbb{Q}$, $\not \in$ $\{-1, 0, 1\}$, $s(x) \ne s(1 - x)$ $\forall$ $x \ne \frac 12$ $\in$ $\mathbb{Q}$. Thus, the function $s$ satisfies all the conditions in the problem, as desired.

The answer is yes, and we construct such a function. First, note that the only condition on $s(0)$ is that $s(0)=-s(1)$, which we can manually enforce at the end. Thus, we assume that the inputs to $s$ are nonzero rationals.

Write $x=a/b$ where $\gcd(a,b)=1$ and $b>0$. We set \[s(x) = \begin{cases}1 &\text{if } (a^{-1}\mod b)\le b/2 \\ -1 &\text{if } (a^{-1}\mod b)>b/2.\end{cases}\] From this definition, it is clear that $s(x)= - s(-x)$ and $s(x)=-s(1-x)$, so it suffices to show that $s(x)=-s(1/x)$. Since we know $s$ is odd, it suffices in fact to show $s(x)=-s(1/x)$ for $x>0$.

[b]Lemma:[/b]  If $a,b\ge 1$ are relatively prime positive integers, then \[a\cdot(a^{-1}\mod b) + b\cdot(b^{-1}\mod a) = ab+1.\]

[i]Proof:[/i]  Indeed, letting $T$ be the left side of the above equation, we see that $T\equiv 1\pmod{a}$ and $T\equiv 1\pmod{b}$, so $T\equiv 1\pmod{ab}$ since $\gcd(a,b)=1$.

Furthermore, we have \[a+b\le T\le a(b-1)+b(a-1)<2ab,\] so we must in fact have $T=ab+1$, as desired. $\blacksquare$

The above lemma implies that $(a^{-1}\mod b)\le b/2$ if and only if $(b^{-1}\mod a)>a/2$, which implies that $s(a/b) = -s(b/a)$.

[b]Remark:[/b]  The condition is basically the same as ISL 2017 N8, which motivates the classification of $s$.

We define $n(x)$ on the positive rational numbers to be the number of terms in the continued fraction of $x$. Then let $s(x)=-1^{n(x)}$ and $s(-x)=-s(x)$. Then, clearly $s(x)s(-x)=-1$. If $xy=1$ and $x>1$ then $y = 0+\tfrac{1}{x}$ and therefore $n(y)=1+n(x)$, so $s(x)s(y)=-1$.

Suppose $x+y=1$. WLOG, $x > \tfrac{1}{2}$. If $x < 1$ then $x=[0;1,a,\dots]$ and $y=[0;1+a,\dots]$ where the $\dots$ is the same. If $x>1$ then $x=(-y)+1$ so they have the same number of terms in the continued fraction, thus $s(x)=s(-y)=-s(y)$ as desired.


function
complex analysis
complex analysis unsolved

Let f be an analytic function in A= {z : |z| < 1 }. There are 2 points a,b (different points ) in A s.t. f(a) = a , f(b) = b.
|f(z)| < 1 in A.
Prove that f(z) = z.

If a,b =0 we are done.
What do we do if not?

Let $ \phi(z)\equal{}\frac{z\minus{}a}{1\minus{}\overline{a}z}$; this is a bijection from the unit disk to itself, which sends $ a$ to zero.

Now let $ g\equal{}\phi\circ f\circ\phi^{\minus{}1}$. We have $ g(0)\equal{}0$, $ |g(z)|<1$ on the unit disk, and $ g(c)\equal{}c$ for some other $ c$ ($ c\equal{}\phi(b)$).

I think you know how to finish it now.


could someone help me write the prgm for the shoelace formula? i found online for TI-82s and others but i need for TI-83s or 84s. thanks! i'm not a master at prgming, but i'm in a contest soon and shoelace formula shows up once in a while! thanks again!  :)

:ClrHome
:Prompt N
:For(C,1,N)
:Prompt X
:Prompt Y
:ClrHome
:X$ \rightarrow L_1(C)$
:Y$ \rightarrow L_2(C)$
:End
:$ L_1(1)L_2(N)\rightarrow S$
:$ L_1(N)L_2(1)\rightarrow T$
:For(D,2,N)
:$ S\plus{}L_1(D)L_2(D\minus{}1)\rightarrow S$
:$ T\plus{}L_1(D\minus{}1)L_2(D)\rightarrow T$
:End
:ClrHome

Where $ N$ is the number of sides, and $ x, Y$ are the coordinates ordered in cyclic order.
:abs(S-T)/2$ \rightarrow A$
:Disp A

The calculations involved in the shoelace theorem are so simple, I can't imagine why you would need need a calculator program to do this.

Perhaps the formula itself is hard to understand. Don't memorize it. [b]Don't memorize it.[/b]
[b][i]Oh my god [u]DON'T memorize it[/u][/i][/b]. If you're going to memorize anything, memorize the process.

Start from one coordinate (one with one or two zeroes is usually best). From that coordinate, go clockwise and write down the $ n$ coordinates in a $ n\times 2$ matrix. Remember to put down the coordinate you started with twice.

From the top, draw lines between the entry $ M_{r,0}$ and the entry $ M_{r\plus{}1,1}$. ($ M$ is your matrix). Sum the products of the things you connected with lines.

Then draw lines the other way and sum the stuff, subtract the two sums, absolute value, half, done. It's way faster on paper than it is in words. (Easier done than said...?)

Does anyone mind proving this awesome theorem?

[url=http://www.mathreference.com/la-det,shoe.html]This[/url] is a nice link.


Solve:

$ \ds\left[\frac{x+1}{3}\right]+\left[\frac{x+2}{3}\right]=\left[\frac{x+1}{2}\right]+\frac{x-1}{2}$

[hide="Solution"]$ \frac{x\minus{}1}{2}$ must be an integer, and therefore so must $ \frac{x\plus{}1}{2}$.

After some simplification, the equation becomes $ \left\lfloor\frac{x\plus{}1}{3}\right\rfloor\plus{}\left\lfloor\frac{x\plus{}2}{3}\right\rfloor \equal{} x$.

[b]Case 1:[/b] $ x\equiv0\bmod3$

The equation becomes $ \frac{x}{3}\plus{}\frac{x}{3}\equal{} x$, and $ x \equal{} 0$. However, this contradicts the fact that $ \frac{x\minus{}1}{2}$ is an integer.

[b]Case 2:[/b] $ x\equiv1\bmod3$

The equation becomes $ \frac{x\minus{}1}{3}\plus{}\frac{x\plus{}2}{3}\equal{} x$, and $ x \equal{} 1$.

[b]Case 3:[/b] $ x\equiv2\bmod3$

The equation becomes $ \frac{x\plus{}1}{3}\plus{}\frac{x\plus{}1}{3}\equal{} x$, and $ x \equal{} 2$. This again contradicts the fact that $ \frac{x\minus{}1}{2}$ is an integer.

Therefore, the solution is $ \boxed{x \equal{} 1}$.[/hide]

[hide="solution"]As the last part $ \frac{x\minus{}1}{2}$ decreases without bound as $ x$ increases. Due to the same part of the equation $ x$ must also be odd.

So far we have that $ x$ is odd and positive.

Also, due to the fact that the rest of the equation [i]increases[/i] without bound as $ x$ increases means that $ x$ should be as low as possible.

The lowest, odd, positive number is $ \boxed{1}$.

I realize that this isn't a formal solution, but it is correct.

$ \left[\frac{\bold{1}\plus{}1}{3}\right]\plus{}\left[\frac{\bold{1}\plus{}2}{3}\right] \equal{}\left[\frac{\bold{1}\plus{}1}{2}\right]\plus{}\frac{\bold{1}\minus{}1}{2}$
$ \left[\frac{\bold{2}}{3}\right]\plus{}\left[\frac{\bold{3}}{3}\right] \equal{}\left[\frac{\bold{2}}{2}\right]\plus{}\frac{\bold{0}}{2}$
$ \left[.6666666\right]\plus{}\left[1\right] \equal{}\left[1\right]\plus{}0$
$ 0\plus{}1 \equal{} 1\plus{}0$
 :lol:
[/hide]

Edit:

Well put, i_like_pie.


A prism has an equilateral triangle with sides of length 10 inches at its base. If the area of one of the lateral sides of the prism is 70 square inches, what is the volume of the prism in cubic inches

Answer:
[hide]
175 sqrt 3
[/hide]

I'm assuming this is a triangular prism with rectangular faces.  It's easy to check (or look up) that the area of an equilateral triangle with side length $ s$ is $ \frac{s^{2}\sqrt{3}}{4}$.  Volume is just $ A\times h$.  Since the area of the side face is 70 and its width is 10 (why?), it's height must be 7.  We know the area of the base so now we just multiply: $ \frac{10^{2}\sqrt{3}}{4}\times 7\equal{}175\sqrt{3}$.


Given a square of square of arbitrary side lengths, find the probability that 3 randomly chosen points on the square will form a triangle that contains the center of the square. (they can all be the same point, they can all be on the same side.)

[hide="My Answer"]$\frac14$[/hide]
[hide="Partial Working"]
Given arbitary points A,B we construct the given traingle if C lies as follows
[img]http://img300.echo.cx/img300/9673/pic9nf.jpg[/img]
I think then the probability of this C being chosen is 1/4. That is kinda of a guess. I presume there is a way to show by integration?[/hide]

[quote="K81o7"]Well it appears that the probability depends on the first two points' locations.  Suppose A and B were really close together near a corner?  The probability is 1/4 if and only is A and B have already been chosen and lie in certain positions, for example, two vertices adjecent.  Suppose that A and B were picked so that AB runs through the center?  Or suppose that A and B were the same point?  Yeah, I think you would have to draw a curve of some sort, might use integration.[/quote]

A & B are arbitary, so somehow we have to 'sum' (integrate I presume) all possibilties

[quote="K81o7"]Yeah...but certain possibilities for A and B have equivalent probability for ABC to contain the center.  I'm not sure how to sum it up though...[/quote]

Neither...


geometry
power of a point
radical axis
geometry solved

Let a triangle ABC . M , N lie on straight line BA , BC . H , K are the orthocentrics of triangle BMN , ABC . I = CM :cap: AN . Prove that : ACNM is a inscribed quadrilateral <-----> H , I , K are in a straight line

I solved it 5 seconds after I drew the sketch. In fact, following [url=http://groups.yahoo.com/group/Hyacinthos/message/10646]the well-known proof of the Bodenmiller theorem[/url], the points H and K lie on the radical axis of the circles with diameters AN and CM. In other words, the line HK is the radical axis of the circles with diameters AN and CM. Hence, the point I lies on the line HK if and only if the point I has equal powers with respect to the circles with diameters AN and CM, i. e. if and only if $IA \cdot IN = IC \cdot IM$. But this is clearly equivalent to saying that the points A, C, N, M lie on one circle, i. e. that the quadrilateral ACNM is inscribed. Hence, the point I lies on the line HK if and only if the quadrilateral ACNM is inscribed. And this completes the proof.

  Darij


function
integration
calculus
real analysis
real analysis unsolved

Find all continuous real functions $ f,g$ and $ h$ defined on the set of positive real numbers and satisfying the relation \[ f(x\plus{}y)\plus{}g(xy)\equal{}h(x)\plus{}h(y)\] for all $ x>0$ and $ y>0$.

[i]Z. Daroczy[/i]

$ f(x + y) + g(xy) = h(x) + h(y)$ (Eq. $ I$)

Integrate between $ \epsilon$ and 1:
$ \int_{y+\epsilon}^{y+1}f(u)\,du + \frac{1}{y}\int_{\epsilon y}^{y}g(u)\,du = \int_{\epsilon}^1 h(u)\, du + (1-\epsilon) h(y)$
which tells us that $ h$ is continuous differentiable.
In the same way, we may substitute $ y = u - x$ in $ I$ and integrate from $ x = \epsilon$ to $ \frac{u}{2}$ and conclude that $ f$ is continuous differentiable.
And so $ g$ has also this property.
$ f'(x+y) + xg'(xy) = h'(y)$
Integration between $ x=\epsilon$ and $ x=1$ shows that $ h$ is twice differentiable.

$ g(x^2-\epsilon^2) - g(x^2) = h(x+\epsilon) + h(x-\epsilon) - 2h(x)$
Divide this equation by $ \epsilon^2$ and take the limit for $ \epsilon \to 0$.
Then $ -g'(x^2) = h''(x)$.

Write $ I$ in the following form:
$ f(u) + g(v) = h(x) + h(y)$, with $ x,y = \frac12(u \pm \sqrt{v^2-4u})$, $ (x \ge y > 0)$.
Differentiate both sides by $ \frac{\partial^2}{\partial u \partial v}$.
Then
$ 0 = h''(x)\frac{dx}{du}\frac{dx}{dv} + h'(x)\frac{d^2x}{du dv} + h''(y)\frac{dy}{du}\frac{dy}{dv} + h'(y)\frac{d^2y}{du dv}$.
Then
$ (h''(x)-h''(y))(x-y)^2 +  (h''(x) + h''(y))(x^2-y^2) = 2(x+y)(h'(x)-h'(y))$.
If $ y=c$ (a constant) and $ h' = j$, then $ j'(x)x(x-c) - j(x)(x+c) = px+q$ for certain constants $ p$ and $ q$.
The general solution of this differential equation is $ j(x) = r + sx + \frac{t}{x}$, where $ r,s,t$ are constants.
Then $ h(x) = a + bx + cx^2 + d\ln(x)$ ($ a,b,c,d$ are constants).
$ g'(x) = -h''(\sqrt{x})$, so
$ g(x) = \alpha -2cx + d\ln(x)$
$ f(x) = -g(\frac{x^2}{4})+2\h(\frac{x}{2}) = -\alpha+2a +bx +cx^2$.


2^2001 * 3^2003 = 2^2001 * 3^2001 * 3^2 = 6^2001 * 3^2

[hide="And the answer is..."]So The top is 6^2001 * 3^2. The bottom is 6^2002. Dividing both the top and botton by 6^2001, the fraction becomes: 3^2/6 = 9/6 = 3/2. THE ANSWER IS E[/hide]

[quote="jli"][quote="Treething"]2^2001 * 3^2003 is approximately $1.085 * 10^{1558}$
6^2002 is approximately $7.224 * 10^{1557}$

When I divide I get 1.5, or $\boxed{E}$.[/quote]

very funny  :dry:

by the way, what calculator did you use?[/quote]

ti 84 using logs lol

[quote="mathgeniuse^ln(x)"][quote="jli"][quote="Treething"]2^2001 * 3^2003 is approximately $1.085 * 10^{1558}$
6^2002 is approximately $7.224 * 10^{1557}$

When I divide I get 1.5, or $\boxed{E}$.[/quote]

very funny  :dry:

by the way, what calculator did you use?[/quote]

I did that too, with a TI 84 Plus Silver Edition.

[hide]But without a calculator, I get

$\frac{6^{2001}*3^{2}}{6^{2002}}=\frac{3}{2}=1.5$[/hide][/quote]

It doesn't cause overflow?!

[quote="noneoftheabove"]how did you go about using logs[/quote]
Using logs would make this unneccesarily complicated, but here's the way:
Suppose the ratio is $R$, then we have \[\log \frac{2^{2001}\cdot 3^{2003}}{6^{2002}}= \log R\]  \[2001 \log 2+2003\log3-2002\log 6 = \log R\]  \[2001(\log 2+\log 3-\log 6)+2 \log 3-\log 6 = \log R\]  \[\log 9-\log 6 = \log R\]  \[\log \frac96 = \log R\]  \[R = \frac96 = \boxed{\frac32}\]

He meant that he used logs to prevent an error on his calculator. :D


[quote="vinskman"]$ 0 < a_1\le a_2\le a_3 \le \cdots \le a_{2005} < 1,\text {for } k \equal{} 1,2,3,\cdots ,2005 \text { and }$

$ \text {Suppose } f(k) \equal{} \frac {k} {1 \minus{} a_1a_2a_3\cdots a_{2005}},\text { prove that}$

$ f(1) < f(2) < f(3) < \cdots < f(2005)$[/quote]
Obviously $ k < k\plus{}1$. So $ \frac {k} {1 \minus{} a_1a_2a_3\cdots a_{2005}} < \frac {k\plus{}1} {1 \minus{} a_1a_2a_3\cdots a_{2005}}$ and $ f(k) < f(k\plus{}1)$. Have you posted the correct problem...?

$ \text{Indeed ,I get error .}$

$ \text{f(k) is revised to }$

$ f(k) \equal{} \frac {1} {1\minus{}a_1a_2a_3\cdots a_k}$

[quote="vinskman"]$ \text{Indeed ,I get error .}$

$ \text{f(k) is revised to }$

$ f(k) \equal{} \frac {1} {1 \minus{} a_1a_2a_3\cdots a_k}$[/quote]
Then $ a_1...a_k > a_1...a_{k\plus{}1}$ and $ 0 < 1\minus{}a_1...a_k < 1\minus{}a_1...a_{k\plus{}1}$ So $ \frac{1}{1\minus{}a_1...a_k} > \frac{1}{1\minus{}a_1...a_{k\plus{}1}}$. So $ f(k) > f(k\plus{}1)$. And then $ f(1) > f(2) > ... > f(2005)$ Which is the opposite of your statement.. Is there a mistake in the problem again..?

Mistakes updated.

$ 0<a_1\le a_2\le a_3 \le \cdots \le a_{2005}<1,\text {for } k\equal{} 1,2,3,\cdots ,2005 \text { and }$

[u]$ f(k) \equal{} \frac {k} {1\minus{}a_1a_2a_3\cdots a_k}$[/u]

$ \text { Prove that }$

$ f(1)<f(2)<f(3)<\cdots <f(2005).$


geometry
parallelogram
3D geometry
pyramid
geometry unsolved

Parallelogram $ABCD$ is the base of a piramid $SABCD$. Planes determined dy triangles $ASC$ and $BSD$ are mutually perpendicular. Find the area of the side $ASD$ , if areas of sides $ASB,BSC$ and $CSD$ are equal to $x,y$ and $z$ respectively.

I think it's quite an easy problem. We consider a perpendicular projection of the pyramid $SABCD$ to the plane $SBD$. Then $A$ comes into $A_1$ and respectively $C$ into $C_1$. Without loss of generality we may assume that $C_1$ ( in this projection) lies under the line $BD$. (this position won't matter in further solution). Of course quadrilateral $A_1BC_1D$ is parallelogram. As $ACS$ and $BDS$ are perpendicular, points $S,C_1,O,A_1$ (where $O$ is the intersection point of $AC$ and $BD$ ) are collinear. In further part of the solution the area of $XYZ$ will be written as $[XYZ]$.  . Consider two points $O_1$ and $C_2$ such that $C_1C_2$ and $O_1O$ are perpendicular to $BDS$ and lie on the same side of $BDS$ and also that $AA_1=CC_2=OO_1$.
Let $C_H,O_H,A_H$ be perpendicular projections of $C_1,O,A_1$ to the line $SD$ and let $C_M,O_M,A_M$ be perpendicular projections of $C_1,O,A_1$ to the line $SB$. Let's write: $C_1C_H=h_1$, $ C_1C_M=h_2$, $ A_1A_H=h_3$, $ A_1A_M=h_4$ and $AA_1=CC_2=OO_1=H$. We also have to write something more $SB=a$, $SD=b$, $[ASB]=x, [BSC]=y, [CSD]=z, [DSA]=t$. Now we can easily write: \[x^2=\frac{a^2(h_4^2+H^2)}{4}\]  \[y^2=\frac{a^2(h_2^2+H^2)}{4}\] \[z^2=\frac{b^2(h_1^2+H^2)}{4}\] \[t^2=\frac{b^2(h_3^2+H^2)}{4}\] We can substract these equations and eliminate $H^2$ which gives \[a^2(h_4^2-h_2^2)=4(y^2-x^2)\] \[b^2(h_3^2-h_1^2)=4(t^2-z^2)\]. But as $[BC_1S]=[DC_1S],  [BOS]=[DOS]$ and $[BA_1S]=[DA_1S]$ so we have \[\frac{h_1}{h_2}=\frac{h_1+h_3}{h_2+h_4}=\frac{h_3}{h_4}=\frac{a}{b}\] Moving forward, using this equation we obtain $\frac{h_3-h_1}{h_4-h_2}=\frac{a}{b}$ which means that \[a^2(h_4^2-h_2^2)=b^2(h_3^2-h_1^2)\] and so on \[y^2-x^2=t^2-z^2\] and \[t=\sqrt{y^2+z^2-x^2}\]


number theory
least common multiple
greatest common divisor

When the least common multiple of 8 and 20 is multiplied by the greatest common factor of 8 and 20, what is the result?

The product of the LCM and GCD of any two positive integers is always equal to those two integers' product.

So the product of the LCM and GCD of 8 and 20 is just $ (8)(20)\equal{}\boxed{160}$.

[quote=Veggieman8][quote=phiReKaLk6781]LCM*GCF=product of the two numbers. Thus, it is 8 times 20, which is $ \boxed{160}$.[/quote]

you really don't have to repeat him, you know[/quote]

phiRe was most likely sniped. next time, just move on and try to avoid reviving posts unless a sol hasn't already been given.


calculus
integration
analytic geometry
linear algebra
real analysis
real analysis theorems

Hi, guys. can somebody explain to me what's the difference between these two text??

Haven't used Stewart but have used Apostol. Apostol's Vol. 1 and Vol. 2 groom you for an introductory analysis course. A lot of thought has been placed into pedagory. Apostol is, I daresay, the sole introductory book containing references to measure theory (and in the first chapter, at that). When I say a lot of thought has been placed into pedagogy I mean, for example, the difference between Calculus and his Mathematical Analysis is vast. The latter assumes you have already been exposed to certain things while the former appreciates intuition and pictures. I daresay you won't find a helpful picture in Mathematical Analysis. Vol. 1 addresses things like integration in polar coordinates without reference to the general change of variables theorem.

In other words, it teaches you maths "the right way". You learned arithmetic without learning what the field axioms were. And now you can learn calculus and a little bit of linear algebra without knowing about countable sets, measure, and so on. Lots of interesting examples are provided.

Stewart is an ordinary calculus text, suitable for the standard course. It's pretty good at what it does.
Apostol is different- it doesn't fit the standard course well, since it spends so much more time on theory.

it i intended to major in mathematics and that I have learn the standard calc would apostol calculus be a good choice? actually I'm trying to prepare myself for the caltech transfer exam. I know they uses apostol calculus for their calc course

Then Apostol would be good for you. Try exercises from Spivak too, for practice.


Και όμως μου φαίνεται διασκεδαστική
Έστω f R->[-1,1] μια συνεχής συνάρτηση στο [1,3] ώστε:
1)αν f(x) ανήκει Ζ τότε x ανήκει Z
2)(f(1)f(3))^2 =- f(3)^4-2f(1)f(3)^3 +1
Έστω Α(1,0) και Β(3,0)
Να αποδείξετε ότι υπάρχει ένα τουλάχιστον σημείο τομής της Cf με [b]το[/b] ΑΒ με την ακόλουθη ιδιότητα: αν φέρουμε κάθετη σε αυτό προς το ΑΒ υπάρχουν σε αυτή 2 τουλάχιστον σημεία ώστε το πολύγωνο που ορίζεται από τα Α,Β,Γ,Δ να είναι κανονικό

Επίσης όπως διαπυστώθεικαι από τον Cretanman πρέπει (εκτός και αν κάποιος αποδείξει το αντίθετο, δεν ερευνήθηκε σχολαστικά) ότι $f(3)\big(f(1)+f(3)\big)$ <>1

Katarxin i sxesi $(f(1)f(3))^{2}=-f(3)^{4}-2f(1)f(3)^{3}+1$ ginetai $f(3)^{2}\big(f(1)^{2}+f(3)^{2}+2f(1)f(3)\big)=1$ ap'opou $f(3)\big(f(1)+f(3)\big)=\pm 1 \ \ (1)$.
Katarxin den mporei logw tis sxesis $(1)$ na exoume $f(3)=0$.
An ipothesoume oti $f(1)=0 \ \big[$ara $f(3)=1$ mono ean $f(3)\big(f(1)+f(3)\big)=1$ (dioti ean $f(3)\big(f(1)+f(3)\big)=-1$ tote den ginetai na exoume $f(1)=0$) $\big]$ tote profanws to zitoumeno simeio sto opoio fernoume tin katheti einai to A kathws mporoume na epileksoume ta Γ,Δ epi tis kathetou (profanws ekaterwthen tis AB) etsi wste ta A,B,Γ,Δ na sximatizoun isopleuro trigwno.

As ipothesoume loipon twra oti $f(1),f(3)\neq 0$ kai $f(3)<1$.

Diakrinoume tis eksis periptwseis
I)Ean $f(3)\big(f(1)+f(3)\big)=-1$ diladi $f(1)f(3)=-1-f(3)^{2}<0$ logw tis $f(3)<1$. Ara $f(1)f(3)<0$ kai apo to thewrima Bolzano h $f(x)$ exei mia riza $x_{0}$ sto diastima $(1,3)$.

Se kathe periptwsi loipon $f(x_{0})=0 \in \mathbb{Z}$ ara anagastika $x_{0}\in \mathbb{Z}$ kai den mporei na einai allo apo to $x_{0}=2$ (to opoio einai kai to meson tou AB). Apo to $(2,0)$ ean feroume tin katheto tote profanws mporoume na epileksoume panw tis ta simeia $(2,1)$ kai $(2,-1)$ wste to A,B,Γ,Δ na einai tetragwno.

II)Ean $f(3)\big(f(1)+f(3)\big)=1$ tote ipothetw oti douleuoume me omoio skeptiko alla den mporw na kataliksw se sigouro apotelesma...
Gia paradeigma vgainei oti sigoura $f(1)f(3)>0$ ara sigoura den mporoume na efarmosoume to thewrima Bolzano sto $(1,3)$ gia na doulepsoume opws parapanw. Episis $f(1)<1/2$ kai $f(3)>1/2$ 'h to anapodo afou sigoura $-1<f(1),f(3)<1$ logw tis ipothesis gia to sinolo timwn. Poli endiaferousa pantws askisi!! Mpravo!
Gia deite to mipws xanw kati!

[color=red]Simeiwsi (prosthethike meta to post): Gia tin 2h periptwsi koitakste to epomeno post tou thatin.[/color]

Alexandros

thanx :)
Ίσως ξέχασα και κάτι...
ας πούμε ότι το $f(3)\big(f(1)+f(3)\big)$ δεν μπορεί να είναι 1 :blush:

Vazw loipon mia simeiwsi gia tous epomenous anagnwstes sto parapanw post me tin lisi.

Alexandros

Y.G. Oudeis alanthastos! Opote den iparxei provlima...

Το πρόσθεσα στην εκφώνηση οπότε δε χρειάζεται :)
Α, και συγχαρητήρια για τη λύση σου!!

PS: δεν είναι θέμα λάθους αλλά μεταφοράς.


Substitute $ u\equal{}x\plus{}y,v\equal{}x\minus{}y$. Solve that. Transform back to the original variables.

hello
[hide="My solution"]
$ dx\minus{}dy\equal{}\frac{dx\plus{}dy}{x\plus{}y}$
$ dx\minus{}dy\equal{}d[ln(x\plus{}y)]$
Integrating the sides
$ x\minus{}y\equal{}ln(x\plus{}y)\plus{}C$
[/hide]
thank u

hello, an explicit solution is given by the form $ y(x)\equal{}\minus{}x\plus{}\text{LambertW}(Ce^{2x})$.
Sonnhard.

[quote="vijaymenon"]Solve the differential equation
$ (x \plus{} y) ( dx \minus{} dy ) \equal{} dx \plus{} dy$

please help me[/quote]

It's a specific case of \[ y'\equal{}\frac{ax\plus{}by\plus{}e}{cx\plus{}dy\plus{}f}\]  So? It's a big deal if you look into it. :)


We consider the set of points given by $P_s:=\left(1,\frac{2s}{s^2-1}\right)$ for all rational $s \neq \pm 1$. Their distances from the origin are rational by $\sqrt{1^2+\left(\frac{2s}{s^2-1}\right)^2} = \sqrt{\frac{s^4-2s^2+1+4s^2}{(s^2-1)^2}}=\frac{s^2+1}{s^2-1}$, and trivially their pairwise distances are rational.
If we invert on the unit circle and $A,B$ are points with images $A^\prime , B^\prime$, we have $\overline{A'B'}=\frac{\overline{AB}}{\overline{A0} \cdot \overline{B0}}$. Especially all $P_s$ have pairwise rational distances again and get, together with the whole line $x=1$, mapped to the circle around $\left(\frac{1}{2},0\right)$ with radius $\frac 12$.
Thus their images solve the problem.

Very nice solution,Zetax!!!!!
there is an agly one:
Lemma:There exist infinite sequence $\beta_i$ such that $cos \beta_i,sin \beta_i$ are both rational.then the distences between any two point of $P_i(cos2 \beta_i,sin 2\beta_i)$ are rational

[quote="Peter"]Does there exist a circle and an infinite set of points on it such that the distance between any two points of the set is rational?[/quote]

[b] Lemma [/b]  There are infinitely many rational numbers $ q$ such that $ \sqrt{1\plus{}q^{2}}$ is also rational.

[b] Proof of Lemma [/b] We need to show that the curve $ 1\plus{}x^{2}\equal{}y^{2}$ has infinitely many rational points. Indeed, for all nonzero rational numbers $ q$, the rational point $ \left(\frac{q^{2}\minus{}1}{2q},\frac{q^{2}\plus{}1}{2q}\right)$ lies on the curve.

Now, we prove the following statement.

   There exist infinitely many concyclic [b]rational points[/b] such that the distance  between any two points of the set is rational.

 [b]Proof[/b] We construct a set $ \mathbb{U}$ satisfying the conditions:
 \[ \mathbb{U}\equal{}\left\{\left(\frac{1\minus{}q^{2}}{1\plus{}q^{2}},\frac{2q}{1\plus{}q^{2}}\right)\;\vert\;\, q\in\mathbb{Q},\,\sqrt{1\plus{}q^{2}}\in\mathbb{Q}\right\}.\]
  One may use [b]Lemma[/b] to see that $ \mathbb{U}$ is infinite. Also, it is easy to check that $ \left(\frac{1\minus{}q^{2}}{1\plus{}q^{2}},\frac{2q}{1\plus{}q^{2}}\right)\in\mathbb{U}$ is a rational point on the circle $ x^{2}\plus{}y^{2}\equal{}1$. It remains to prove that any two points in $ \mathbb{U}$ has rational length. Let  $ \mathbf{P}\left(\frac{1\minus{}p^{2}}{1\plus{}p^{2}},\frac{2p}{1\plus{}p^{2}}\right)\in\mathbb{U}$ and $ \mathbf{Q}\left(\frac{1\minus{}q^{2}}{1\plus{}q^{2}},\frac{2q}{1\plus{}q^{2}}\right)\in\mathbb{U}$. It is not difficult to check that
  \[ \overline{\mathbf{P}\mathbf{Q}}\equal{}\frac{2\,\vert p\minus{}q\vert}{\sqrt{1\plus{}p^{2}}\sqrt{1\plus{}q^{2}}}.\]
  Since $ p$, $ q$, $ \sqrt{1\plus{}p^{2}}$ and $ \sqrt{1\plus{}q^{2}}$ are all rational, it is clear that $ \overline{\mathbf{P}\mathbf{Q}}$ is also rational.

[b]Note[/b] In the solution, we used the formula
 \[ \overline{\mathbf{P}\mathbf{Q}}\equal{}\sqrt{\left(\frac{1\minus{}p^{2}}{1\plus{}p^{2}}\minus{}\frac{1\minus{}q^{2}}{1\plus{}q^{2}}\right)^{2}\plus{}\left(\frac{2p}{1\plus{}p^{2}}\minus{}\frac{2q}{1\plus{}q^{2}}\right)^{2}}\equal{}\frac{2\,\vert p\minus{}q\vert}{\sqrt{1\plus{}p^{2}}\sqrt{1\plus{}q^{2}}}.\]
  Here, we offer an outline of a trigonometric proof. After setting $ p\equal{}\tan{\alpha}$ and $ q\equal{}\tan{\beta}$, where $ \alpha,\beta\in\mathbb{R}$, we obtain $ \mathbf{P}(\cos 2\alpha,\sin 2\alpha)$ and $ \mathbf{Q}(\cos 2\beta,\sin 2\beta)$. Since $ \overline{\mathbf{O}\mathbf{P}}\equal{}\overline{\mathbf{O}\mathbf{Q}}\equal{}1$ and since $ \angle\mathbf{P}\mathbf{O}\mathbf{Q}\equal{} 2\,\vert\alpha\minus{}\beta\vert$, we obtain $ \overline{\mathbf{P}\mathbf{Q}}\equal{}2\,\left\vert\sin\left(\alpha\minus{}\beta\right)\right\vert$. It now remains to observe that
 \[ \left\vert\sin\left(\alpha\minus{}\beta\right)\right\vert \equal{}\left\vert\cos\alpha\cos\beta\left(\frac{\sin\alpha}{\cos\alpha}\minus{}\frac{\sin\beta}{\cos\beta}\right)\right\vert \equal{}\frac{\vert\tan\alpha\minus{}\tan\beta\vert }{\sqrt{ 1\plus{}{\tan\alpha}^{2}}\sqrt{ 1\plus{}{\tan\beta}^{2}}}.\]


geometry
parallelogram
geometric transformation
rotation
homothety
geometry solved

Circle passing through vertices $A$, $B$ of triangle $ABC$ cuts side $BC$ at $D$. Circle passing through vertices $B$,$C$ cuts side $AB$ at $E$ and the first circle at $F$ for the second time. If point $A$, $E$, $D$, $C$ lie on a circle with center $O$,prove that angle $BFO$ is a right angle.

Please anyone draw it to me. I can drow it but I'm nut sure about this because...

THANKS Jimbo

sorry guys I made mistake it is triangle $ABC$ not $ABS$
thanks Jimbo

[quote="Muhammadaminov"]Circle passing through vertices $A$, $B$ of triangle $ABS$ cuts side $BC$ at $D$. Circle passing through vertices $B$,$C$ cuts side $AB$ at $E$ and the first circle at $F$ for the second time. If point $A$, $E$, $D$, $C$ lie on a circle with center $O$,prove that angle BFO is a right angle.

Please anyone draw it to me. I can drow it but I'm nut sure about this because...

THANKS Jimbo[/quote]

what is $C$, another arbitrary point with nothing to do with triangle $ABS$

[quote="Andreas"]No, it is the third vertices.  :)[/quote]

Then, what is $S$? :)

[quote="Andreas"]It should be only a typo because the problem works without it.[/quote]

Oh! OK. It was me, drawing incorrectly the figure, because at first I thought it was a typo but made a mistake and took point $E$ as the other point where the two circumferences meet and then, of course, everything goes wrong

Daniel

[quote="Muhammadaminov"]Circle passing through vertices $A$, $B$ of triangle $ABC$ cuts side $BC$ at $D$. Circle passing through vertices $B$,$C$ cuts side $AB$ at $E$ and the first circle at $F$ for the second time. If point $A$, $E$, $D$, $C$ lie on a circle with center $O$,prove that angle $BFO$ is a right angle.

[/quote]

We can obtain this result if we note $O_1BO_2O$ -parallelogram
This is from equality of angles $\angle AEC=\angle ADC$ so $\angle O_1AB=\angle O_1BA=\angle O_2BC=\angle O_2CB=\angle OCA= \angle OAC$
So triangles $AO_1O$ and $ABC$ are similar. So using rotation and homothety with common centre A we can get $BO_2 \parallel OO_1$
Similarly $BO_1 \parallel OO_2$ and thus $OO_1O_2B$ -parallelogram.
If $O_1$ is a midle of $BA_1$ and $O_2$ -the midle of $BC_1$
then $O$ is the midle of $A_1B_1$.  $BF$  is the altitude on $A_1C_1$
So $\angle OFB = 90^\circ$
V:)

The problem is exactly the one from http://www.mathlinks.ro/Forum/viewtopic.php?t=60787 , applied to the triangle ABD with the circle with center O passing through the points A and D and intersecting the sidelines AB and BD again at the points E and C, respectively.

  Darij


A little random but are pens allowed? I see no reason why not but who knows.


For mathcounts, how many people advance from chapter to state in each chapter?
No 1 and No 2 teams plus 4 individuals for a total of 12?

It depends on the chapter - email your chapter coordinator for more information.

What does your chapter do?
I like to know what is a typical chapter choice.

My chapter (Charlotte, NC) takes the top 5 teams to state (20 individuals), and if any of the top 5 individuals arent on the top 5 teams then they get to go to state.

My chapter takes top 12 to state and the top team.

i think top 6 teams and top 6 individuals in my chapter (central Ohio)

Since California is enormous compared to Massachusetts or North Carolina only the very best individuals and teams go to the state competition.
For mine it was the top two teams only

North Carolina is a pretty good state, but not as good as CA. Our chapter fails, but Raleigh chapter is intense. Only like top 5 individuals go

Our chapter (Lehigh Valley, PA) takes the top team and the top 4 individuals but if the top team is the top 4 individuals I think they also take the next 2 individuals

in my california chapter it took top 2 teams and 2-4 individuals

My chapter takes 36 people in total, including top 6 teams (24), and the top 12 people who are not on any team.

Mine takes top 2 teams (8) and the next 8 individuals not already on a team, for a total of 16. My chapter includes 2-3 counties that take the test at different times and places, so qualification isn't based on another event as well.

How many usually go to state? For mine, there are 10 chapters with 16 qualifiers from each (see above), for a total of 160.

For my chapter, we send the top 3 teams and the top 10 individuals, however it might be less than 22 total because the top 10 could also be on the team.

I believe that per national MATHCOUNTS requirements, the top 4 individuals of a chapter competition must qualify for state along with a number of teams based on a percent of the students at the event (lots of schools/students = more schools qualify).  Nationals sets the minimums and then the state directors can decide to accept more than the minimum.

In NC they accept the minimum (4 individuals and a number of teams based on the size of the event) and then they invite "wildcard" students after all chapter competitions are held ... in past years they have set a score (42 last year I believe), above which students were invited to states if they didn't qualify by being on a team or being one of the top 4.  This really only matters to the central chapter in NC, where a score of 42 would not necessarily put you in the top 4.  I believe there was only one 'wildcard' student last year ... he made nationals and finished in the top 40 after placing 6th in our chapter.

Note that I am not an official ... but I speak with our state coordinator occasionally and this is the information I got from him.


probability
modular arithmetic
algebra theorems
algebra

I am trying to understand the following argument of Rabin's theorem using groups. Rabin's theorem is interesting because, whereas you would normally find if a number is a pseudoprime manually using the $ a$-pseudoprime test, one $ a$ after the other, Rabin's theorem finds the probability of certain $ a's$ being false witnesses in a nice way. But i am not quite sure what is going on.

I have [color=red]highlighted [/color] things which I don't understand.

Consider the number $ 8911\equal{}7\cdot 19 \cdot 67$, which is Carmichael. Here $ s\equal{}s_1\equal{}s_2\equal{}s_3\equal{}1, q\equal{}4455$. The strong $ a$-pseudoprime sequence for any $ [a]$ in $ U_{8911}$ looks like $ \{a^{4455}, a^{8911}\}$.

Every invertible $ a \pmod{8911}$ satisfies $ a^{8911}\equiv 1 \pmod{8911}$ [color=red](?)[/color]. But exactly half of the numbers $ a$ coprime to $ 7$ have $ a^{4455} \equiv a^3\equiv 1 \pmod{7}$, namely the squares $ \mod{7}$, which are 1,4, and 2. Similarly, exactly half of the numbers $ a$ coprime to $ 19$ have $ a^{4455}\equiv a^9\equiv 1 \pmod{19}$, namey the squares $ \pmod{19}$, and exactly half of the numbers $ a$ coprime to 67 have $ a^{4455}\equiv ^{33}\equiv 1 \mod{67}$, namely the squares $ \pmod{67}$.

Thus under the one-to-one correspondence between $ U_{8911}$ and $ U_{7}\times U_{19}\times U_{67}$, if $ \alpha$ in $ U_{8911}$ corresponds to

    $ ([a_1]_7, [a_2]_{19}, [a_3]_{67})$,

then $ \alpha^{2^{s\minus{}1}q}\alpha^{4455}$ corresponds to one of the following eight tuples in $ U_7\times U_{19}\times U_{67}$, and each is equally likely:

([1],[1],[1]),
([1],[1],[-1])
([1],[-1],[1])
([1],[-1],[-1])
([-1],[1],[1])
([-1],[1],[-1])
([-1],[-1],[1])
([-1],[-1],[-1])

If $ \alpha^{4455}$ in $ U_{8911}$ corresponds to ([1],[1],[1]), then $ \alpha^{4455}\equal{}[1]$. This occurs for $ 1/8$ of all $ \alpha$ in $ U_{8911}$. If $ \alpha^{4455}$ in $ U_{8911}$ corresponds to ([-1],[-1],[-1]), then $ \alpha{4455}\equal{}[\minus{}1]$. This also occurs for $ 1/8$ of all $ \alpha$ in $ U_{8911}$. The other $ 3/4$ of all $ \alpha$ in $ U_{8911}$ have $ \alpha^{4455}$ neither $ [1]$ nor $ [\minus{}1]$, [color=red]hence all these $ \alpha$ are witnesses to the compositeness of 8911.[/color]

Thanks in advance, and I appreciate any explanation of this, since I am reading it without any teacher's help.

Los

The statement with a question mark next to it should read $ a^{8910} \equiv 1 \bmod 8911$ and follows from the fact that $ 8911$ is Carmichael.  You can prove it with CRT.

The red statement is a question you have already asked.  $ \alpha^{4455}$ is a square root of $ 1 \bmod 8910$.  If $ 8910$ is prime, then it can only have the values $ 1, \minus{}1$, so if it is anything else, $ 8910$ is composite.


AMC
AMC 12
LaTeX
trigonometry
complex numbers

Hi, does anyone have the answers and solutions to AMC 2002 A?

I can't find it anywhere online.

Thanks

Does anyone have question 18 from the 2002 AMC A?  It's cut off on my copy.

Thanks

Assuming you're talking about the AMC12, #18 is

Let $C_1$ and $C_2$ be circles defined by
\[(x-10)^2 + y^2 = 36\]
and
\[(x+15)^2 + y^2 = 81,\]
respectively.  What is the length of the shortest line segment $\overline{PQ}$ that is tangent to $C_1$ at $P$ and $C_2$ at $Q$?

How does one do #24?

Find the number of ordered pairs of real numbers (a,b) such that
[tex]
 (a + bi)^{2002} = a - bi.
[/tex]

You need to put brackets around the 2002, i.e. (a + bi)^{2002} = a - bi, or the LaTeX won't be placed properly:

$(a + bi)^{2002} = a - bi.$

First of all, are you familiar with the polar form of complex numbers?  It can be used to solve this problem quickly.

Multiply $a+bi$ to both side of the equation we get

$(a+bi)^{2003}=a^2+b^2=r^2$, for some real number $r$

Rewrite $a=r\cos\theta$, $b=r \sin \theta$

Apply Demorvies Theorem

we have

$r^{2003}(\cos 2003\theta+ i\sin 2003\theta)=r^2$

Note that on the RHS is real, so

$r^{2003} \sin 2003 \theta=0$, or $\sin 2003 \theta=0$, $\cos 2003 \theta=1$ .

Substitute that in we have

$r^{2003}=r^2$, Note that $r$ has to be real, therefore r=0, 1. substitute for $r^2$, we have
$(a+bi)^{2003}=1 or 0$
When $(a+bi)^{2003}=1$, we have 2003 possible values of $a+bi$. When $(a+bi)^{2003}=0$, we have 1 more possible value.

So there are 2004 pairs of (a, b) in total.

Hope I didn't mess up anywhere :)


I have some real data, basically several x and y coordinates. I am trying to come up with a formula for the data, but I am not sure where to begin. Based on looking at a graph, the data seems to be a polynomial, more specifially a power of two or three. Is there a piece of software that can guess a function based on the data? Any help is appreciated.

If there existed software that could tell you the underlying function from a set of data, well, whoever owned it would be very rich.

Obviously, as you increase the degrees of freedom in the function you are trying to fit to the data, you will achieve progressively smaller error until you reach zero.  Depending upon the application, however, a polynomial of degree 100 may be meaningless.

For basic fitting, any of the TI-8x graphing calculators will do regression fits for quadratic, cubic, sinusoidal, quartic, and linear functions.

For anything more complicated, you're probably going to have to play around with excel or a programming language to specify the type of function you want, the range of coefficients desired, and the error function to minimize.


geometry
incenter
circumcircle
geometry proposed

given a triangle ABC with incenter I , incircle touches sides AB and AC respectivetily  at points P and Q. BI and CI meet PQ at points K and L respectivetily. prove that : circumcircle of ILK touches incircle of ABC if and only if    AB+AC=3BC

Since $ I$ is center of incircle, circumcircle of $ ILK$ touuches incircle iff diameter of $ ILK$ is $ r$ where $ r$ is radius of incircle of $ ABC$. Let incircle tangent $ BC$ at $ R$ and $ PI,QI$ meet incircle at $ X,Y$. Since $ BI\perp PR,\angle PRX\equal{}90$ , $ IK\parallel RX$ similarly $ IL\parallel RY$. Since $ IP\equal{}IX,IQ\equal{}IT,$ $ KL\parallel XY$.Hence $ \triangle IKL$ and $ \triangle RXY$ are similar. Since diameter of $ RXY$ is 2r, diameter of $ IKL$ is $ r$ iff $ \frac{KL}{XY}\equal{}\frac{1}{2}$. But $ XY\equal{}PQ$. So it is suffice to show $ \frac{LK}{PQ}\equal{}\frac{1}{2}$ iff $ b\plus{}c\equal{}3a$. Let $ M$ be a midpoint of $ BC$ and $ S\equal{}BL\cap CA$. It is not hard to show $ \angle BLC\equal{}90($since $ R,Q$ are symmetric wrt $ LC,\angle LRI\equal{}\angle LQI\equal{}\angle LPI$. Hence $ I,L,P,R$ are cyclic. And $ I,P,B,R$ are cyclic. Hence $ \angle BLC\equal{}\angle BLI\equal{}90)$Since $ M,L$ is midpoint of $ BC,BS$,$ ML\parallel AQ$ similarly $ MK\parallel AP$. Hence $ \triangle MLK$ and $ \triangle AQP$ are similar. $ \frac{LK}{PQ}\equal{}\frac{LM}{AQ}\equal{}\frac{\frac{a}{2}}{\frac{b\plus{}c\minus{}a}{2}}\equal{}\frac{a}{b\plus{}c\minus{}a}$. Hence$ \frac{LK}{PQ}\equal{}\frac{1}{2}$ iff $ b\plus{}c\equal{}3a$.


1)  Let $ a,b,c$ are positive such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
Prove

$ 3 \plus{} \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \geq 2 \left( c\sqrt {a} \plus{} a\sqrt {b} \plus{} b\sqrt {c}\right)$

2) Let $ a,b,c$are positive that satisfy $ a^2\plus{}b^2\plus{}c^2\equal{}a\plus{}b\plus{}c$
Prove $ \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2} \geq \frac{a}{\sqrt{b}}\plus{}\frac{b}{\sqrt{c}}\plus{}\frac{c}{\sqrt{a}}$

I can try #1:
[hide]By $ AM \minus{} GM$, $ \sum_{cyc}\frac {a^2}{b^2} \ge 3$, so the question can be rephrased as:
$ 3 \plus{} 3\ge2\sum_{cyc}c\sqrt {a}$
$ 3\ge\sum_{cyc}c\sqrt {a}$, which can be proven by the following:

By Cauchy's, $ (\sum_{cyc}c\sqrt {a})^2\le(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)$.
We are given that $ (a^2 \plus{} b^2 \plus{} c^2) \equal{} 3$. By $ QM \minus{} AM$, $ \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{3}} \equal{} 1\ge\frac {a \plus{} b \plus{} c}{3}$, so $ (a \plus{} b \plus{} c)\le3$.

Thus,  $ (\sum_{cyc}c\sqrt {a})^2\le 3\cdot3$
$ \sum_{cyc}c\sqrt {a}\le3$, which is what we want.

Is this correct?
[/hide]

and #2:
[hide]edit: my solution is incorrect...[/hide]

No.1 is easy. Here is my sols:

$ \sum_{cyc} \left( \frac{a^2}{b^2}\plus{}b^2\plus{}c^2 \right) \geq \sum_{cyc}  \left( 1\plus{}bc\plus{} \frac{ac}{b} \right) \geq \sum_{cyc} \left (1\plus{}2c\sqrt{a} \right)$

$ \sum \frac{a^2}{b^2}\ge \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2}\equal{}a\plus{}b\plus{}c$

Thus

$ (\sum\frac{a^2}{b^2})^2\ge(\sum\frac{a^2}{b^2})(\sum a)\ge (\sum\frac {a}{\sqrt {b}})^2$


Solve in natural numbers:
$ n^3\minus{}5n\plus{}10\equal{}2^k$.

[quote="Erken"]Solve in natural numbers:
$ n^3 \minus{} 5n \plus{} 10 \equal{} 2^k$.[/quote]
We observ that $ n\equal{}2,k\equal{}3$ solution, and have no solutions for $ k\equal{}0,1,2$.
For $ k>3$ we have
\[ (n\minus{}2)(n^2\plus{}2n\minus{}1)\equal{}2^3(2^{k\minus{}3}\minus{}1)\]
If $ k\equal{}4,5,6$ have no solutions and if $ k>6\Rightarrow (n\minus{}2)|2^3\Rightarrow n\equal{}3,4,6,10$ but for these values have no solutions.

[quote="cckek"]

 $ k > 6\Rightarrow (n \minus{} 2)|2^3$[/quote]
can you explain this step in more details,please?!!!!


THEOREM POCKINGTON: [tex]x^4-x^2y^2+y^4=z^2[/tex] has no solution positive integer [tex]x\neq{y}[/tex]
THEOREM CELL: [tex]x^4+9x^2y^2+27y^4=z^2[/tex] has no solution positive integer
goodluck

yes, robertux, how can you tell me: can you solve new pro :D
For this pro, i used Pitago-equation: [tex]x^2+y^2=z^2[/tex]

[quote="Pascual2005"]The first problem has already been solved in the forum.[/quote]
two pro are similar
okie?

[quote="RobertuX"][quote="nguyenquockhanh"]THEOREM POCKINGTON: [tex]x^4-x^2y^2+y^4=z^2[/tex] has no solution positive integer
THEOREM CELL: [tex]x^4+9x^2y^2+27y^4=z^2[/tex] has no solution positive integer
goodluck[/quote]
Well, a solution of first equation is $x=y=z=1$.  :D  :D[/quote]

x and y are distinct integers  :)


The Pentasect is an ancient cult that counts in base five and believes that numbers that contain a 2 in their base five representation are evil. All other numbers are good. In addition, a number n is balanced if the number of good and evil numbers between 1 and n (inclusive) are equal. The smallest balanced number is 2 (since 1 is good and 2 is evil). Compute the largest balanced number in base 10.

[hide]From 1-29, there's 29 numbers and 12 numbers that are evil.  Since that's less that half and from 20-29 are also evil, the only balenced number in that realm is 2.  From 30-99, we have 70 numbers and one tenth or 7 of them are evil.  That only increases the balence in favor of the good numbers so after we go from 1-100, we have 80 good numbers and 19 evil numbers so we have a favor of 80-19 or 61 for the good numbers.  (Good and evil numbers cancel out when we measure bias.)  From 100-199 we have 61+1 numbers since 100 is counted but 0 isn't.  That puts the bias in favor of the good numbers to 123.  We subtract 100 from 200-299.  This results in 23.  300-999 is 700 numbers and since the first digit isn't a 2, that's 62*7+23.  That's 457.  Next, we ponder the interval 1000-1999.  This is just like 1-999 except it's one more since 1000 is allowed but 0 isn't.  This gives 457*2+1 or 915.  But 2000-2999 has 1000 numbers so we have -85 at 2999.  The next balenced number is therefore, 2914.  The next such number when the evil balence is counteracted.  3099 only puts the balence to -85+62 or -23.  3129 puts it to -17.  3132 puts it to -16.  3141 is 9 good numbers in a row and puts it to -7.  3142 is evil and puts it to -8.  But 3150 resets it to 0 and is a balenced number.  So is 3152.  From 3153-3199, we have 4 evil numbers and 47 good numbers so the good bias is 43.  3200-3299 yields -100 so at 3242 we have another balenced number.  Since at 3299 yields -57, By 3329, we've whittled that down to -51.  By 3361, we've whittled that down to -25.  By 3381, we've whittled that down to -9.  By 3391 we have -1 so at 3394 we have another balenced number.  By 3399, we have a good bias of 5.  From 3400-3999 we add a good bias of 62*6.  From 4000-9000 we add a good bias of 458*6.  These add up to be 5+372+2748=3125.  10000-19999 has that much plus one.  We add a positive bias of 6251, but notice that 20000-29999 has a bias of -10000 so there's another balenced number in 30000-39999.  Actually, this is a combinatorics problem.  1000...0-1999...9 with n 9s and 0s has (9^[n-1])(n choose 1) ways of having an evil number with 1 two, (9^[n-2])(n choose 2) ways of having an evil number with 2 twos...  We add all these up and subtract them from 10^n twice since once is to seperate good numbers from evil and the second is two simulate the cancelling out for a good bias of 0.  From 1-999...9 with n 9s is the same thing except 1000...0 counts and 0 doesn't so that's one less.  Also, notice that (n choose 0)+(n choose 1)+...(n choose n)=2^n.  Then of course, at 2999...9, we subtract 10^n.  So we have 10^n-4(9^[n-1]{n choose 1}+...+1{n choose n})-1, where n is a positive integer.  For n greater than 3, n is negative and the negativity is increasing.  However, that doesn't necessarily ensure that there's infinitely many balenced numbers because if the negative good bias is so big that it can't be balenced by 3000...0-1999...9 where there's n+1 9s, then there would be a largest balenced number.  That would be the case before we subtracted the 10^n from 2999...9, since we can make n as large as needed.  2*10^n-4(9^[n-1]{n choose 1}+...+1{n choose n}-1.  I think this number is always a balenced number for n greater than or equal to 3.  Any way, this is always positive, I think.  [/hide]

Actually, I don't think 2*10^n-4(9^[n-1]{n choose 1}+...+1{n choose n})-1 is always a balenced number, because balenced numbers must be even and this is always odd.


let $ x_0 \equal{} a, x_1 \equal{} b$ and for $ 2\leq n : x_{n} \equal{} \frac {x_{n \minus{} 1} \plus{} x_{n \minus{} 2}}{2}$ . Find $ lim x_{n}$?

observe that
$ x_n - x_{n - 1} = \left( - \frac {1}{2}\right)(x_{n - 1} - x_{n - 2})$
we get a recurrence relation and it directly gives
$ x_n = \left( - \frac {1}{2}\right)^{n - 1}(b - a) + x_{n - 1}$
if we proceed, it can be seen that
$ x_n = (b - a)\left(1 + ( - 0.5) + \cdots + ( - 0.5)^{n - 1}\right) + a$
and hence
\begin{align*}\lim_{n\to\infty}x_n & = (b - a)\frac {1}{1 - \left( - \frac {1}{2}\right)} + a \\
& = \frac {a + 2b}{3} \end{align*}

otherwise, like solving differential equation, we set a characteristic equation
$ 2\lambda^2 - \lambda - 1 = 0$
and by initial condtions, we get
$ x_n = A\left( - \frac {1}{2}\right)^n + \frac {a + 2b}{3}$
for some constant $ A$ (it can be found and I don't solve for it), that gives we need.


modular arithmetic
number theory
algebra
system of equations

show the system of equations:
$11x - 5 =7$
$9x + 10y = -3$ has no integer solutions.

My solution ( I don't know my modular arithmetic well enought to know if I'm doing this right ) :
[hide]

taking both equations mod 5, we get

$x \equiv 2 \mod{5}$ and
$4x \equiv 2 \mod{5}$

here is the step I am hazy on:

subing the first into the second, we get $8 \equiv 2 \mod{5}$ which is obviously false, implying that both equations can't simltaniously hold, so there is no solutions for the system.

[/hide]

Wait...if you have a system with two variables...there's only one solution if they're not parallel...and the only solution isn't an integer solution...

In your second step, what you need to do is add 5 into it until you can divide out the 4.  By adding 5, it's still congruent to the same value mod 5, but allows you to get rid of the 4.
$4x \equiv 2 \implies 4x \equiv 12 \implies x \equiv 3 \pmod 5$
Then, we have $x \equiv 2$ and $x \equiv 3$, which is impossible.

[quote="Altheman"]he is right...just solve the system, and the solution is not an integer[/quote]

I can solve it that way, but i was doing these question to practice mod arimetic, not solving silly linears.....

If you add those two mod equations (if practice is what you want) then
   $(x \equiv 2 \pmod 5)+(4x \equiv 2 \pmod 5)\implies 5x \equiv 4 \pmod 5$
which has no integer solutions.

The first equation becomes $x\equiv 0\pmod{2}$ and the second equation becomes $x\equiv 1\pmod{2}$.  These equations cannot simultaneously be true, so there are no integer solutions.  :)

The nice thing about using mods here is that you can choose a mod that gets rid of the $y$ completely (like $\pmod{2}$, $\pmod{5}$, or $\pmod{10}$)


calculus
integration
logarithms
real analysis
real analysis unsolved

hope i did not ask this before...  :oops: prove that

$ \sum_{k\equal{}1}^{\infty}\;\frac{(\minus{}1)^{k\plus{}1}}{k^{2}\cdot\dbinom{2k}{k}}\;\equal{}\;\boxed{2\left(\sinh^{\minus{}1}\left(\frac{1}{2}\right)\right)^{2}}$

Use Lehmer's formula, (1985):

$ \sum_{m\geq 1}\frac{(2x)^{2m}}{n\binom{2m}{m}}\equal{}\frac{2x\arcsin x}{\sqrt{1\minus{}x^{2}}}$, $ |x|<1$.

It follows by integration that

$ \sum_{m\equal{}1}^{\infty}\frac{(2x)^{2m}}{m^{2}\binom{2m}{m}}\equal{}2(\arcsin(x))^{2}$, $ |x|<1$.

Let $ x\equal{}i/2$.  After simple calculations you get that your sum equals $ 2\ln^{2}\frac{\sqrt{5}\minus{}1}{2}$.


modular arithmetic
quadratics
number theory proposed
number theory

Let's consider $f(x)=x^8+4x^6+2x^4+28x^2+1$ and a prime $p>3$.

If there exist an integer $z$ s.t. $p|f(z)$, prove that

there exist integers $z_1, z_2, \cdots , z_8$ s.t. all coefficients of

$f(x)-(x-z_1)(x-z_2)\cdots (x-z_8)$ are divisible by $p$.

Let $y=x^2+1$.

\[ h(y)=y^4-4y^2+32y-28=(y^2+2)^2-2(2y-4)^2 \]

Since $p\mid f(z)=h(t)$ (where $t=z^2+1$), one has

\[ (t^2+2)^2\equiv 2(2t-4)^2\pmod p \]

Now, $2t-4$ is not divisible by $p$ (otherwise one would have $p=2$ or $t\equiv 2\pmod p$, and that would mean $p\mid 6^2$ which is not possible since $p>3$). So, there is $u$ such that $u(2t-4)\equiv 1\pmod p$, from where one finds

\[ \big(u(t^2+2)^2\big)^2\equiv 2\pmod p \]

Let $a=u(t^2+2)$. Now one has:

\[ h(y)\equiv(y^2+2)^2-(2a(y-2))^2\pmod p \]

After factoring, one of the factors is

\[ y^2+2-2a(y-2)=y^2-2ay+a^2-a^2+4a+2=(y-a)^2+(2+4a-a^2) \]

what is, by modulo $p$, congruent to $(y-a)^2+4a$. Similarly, the second factor is, by modulo $p$, congruent to $(y+a)^2-4a$, and so

\[ h(y)\equiv\big((y-a)^2+4a\big)\big((y+a)^2-4a\big)\pmod p \]

Since $h(t)$ is divisible by $p$, on of the numbers

\[ (t-a)^2+4a,\quad (t+a)^2-4a \]

is divisible by $p$. Because of $p\neq 2$, at least one of the numbers $a,-a$ is congruent to a square (mod $p$), i. e.  there is $b$ such that

\[ b^2\equiv\pm a\pmod p \]

In any case, one finds out that for an integer $b$ holds

\[ h(y)\equiv\big((y-b^2)^2+4b^2\big)\big((y+b^2)^2-4b^2\big)\pmod p\quad (*) \]

and $p$ divides the number

\[ \big((t-b^2)^2+4b^2\big)\big((t+b^2)^2-4b^2\big)=\big((t-b^2)^2+4b^2\big)(t+b^2-2b)(t+b^2+2b)= \]
\[ =\big((t-b^2)^2+(2b)^2\big)\big(z^2+(b-1)^2\big)\big(z^2+(b+1)^2\big) \]

(Recall that $t=z^2+1$). It follows that $p$ divides a sum of two squares, and therefore $-1$ is a quadratic residue modulo $p$, i. e.

\[ c^2\equiv -1\pmod p\quad (**) \]

Putting this into $(*)$, $h(y)$ factorises as follows (mod $p$):

\[ h(y)\equiv (y-b^2-2bc)(y-b^2+2bc)(y+b^2-2b)(y+b^2+2b)\pmod p \]

Putting back $y=x^2+1$, it follows (having $(**)$ in mind):

\[ f(x)\equiv_p\big(x^2-(b+c)^2\big)\big(x^2-(b-c)^2\big)\big(x^2-(c(b-1))^2\big)\big(x^2-(c(b+1))^2\big) \]

Therefore, we could take $z_1,\dots,z_8$ to be $\pm(b+c),\pm(b-c),\pm(bc-c),\pm(bc+c)$.

Wow, Wonderful!  :D

I'll post my solution soon.  :)

But yours is better than mine.


geometry perimeter trigonometry rotation vector AMC AIME

Let ABCD be a square with triangle APQ inscribed such that P is on side CD, Q is on side BC, and $\angle PAQ = 45^o$. Prove that the perimeter of PQC is constant for any placements of P and Q which satisfy the given conditions.

[b]Trigonometry method[/b]
With reference to the figure, denote$ \angle BAQ = \theta$ and $\angle DAP = \phi$ and $AB = a$.
WLOG, we may assume $\theta \geq \phi$
$tan(\theta+\phi) = tan 45^0 \Longrightarrow tan\theta + tan\phi = 1-tan\theta tan\phi$---(1)
With some calculations,
$PC=a-a tan\theta$ and $QC = a - a tan\phi$.
Therefore,
$PQ = \sqrt{PC^2 + QC^2}\\
=a\sqrt{(1-tan\theta)^2+(1-tan\phi)^2}\\
=a\sqrt{1-2tan\theta+tan^2\theta+1-2tan\phi+tan^2\phi}\\
=a\sqrt{tan^2\theta + 2 tan\theta tan\phi + tan^2\phi}$
(by equation (1))
$=a(tan\theta +tan\phi)$
Hence,
$PQ+QC+PC=a(tan\theta + tan\phi)+(a - a tan\phi)+(a-a tan\theta )= 2a$ which is a constant.

Remark: The constant should be 2a as expected because we will get this value in the limiting case when $\phi$ approaches zero.

[b]Geometry method[/b]
---still thinking-----

Just got this proof on the way home by KCR train.
In view of the trigonometric method, we only need to prove that PQ + QC + P C = 2a,
this is equivalent to showing that BQ + PD = PQ. ( crucial observation)
Next, rotate $\Delta ABQ$ through $90^0$ in the anti-clockwise sense around point A.
So, A is fixed, B maps to D, and Q maps to Q', say.
We got the $\Delta ADQ'$ which is congruent to $\Delta ABQ$.
Note that $\angle Q'AP=45^0$ and Q'DP is then a striaght line.
Then it can be shown that $\Delta APQ$ and $\Delta AQ'P$ are congruent. ( using SAS argument)
Here comes the result:
BQ + PD = Q'D + DP =Q'P= PQ. (Q.E.D.)

I think we can also prove the statement by vectors approach and co-geom approach. But I am not able to do so yet. May I invite other maths  fans to attack the problem with more other approaches?

Similar to AIME II 2005 #12? (Consider A as the center of a large square with 2x sides)

I dont know if this is correct but i wanna try though. It's proving by reasoning.

Since Angle A is 45 degrees and should be maintained, therefore the change in PC would also correspond to the change in QC. Thus, the Perimeter is constant.


In how many ways can the set $N ={1, 2, \cdots, n}$ be partitioned in the form $p(N) = A_{1}\cup A_{2}\cup \cdots \cup A_{k},$ where $A_{i}$ consists of elements forming arithmetic progressions, all with the same common positive difference $d_{p}$ and of length at least one? At least two?

[hide="Solution"]

[b]Part 1[/b]

Claim: There are $2^{n}-2$ ways of performing the desired partitioning.

Let $d(k)$ equal the number of ways $N$ can be partitioned as above with common difference $k.$ We are thus trying to evaluate

$\sum_{i=1}^{n-1}d(i)$

[b]Lemma: $d(i) = 2^{n-i}$[/b]

We may divide $N$ into various rows where the first term of each row denotes a residue $\bmod{i}.$ The only exception is the last row, as no row starts with $0$; the last row will start with $i.$ We display the rows as such:

$1, 1+i, 1+2i, 1+3i, \cdots$
$2, 2+i, 2+2i, 2+3i, \cdots$
$\cdots$
$i, 2i, 3i, \cdots$

Since all numbers have one lowest remainder $\bmod{i}$ and we have covered all possible remainders, all elements of $N$ occur exactly once in these rows.

Now, we can take $k$ line segments and partition a given row above; all entries within two segments would belong to the same set. For example, we can have:

$1| 1+i, 1+2i, 1+3i | 1+4i | 1+5i, 1+6i, 1+7i, 1+8i,$

which would result in the various subsets: ${1},{1+i, 1+2i, 1+3i},{1+4i},{1+5i, 1+6i, 1+7i, 1+8i}.$ For any given row with $k$ elements, we can have at most $k-1$ segments. Thus, we can arrange any number of segments where the number lies between $0$ and $k-1$, inclusive, in:

$\binom{k-1}{0}+\binom{k-1}{1}+\cdots+\binom{k-1}{k-1}= 2^{k-1}$

ways. Applying the same principle to the other rows and multiplying, we see that the result always gives us $2^{n-i},$ as desired.

We now proceed to the original proof.

Since $d(i) = 2^{n-i}$ by the above lemma, we have:

$\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}= 2^{n}-2$

Thus, there are $2^{n}-2$ ways of partitioning the set as desired.

[b]Part 2[/b]

Everything is the same as above, except the lemma slightly changes to $d(i) = 2^{n-i}-i.$ Evaluating the earlier sum gives us:

$\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}-i = 2^{n}-\frac{n(n-1)}{2}-2$

[/hide]


Here's an interesting problem that I generalized from a textbook-ish problem:

A boy is initially seated on the top of a hemispherical ice mound of radius $ R$.  He begins to slide down the ice with a negligible initial speed.  Approximate the ice as being frictionless.  At what height does the boy lose contact with the ice?

Enjoy!

[hide]If we assume that boy is a point mass then
$ h \equal{}\frac{2}{3}R$[/hide]

yes that's correct $ @petko$ assuming the boy is a point particle :rotfl:

$ h \equal{}\frac{2}{3}R$ is the answer I get too..... but , pardesi, what else could one assume the boy to be that would still make the problem easily solvable?

no i didn't mean to go against what $ @petko$ just thinking of the magnification  :wink:

As a personal opinion, this post should belong to the Introductory Physics Forum: at least im my country problems like this are solved in the 11th/12th grade.

[quote="Zakary"]what else could one assume the boy to be that would still make the problem easily solvable?[/quote]

The assumptions are that the boy is a point mass, and there is no friction or air resistance.

Here are other problems related to this one that might be more challenging:

1) Instead of ice, consider an hemisphere made of other material, so that friction cannot be ignored.

2) Instead of an hemisphere, consider the boy to slide, without inital velocity, from the top of a sphere of radius R that is glued to the ground.

3) Consider that the ice mound can also move.

adding to Carcul's List of problems
4)u can also have a sphere of radius $ r$ on top of it which is rolling without slipping


Describe a solid that has 1 face, 2 edges, and 4 vertices.

Perhaps it looks like a candy wrapper (the kind that might be for a jawbreaker or something).  But the "face" isn't really a face...

Well it doesn't satisfy euler's formula, so it's got at least one hole in it.

I'm thinking of some sort of jacked up donut....

And now I'm hungry....

lol i just thought of a square.. but then that's not 3 dimensional... and if you thought of it like a piece of paper, it would still have some defined height.... so it would be a rectangular prism...... or maybe the solid is not even in 3 dimensions... *shrug, he never specified, did he?

I thought of that^^^, but couldn't imagine one with 4 vertices....

[hide="Hint"]It has a [url=http://mathworld.wolfram.com/]Mathworld[/url] site dedicated to it :P
Bonus point for the URL[/hide]

Well I don't know the name of the figure but it could perhaps be like a candy wrapper as I proposed earlier.  Like a round shape with two ends clasped together to form two edges.


linear algebra
matrix
induction
linear algebra unsolved

Let $ P \equal{}${$ A_{1} , A_{2},...A_{k}$} a finit family of matrix ( $ A_{i} \in M_{n}(C)$ ) with property : exist $ A \in P \minus{} I_{n}$,A invertible,s.t $ AB \in P$ for every $ B \in P$.
Prove that :
1) $ A^{ \minus{} 1} \in P$
2)$ A_{1} \plus{} A_{2} \plus{} ... \plus{} A_{k}$ is not invertible

I believe (2) but I don't believe (1).

Let $ P\equal{}\left\{\begin{bmatrix}1&1\\\minus{}1&0\end{bmatrix}, \begin{bmatrix}\minus{}1&0\\0&\minus{}1\end{bmatrix}, \begin{bmatrix}0&\minus{}1\\1&1\end{bmatrix}\right\}$

and let $ A\equal{}\begin{bmatrix}0&1\\\minus{}1&\minus{}1\end{bmatrix}.$

Then $ A\in P\minus{}I$ but $ A^{\minus{}1}\not\in P$

Of course, I was interpreting $ P\minus{}I$ as algebraic subtraction. Did you mean $ P\setminus\{I\}?$ That would be different.

I am pretty sure that $ A \minus{} I$ must mean $ A\backslash I$.

  darij

darij: \setminus has better spacing than \backslash .

$ B\rightarrow A\Rightarrow A^2\in P$. By induction we get that $ A^n\in P,(\forall) n\geq 1$. Since $ P$ is finite there will exist $ i,j\in N^*$ so that $ A^i\equal{}A^j\Rightarrow  A^{\minus{}1}\equal{}A^{i\minus{}j\minus{}1}.$
Let's assume $ i>j$. This implies $ i>j\plus{}1$, since otherwise we get $ A\equal{}I$.So $ A^{i\minus{}j\minus{}1}\in P$ and this solves 1).

for 2/

A=A(1)+...+A(k)

find relation with A^2 and A


A cylinder has a radius of 3 centimeters and a height of 2 centimeters, while a second cylinder has a radius of 2 centimeters and a height of 3 centimeters. What is the ratio of the volume of the first to the volume of the second?

You forgot pi, but since this is a ratio the pi will cancel out. A cylinders volume is actually pi r^2 h but JOngaos answer of 3/2 is indeed correct.


I just wonder whether there is any plan to make Asymptote functioning on AoPSWiki the same way as on the Forum in the near future. Or is it technically impossible for some reason? :?

On the todo list - right there with integrating the Wiki with the Forum.  We hope to knock these off this summer.


[quote="star99"]I've just been wondering where you go to school, cuz the Bay area is pretty big.  You could also just tell your school district.  I go to school in the Cupertino Union School District (CUSD).[/quote]
This is an old topic but... Me too. I'm in 8th grade now.

[quote="star99"]what school?[/quote]
Cupertino Middle.

whoah FUHSD CUSD is more open than i thought

more people go there than i would've guessed

Evergreen Valley High School (it is new  :) )

[quote="DiscreetFourierTransform"]Menlo School, Atherton, CA
We played GUNN in basketball and owned you guys ;-) I love being a math nerd and a baller ;-)[/quote]

lol, did you beat PALY? i heard that they were good

No, we play them Friday. Honestly, I'm quite scared.

I go to Gunn for Chinese school!
I go to Redwood. Which feeds into Saratoga High.

If I lived in California....I would go to Henry M. Gunn High School...but I don't...

[quote="DiscreetFourierTransform"]Results are in on our Menlo vs. Paly Bball game  :(
We lost by four. It was a *really* close game the whole time, and I'm proud of my team for playing with a good atitude and spirit. If any Paly team players are browsing, just want to say, Great Game![/quote]

nice job. what was the final score? :P Are there paly people here?


inequalities
search
inequalities proposed

Let $ x$, $ y$, $ z$ be real numbers satisfying $ x^2\plus{}y^2\plus{}z^2\plus{}9\equal{}4(x\plus{}y\plus{}z)$. Prove that
\[ x^4\plus{}y^4\plus{}z^4\plus{}16(x^2\plus{}y^2\plus{}z^2) \ge 8(x^3\plus{}y^3\plus{}z^3)\plus{}27\]
and determine when equality holds.

[quote="FelixD"]Let $ x$, $ y$, $ z$ be real numbers satisfying $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 9 \equal{} 4(x \plus{} y \plus{} z)$. Prove that
\[ x^4 \plus{} y^4 \plus{} z^4 \plus{} 16(x^2 \plus{} y^2 \plus{} z^2) \ge 8(x^3 \plus{} y^3 \plus{} z^3) \plus{} 27\]
and determine when equality holds.[/quote]
Notice that
$ x^4\minus{}8x^3\plus{}16x^2\minus{}9\plus{}6(x^2\minus{}4x\plus{}3)\equal{}(x\minus{}1)^2(x\minus{}3)^ 2\ge 0.$
Therefore
$ x^4\minus{}8x^3\plus{}16x^2\minus{}9 \ge \minus{}6(x^2\minus{}4x\plus{}3).$
Adding this and its analogous inequalities, we get
$ \sum x^4\minus{}8\sum x^3\plus{}16\sum x^2\minus{}27 \ge \minus{}6\left( \sum x^2\minus{}4\sum x \plus{}9\right) \equal{}0.$
This proves the desired inequality.

A quick search would help you, but anyway: MEMO = Middle European Mathematical Olympiad.


2 vertices of an octagon are chosen at random.  What is the probability that they are adjacent?

I asked this because I put $ \frac{4}{14}$ and I got it right. Does it have to be in simplified form? (not sure whether to report or not.)

Can someone also post a fast way of solving the problem?

There are C(8,2) ways to pick 2 vertices of an octagon. There are 8 ways to pick 2 consecutive vertices; because there are 8 vertices. This is also the # of ways to pick 3,4... vertices of an octagon. Anyway, the answer is thus 8/28 = 2/7 :lol:

[quote="myyellowducky82"]I asked this because I put $ \frac {4}{14}$ and I got it right. Does it have to be in simplified form? (not sure whether to report or not.)

Can someone also post a fast way of solving the problem?[/quote]

I don't think you need to put it in simplest form on Alcumus unless it asks for it..
And Ritesh answered your other question.

[hide="Different Solution From Alcumus"]There are $8$ pairs of points that are adjacent to each other.  (this can be thought of as each side of the octagon)  There are $\dbinom{8}{2}$ ways to select any pair of points.  The probability is $\dfrac{8}{\dbinom{8}{2}}=\boxed{\frac{2}{7}}$.[/hide]

Hmmm... I think the faster way is just to use logic.
[hide = "Logical Solution"]
We choose 1 point. For second point, there are 7 choices left. Every point has 2 vertices adjacent to it. So answer is $\boxed{\dfrac{2}{7}}$ :)
[/hide]


If $ ABCDEFG$ is a regular $ 7$-gon with side $ 1$, show that: $ \frac{1}{AC}\plus{}\frac{1}{AD}\equal{}1$.

Already discussed here http://www.mathlinks.ro/Forum/viewtopic.php?t=281985


Are any of you guys worried about the bird flu? Its in Asia right now, and can migrate to America somehow, but alot of people are really worried. Any opinions? :huh:
[url=http://cagle.msnbc.com/news/BirdFlu05/main.asp]BIRD FLU![/url]

[quote="G-UNIT"]CNN estimates 10 million people will die in North America from flu's of that sort.  That comes out to, roughly, 1 in 43 people.[/quote]

Waaaahhh!!!
yeah i heard about that...really scary to think actually...

yes i am because it is in greece and italy is near greece............ :ninja:


function
integration
calculus
inequalities
real analysis
real analysis unsolved

Let $ f: [0,\infty) \to [0,\infty)$ be a continuous function for which there is a positive integer $ a$ such that $ f(f(x))\equal{}x^a$, $ \forall x \in [0,\infty)$.Prove that:

$ \int_{0}^1 (f(x))^2 \geq \frac{2a\minus{}1}{a^2\plus{}6a\minus{}3}$

[i]Mihai Piticari[/i]

From the fact that $ f(f(x)) \equal{} x^{a}$ we know that $ f$ is bijective . Since $ f$ is continuous and bijective we know that $ f$ is strictly monotonic.

Thus, $ f$ cannot be strictly decreasing, because in that case $ f(x) < f(0)$, for all $ x \in (0 ; \infty)$, and since $ f(0)$ is finite we have a contradiction with the fact that $ f$ is surjective.

Since $ f$ is strictly increasing we can observe that $ f(0) \equal{} 0$ and $ f(1) \equal{} 1$. With the substitution $ x \equal{} f(t)$, the hypothesis becomes:

$ \int_{0}^{1} f^2(x) dx \equal{} \int_{0}^{1} f^2(f(t)) d f(t) \equal{} \int_{0}^{1} t^{2a} d f(t)$.

Now, using integration by parts
\[ \int_{0}^{1}t^{2a} d f(t) \equal{} t^{2a}f(t) |_{0}^{1} \minus{} 2a \int_{0}^{1} f(t) t^{2a \minus{} 1} dt \geq 1 \minus{} 2a\sqrt {\int_{0}^{1} t^{4a \minus{} 2} dt \int_{0}^{1} f^2(t) dt }\]

\[ \equal{} 1 \minus{} 2a \sqrt {\frac {1}{4a \minus{} 1} \int_{0}^1 f^2(t) dt}\]
.

 Now, if we denote by $ A \equal{} \sqrt {\int_{0}^{1} f^2 (t) dt}$, the last inequality is :
\[ A^2 \geq 1 \minus{} 2a\sqrt {\frac {1}{4a \minus{} 1}} A\]
.

I think with a little bit of calculation, using the sign of the second degree function from the above, we can prove our inequality.


search
inequalities unsolved
inequalities

Go to Balkan mathematical olympiad place and then search the third problem. Or put directly into search button.

Or, just from this $\frac{a^2}{b}=2a-b+\frac{(a-b)^2}{b}$ and Cauchy-Schwarz we are done. ;)


stronger than vasc's.but I think it can also be sloved by my method with root.
do you have a nicer solution? :)

I wonder if the strongest inequality is:
$3+3xyz \geq 2(xy+yz+zx)$
if it is why didn't you post it ,VASC?
is it your next step? :)

[quote="Vasc"]Arqady, you anticipated my next step. :lol:
Actually, my next step is sharper:
$21+18xyz \ge 13(xy+yz+zx)$;[/quote]
I proved :
If $k\geq\frac{3\sqrt6}{(1+\sqrt[6]4)\sqrt{1+\sqrt[3]2}}-2=0.163002288...$
then $3+3k+3xyz\geq(2+k)(xy+xz+yz)$ ;)

[quote="zhaobin"]
do you have a nicer solution? :)[/quote]
No. :|
[quote="Vasc"]Of course.....
But for the inequality
$21+18xyz \ge 13(xy+yz+zx)$
there is solution without square root. :lol:[/quote]
WOW! :o

[quote="arqady"][quote="zhaobin"]
do you have a nicer solution? :)[/quote]
No. :|

there is solution without square root. :lol:[/quote]
WOW! :o[/quote]
believe me :) !!
I think all of this can sloved by
$f(x,y,z) \geq f( \sqrt{\frac{x^2+y^2}{2}},\sqrt{\frac{x^2+y^2}{2}},z)$,$z \le 1$ .

Can you tell how you figure out the number $k\geq\frac{3\sqrt6}{(1+\sqrt[6]4)\sqrt{1+\sqrt[3]2}}-2=0.163002288...$?with computer?

[quote="zhaobin"]
Can you tell how you figure out the number $k\geq\frac{3\sqrt6}{(1+\sqrt[6]4)\sqrt{1+\sqrt[3]2}}-2=0.163002288...$?with computer?[/quote]
Mixing variables method. :P :)

[quote="Vasc"]
... for the inequality
$21+18xyz \ge 13(xy+yz+zx)$
there is solution without square root. :lol:[/quote]
I capitulate and wait your solution.Thank you.

[quote=szl6208][quote=hoanglong2k]Let $a,b,c$ are positive real number such that $a^2+b^2+c^2=3$
 Prove that $21+18abc\geq 13(ab+bc+ca)$[/quote]

We have
\[21+18abc-13(ab+bc+ca)=\frac{1}{2}\sum{(a+b)(2c-c^2-b-a+ab)^2}+\frac{1}{2}\sum{(a-b)^2(a+b-c)^2}+\sum{(a-1)^2(b-c)^2}\ge{0}\][/quote]

[quote=arqady]
I capitulate and wait your solution.[/quote]
It does not quite fit to you  :-D


A cyclist is travelling along a stright horizontal road.as she passes a bus stop she sees a red traffic light ahead of her. she continues to travel with a constant speed of $ 3ms^{\minus{}1}$ for $ 20$ seconds and then decelerates at a constant rate of $ 0.2 ms^{\minus{}2}$ untill coming to rest at the traffic light.

[size=150](1)[/size]:Calculate the distance between the bus stop and the traffic light.

[size=150](2)[/size]:Calculate the time the cyclist takes to travel from the bus stop to the traffic light.

[size=150](3)[/size]Calculate the average speed of the cyclist between the bus stop and the traffic light.

This looks just like a problem I just did in my textbook.

Are the quantities meant to be $ 3^{\minus{}1}\text{m/s}$ and $ \left(0.2\right)^{\minus{}2}\text{m/s}$?

He's using $ \text{ms}^{\minus{}1}$ to denote $ \text{m/s}$ and $ \text{ms}^{\minus{}2}$ to denote $ \text{m/s}^{2}$.

[quote="t0rajir0u"]He's using $ \text{ms}^{-1}$ to denote $ \text{m/s}$ and $ \text{ms}^{-2}$ to denote $ \text{m/s}^{2}$.[/quote]
Thank you, t0rajir0u.

[hide="solution to (1)"]
Now, for traveling at $ 3\text{m/s}$ for $ 20\text{s}$ is $ 60\text{m}$. After these sixty meters the bicyclist decelerates at $ .2\text{m/s}^{2}$ to reach $ 0\text{m/s}$ at the light. This can be represented by a right triangle with height (and leg) $ 3\text{m}$ and gradient $ -\frac{1}{5}$ along the hypotenuse. So the bottom (or other leg) is $ 15\text{m}$ long. The area of the triangle is equal to the distance traveled. Thus, $ d=\frac{3\cdot15}{2}\Rightarrow d=22.5\text{m}$. So, the answer is $ \boxed{82.5\text{m}}$
[/hide]

I had to get off last time, so here are the other two answers.

[hide="solution to (2)"]
From my other solution we have that the bottom of our distance triangle during the deceleration is $ 15\text{m}$. This is technically incorrect, it is more like $ 15\text{s}$  :lol: . Which means that the time traveled was $ \boxed{35\text{s}}$
[/hide]
[hide="solution to (3)"]
From my other two solutions we have that our cyclist traveled $ 82.5\text{m}$ in $ 35\text{s}$. So, her average speed is $ \frac{82.5\text{m}}{35\text{s}}$ which is approximately $ \boxed{2.4\text{m/s}}$ which is a decent speed.
[/hide]


function
Euler
number theory
totient function
relatively prime

1-a number theoretic function is called a completely number theoretic function if f(rs)=f(r)*f(s).
Demonstrate that the sum of the divisors, the number of the divisors and the toitent of n are not completely number theoretic functions.

Hi,

[hide]I guess the correct term is 'completely multiplicative' function.

That is, a number theoretic function f is called a completely multiplicative number theoretic function if f(rs)=f(r)*f(s). But if the condition holds only for coprme r and s, then f is just a multiplicative function.

Out of sum of divisors, no.of divisors, and totient function :

One can see that if r and s are not coprime, then each of the functions takes into account some constituents more than once, which results in f(rs) not equal to f(r)*f(s). eg, in the no. of divisors case, if d is a common divisor of r and s, then d is counted more than once. (Similarly for other functions.)

Thus, it can be seen that the above three functions are just multiplicative functions, and not completely multiplicative functions.

[/hide]

**********************************************************************************

Sambit

Hi Peter,
The number of positive integers less than or equal to n and relatively prime to n is called the toitent of n.It is also called the Euler function.


Find all integers $ n$ such that $ 3n\plus{}4$ is divisible by $ 7n\plus{}1$.

Hi. [hide="Solution"]

[CTRL key not working again, so can't LaTeX.]

Did you mean to write $ 7$  divides $ 3n\plus{}4$?

Then, since $ 3n\plus{}4$ divides $ 7n\plus{}1$,

$ 3n\plus{}4$ divides $ 7(3n\plus{}4) \minus{}3(7n\plus{}1)$,

that is $ 3n\plus{}4$ divides $ 25$.

$ 3n\plus{}4 \equal{} 1, \minus{}1, 5, \minus{}5, 25$ or $ \minus{}25$

$ n \equal{} \minus{}3, \minus{}1$ or $ 7$.

[/hide]


I was wondering how the polygon() function worked and I realized that it can get all the points by simply finding $ 360/n$ where $ n$ is the number of sides of the polygon, and using s*dir((360/n)i) in a for loop where $ i$ is the $ i\plus{}1$th point of the $ n$-gon and $ s$ is the side length of the $ n$-gon. Anyways, I was wondering if there was a function that when given an array of pairs, it would concatenate consecutive pairs in a path that would end with the cycle instruction. If there isn't a built-in one, how would I go about doing that?

Just define

path array2polygon(pair[] p)
{
path P=nullpath;
for(int i=0;i<p.length;++i) P=P--p[i];
P=P--cycle;
return P;
}

Oh cool, I didn't know you could do that with paths.

[quote="fedja"]
path array2polygon(pair[] p)
{
path P=nullpath;
for(int i=0;i<p.length;++i) P=P--p[i];
P=P--cycle;
return P;
}[/quote]

Optimized code:
[code]path array2polygon(pair[] p){return operator --(...p)--cycle;}[/code]


[u][b]Problem[/b][/u] [b](zaizai-hoang)[/b]
Let $a,b,c\ge 0$. Prove that:
$\sqrt{\frac{a^{2}}{a^{2}+26bc}}+\sqrt{\frac{b^{2}}{b^{2}+26ca}}+\sqrt{\frac{c^{2}}{c^{2}+26ab}}\ge 1$

 :D Now I want find nice solution. Who is help me? :P

[quote="zaizai-hoang"][u][b]Problem[/b][/u] [b](zaizai-hoang)[/b]
Let $a,b,c\ge 0$. Prove that:
$\sqrt{\frac{a^{2}}{a^{2}+26bc}}+\sqrt{\frac{b^{2}}{b^{2}+26ca}}+\sqrt{\frac{c^{2}}{c^{2}+26ab}}\ge 1$
[/quote]
But it's wrong. Try $a=b=c=1.$ :wink:


You guys know about the weird zgzag thing with like i=3, and i+10, right?  How do you find out the answer?
The same thing for the wierd squiggly thing with dx or something.  HELP!! :help:  :help: [/i]

You mean $ \int$ and $ \approx$

The $ \int$ is not for Classroom Math.

By "weird zigzag thing" I think he actually means $ \sum$.  This is a capital sigma, but is just read "sum."  In particular,

$ \sum_{i\equal{}m}^{n} a_i$

is shorthand notation to mean the sum

$ a_m \plus{} a_{m\plus{}1} \plus{} a_{m\plus{}2} \plus{} ... \plus{} a_n$.

For example, $ \sum_{i\equal{}3}^{10} i^2 \equal{} 3^2 \plus{} 4^2 \plus{} 5^2 \plus{} ... \plus{} 10^2$.

Thanks for your help :D
Then where does the squiggly go?

[quote="MathIsGreat"]Then where does the squiggly go?[/quote]

The "squiggly" is an [url=http://mathworld.wolfram.com/Integral.html]integral[/url], used in calculus.

to tell you the truth, im bad at math.  What's an integeral??

Well if you're bad at math, I don't think you can handle integrals if you haven't taken a bit of precalc if not calculus yet.

I remember the first time I looked at the "squiggly thing" and didn't know what it was. I followed up by looking in a calculus book, and finding the definition of an integral. By the time I finished the sentence, I was total lost, not knowing what a [url=http://mathworld.wolfram.com/Derivative.html]derivative[/url] was. I decided it was best until high school, to persue concepts far beyond my reach.  :)


vector
linear algebra
matrix
linear algebra unsolved

Something realted to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=104862]this one[/url].
Suppose $e_{1},e_{2},\dots,e_{n+1}\in\mathbb R^{n}$ that for each $i\neq j$ we know that $(e_{i},e_{j})<0$. Prove that every $n$ of these vectors are linearly independent.

I solved it in a problem posted by alekk in which we had to prove that the rank of a positive matrix with non-diagonal entries negative is at least n-1.

Otherwise, $n$ linearly dependent vectors together with the orthogonal projection of the $n+1$'th onto a hyperplane containing the first $n$ will form a system of $n+1$ vectors with negative pairwise scalar products in an $n-1$-dimensional Euclidean space. According to the problem you gave a link to, this is impossible.

If $e_{1},\dots,e_{n}$ are not linearly independent, there exist $\alpha_{1},\alpha_{2},\dots,\alpha_{n}\in\mathbb R$ that $\sum_{i=1}^{n}\alpha_{i}e_{i}=0$. Thus $\sum_{i=1}^{n}\alpha_{i}(e_{i},e_{n+1})=0$, so all $\alpha_{j}$ does not have the same sign. WLOG you could write $\lambda_{1}e_{1}+\dots+\lambda_{k}e_{k}=\lambda_{k+1}'e_{k+1}+\dots+\lambda_{n}'e_{n}$  that $\lambda_{1},\dots,\lambda_{k},\lambda_{k+1}',\dots,\lambda_{n}'\geq 0$ and all of them are not zero. Suppose $\lambda_{1},\lambda_{k+1}'>0$. So \[0>\sum_{i,j}\lambda_{i}\lambda_{j}'(e_{i},e_{j})=(\sum_{i=1}^{k}\lambda_{i}e_{i},\sum_{i=k+1}^{n}\lambda_{i}'e_{i})=(\sum_{i=k+1}^{n}\lambda_{i}'e_{i},\sum_{i=k+1}^{n}\lambda_{i}'e_{i})\geq0\] Contradiction solves the probelm. :D


Determine all pairs $ (a,b)$ of real numbers such that $ 10, a, b, ab$ is an arithmetic progression.

[hide]We have a system:
\[ ab \minus{} b \equal{} b \minus{} a \\
b \minus{} a \equal{} a \minus{} 10
\]Solving the latter equation for b and plugging into the first equation, and simplifying, we find that:
\[ 2a^2 \minus{} 13a \plus{} 20 \equal{} 0
\]This quadratic has solutions $ a \equal{} 4,\ \frac {5}{2}$. Evaluating these values of $ a$ in either equation yields $ b \equal{} \minus{} 2$ or $ b \equal{} \minus{} 5$, respectively.
Thus, the pairs are $ \displaystyle \boxed{\left(4,\ \minus{} 2\right) \text{ or } \left(\frac {5}{2},\ \minus{} 5\right)}$.[/hide]


If I go one way at 2 minutes per mile and return the same route home at 2 miles per minute, what's my average speed in miles/hour?

[hide]$\frac{5}{4}$
2 minutes per mile = 1/2 miles per minute
1/2 miles per minute one way, 2 miles per minute the other. Total distance traveled is $\frac{5x}{2}$, total time is $2x$, so $\frac{\frac{5x}{2}}{2x}=\frac{5}{4}$[/hide]

Nope, that's not correct...
Since I'm going at $30$ miles per hour and returning at $120$ miles per hour, we have $\frac{30t+120(\frac{t}{4})}{t+\frac{t}{4}}=48$

[hide]You go at 30 miles per hour and return at 120 miles per hour. Thus, the average speed is the harmonic mean:

$\frac{2}{\frac{1}{30}+\frac{1}{120}}=48$.[/hide]

hey iversonfan were you at the scjas workshop this saturday at spring valley?

yeah, but i'll pm you since we can't talk about personal stuff on the boards.


limit
inequalities
logarithms
integration
calculus
calculus computations

Prove that:

\[ \lim_{n \to \infty}\left(1+\frac 1 n\right)\left(1+\frac 2 n\right)^{\frac{1}{2}}...\left(1+\frac n n\right)^{\frac{1}{n}}=e^{\frac{\pi^2}{12}}  \]

I think there is a mistake in your exercise, because $R_n=\frac{\sum\limits_{k=1}^n \frac 1k}{n}\rightarrow 0$ and

$\ a_n=\prod\limits_{k=1}^n \left( 1+\frac 1n\right)^{\frac 1k}=\left(1+\frac 1n\right)^{\sum\limits_{k=1}^n \frac 1k }=\left[\left(1+\frac 1n\right)^n\right]^{R_n}\rightarrow e^0=1$

Sorry I've edited it. I have another problem which looks like this:
\[ \lim_{n\to \infty} \left( \sqrt{1+\frac 1 n}+\sqrt[3]{\left(1+\frac 1 n\right)^2}+\sqrt[4]{\left(1+\frac 1 n\right)^3}+...+\sqrt[n+1]{\left(1+\frac 1 n\right)^n}-n\right) \]

Nice exercise !. Answer: $\lim_{n\to \infty} a_n =1.$[hide="Here is a  solution through a corseting."]$\blacktriangleright$ [b]Denote[/b] $\boxed {\ a_n =\sum \limits_{k=1}^n \left(1+\frac 1n\right)^{\frac{k}{k+1}}-n\ },\ n\in N^*.$
$($ I used to the left the $A.M-G.M.$ - inequality and to the right $\sum\limits_{k=1}^n x_k\le n\cdot x_n).$

$n\left[\left(1+\frac 1n\right)^{\frac 1n\sum\limits_{k=1}^n \frac{k}{k+1}}-1\right]=b_n\le a_n\le c_n =n\left[\left(1+\frac 1n\right)^{\frac{n}{n+1}}-1\right]\ .$

$\blacktriangleright\ b_n=\frac 1n\sum\limits_{k=1}^n \frac{k}{k+1}\cdot \ln \left(1+\frac 1n\right)^n\cdot \frac{e^{v_n}-1}{v_n}\rightarrow 1$ because
$v_n=\frac 1n\sum \limits_{k=1}^n\frac{k}{k+1}\cdot \ln \left(1+\frac 1n\right)\rightarrow0,\ \frac{e^{v_n}-1}{v_n}\rightarrow 1,$
$\frac 1n\sum\limits_{k=1}^n\frac{k}{k+1}\rightarrow 1,\ \ln \left(1+\frac 1n\right)^n\rightarrow 1.$

$\blacktriangleright\ c_n=\frac{n}{n+1}\ln \left(1+\frac 1n\right)^n\cdot \frac{e^{u_n}-1}{u_n}\rightarrow 1$ because
$u_n=\frac{n}{n+1}\ln \left(1+\frac 1n\right)\rightarrow 0,\ \frac{e^{u_n}-1}{u_n}\rightarrow 1,\ \frac{n}{n+1}\rightarrow 1.$

$\blacktriangleright$ Therefore, $a_n\rightarrow 1\ !$[/hide]

Very nice solution!! Mine includes aplying corseting theorem twice, lagrange theorem, and Cesaro-Stolz lemma. I think that no one is interested in reding it  :D Thank you very much.

Virgil you did not say what is $b_n$ in your proof

hint for the first problem.
[hide] Extract logharitm and prove that the limit is equal to $\int_0^1 \frac{\ln(1+x)}{ x}dx$.Then compute the integral by expanding in Taylor series. And use the uniform convergence of that series to change the summation with the integral sign.[/hide]

Good luck ;)