inequalities function calculus derivative

I still don't get how to determine if an inequality is Convex or Concave. Anyone willing to reply i will really apprecaite.

Inequalities aren't convex/concave. Functions are.

A function $f(x)$ is convex on an interval $I$ iff $f''(x)>0$ $\forall x \in I$. And if you don't know Calculus (or just derivatives, see [url=http://mathworld.wolfram.com/ConvexFunction.html]here[/url]). Usually Calculus is the easiest way to show convexity, however.

so one more question: does Jensen's have to deal with Calculus? I dont seem to understand much about it since it deals with Convexity and all.

Usually it does, just because by nature the second derivative is used to prove convexity/concavity a lot. But theoretically it doesn't I suppose.


algebra
polynomial
inequalities solved
inequalities

You can have something greater.It is
[(ab+ac+bc+cd+da+bd)/6]^1/2 \geq [(abc+cda+bcd+dab)/4]^1/3 by use a polynomial (x-a)(x-b)(x-c)(x-d) and Viete.... :)


How can I add a footer that says, for example, "Typeset in \LaTeXe," just on the first page? I've only managed to do this using \footnote, which gives me a superscript number that I don't need. I looked up the package fancyhdr, but I couldn't figure out if that would let me specify the footer for just the first page.

Thank you.

preamble: \usepackage{fancyhdr}\pagestyle{fancy}

first page somewhere: \cfoot{blah}

second page: \cfoot{}

That didn't get me what I wanted, but based on that and with some help from the documentation, I figured out what to do. According to the documentation, I had to call \thispagestyle{fancy} after \maketitle on the first page to avoid from being set to \thispagestyle{plain}.

first page:
\maketitle

\thispagestyle{fancy}
\renewcommand{\headrulewidth}{0pt}
\lhead{}
\chead{}
\rhead{}
\lfoot{{\small{Typeset in $ \LaTeXe$}}}
\cfoot{}
\rfoot{}

second page:
\pagestyle{plain}

Thanks!


Would anyone happen to have old tests or examples of tests from the Stanford EPGY Multivariable Calculus Course, my friend left me his book,notes & course schedule so that I could independently study it without enrolling in the course and paying the expensive fees. And I was hoping somebody had the old tests so I could study and take them to see if I mastered the material......

I do not believe that we are allowed to give you the past tests, and even so, who will grade them?

I was recently informed that I cannot ask for these as it is a violation of the Stanford Honor Code, so ignore this post please.


find all function $f: R\rightarrow R$ such that:
$f(x-y)+f(y-z)+f(z-x)=f(x)+f(y)+f(z)$

AYMANE is right. The problem is too easy. Maybe it was stated incorrectly.

If the problem was:
$f: \mathbb{R}\to\mathbb{R}$ such that, for every $x,y,z\in\mathbb{R}$ with $x\neq y\neq z\neq x$,
$f(x-y)+f(y-z)+f(z-x)=f(x)+f(y)+f(z)$,
the solution is still $f(x)=f(0)$.[hide]Subst. $x=0$ and $z=y-t$ when $t\neq 0\neq y\neq t$,
$f(t)-f(0)=f(y)-f(-y)$.

Since $y$ and $t$ are independent,
$f(y)-f(-y)=g(t)=C$ and
$f(t)-f(0)=h(y)=C$,
when $0\neq t\neq y$

It's valid for every $y$ so we can say
$f(t)=f(0)+C$, constant for every $t\neq 0$.
But then, $f(y)+f(-y)=0$ for every $y\neq 0$, so $C=0$.
Therefore, $f(t)=f(0)$.[/hide]
If the problem was:
$f: \mathbb{R^{*}}\to\mathbb{R}$ such that, for every $x,y,z\in\mathbb{R^{*}}$ with $x\neq y\neq z\neq x$,
$f(x-y)+f(y-z)+f(z-x)=f(x)+f(y)+f(z)$,
it's more difficult.
($\mathbb{R^{*}}\equiv\mathbb{R}\setminus\{0\}$)


calculus
integration
trigonometry
logarithms
calculus computations

What level do you think about the problem? I can't calculate the improper integral.

It likely requires re-expression in terms of infinite summations, since the answer appears very:

[hide="series-based"]

$2 \pi c-\frac{7}{2}\zeta (3)$

where $c$ is the Catalan constant

[/hide]

[quote="Dr. Doe"]$ \int_{0}^{\frac{\pi}{2}}x^{2}\csc x\, dx=\underbrace{\left[x^{2}\ln\left(\csc x-\cot x\right) \right]_{0}^{\frac{\pi}{2}}}_{=0}-\int_{0}^{\frac{\pi}{2}}2x \ln(\csc x-\cot x)\, dx$

$ =-\int_{0}^{\frac{\pi}{2}}2x\, \ln\left(\frac{1-\cos x}{\sin x}\right) dx =-\int_{0}^\frac{\pi}{2}2x\, \ln\left(\frac{2^{2}\sin^{2}\frac{x}{2}}{2 \sin x}\right) dx$

$ =\underbrace{\int_{0}^\frac{\pi}{2}4x \left[-\ln\left(2 \sin \frac{x}{2}\right)\right] dx}_{=A}-\underbrace{\int_{0}^\frac{\pi}{2}2x \big[-\ln\left(2 \sin x\right)\big] dx}_{=B}$

$ A=\int_{0}^\frac{\pi}{2}4x\sum_{k=1}^\infty \frac{\cos kx}{k}dx =\sum_{k=1}^\infty \int_{0}^\frac{\pi}{2}\frac{4x \cos kx}{k}dx$

$ =\sum_{k=1}^\infty \left[\frac{4 \cos\left(\frac{k\pi}{2}\right)}{k^{3}}-\frac{4}{k^{3}}+2\pi \frac{\sin\left(\frac{k\pi}{2}\right)}{k^{2}}\right]=-\frac38 \zeta(3)-4\zeta(3)+2\pi G$

$ B=\int_{0}^\frac{\pi}{2}2x\sum_{k=1}^\infty \frac{\cos 2kx}{k}dx =\sum_{k=1}^\infty \int_{0}^\frac{\pi}{2}\frac{2x \cos 2kx}{k}dx$

$ =\sum_{k=1}^\infty \left[\frac{\cos\left(k\pi\right)}{2 k^{3}}-\frac{1}{2\, k^{3}}+\frac{\pi}{2}\frac{\sin(k\pi)}{k^{2}}\right]=-\frac38 \zeta(3)-\frac12 \zeta(3)$

Therefore $ A-B=2\pi G-\frac72 \zeta(3)$[/quote]

very nice !  :bye:


Prove that $\left(\frac{a^2+b^2+c^2}{3}\right)^{\alpha}\geq\frac{1}{6}\cdot\sum_{sym}a^{2\alpha-1}b,$
where $a\geq0,$ $b\geq0,$ $c\geq0,$ $\alpha=\frac{5+\sqrt{13}}{4}.$

In Israel, that is. Come on, 6 more hours for me in Oregon (alas I am leaving for a New Years party so can't do it)


superior algebra
superior algebra unsolved

Hello,
Can someone please help me solve these questions with full working out please ?

1. Show that 5 is irreducible in Z[sqrt(-2)] and Z[sqrt(3)] but not in Z[i].

2. Find a factorisation of 9 + 4(sqrt(3)) into irreducibles, in Z[sqrt(3)].

3. Find four greatest common divisors of 13 + 39i and −17 + 19i in Z[i].

Thank You.

http://www.mathlinks.ro/Forum/viewtopic.php?t=167373
Do not post it more than one time!


The locus (set of points) equidistant from the circle $ x^{2}\plus{}y^{2}\equal{} 1$ and the line $ y \equal{}\minus{}3$ is given by the graph of
$ (a) y\plus{}2 \equal{} 3x^{2}$
$ (b) x^{2}\plus{}y^{2}\minus{}1 \equal{} (y\plus{}3)^{2}$
$ (c) x^{2}\plus{}y^{2}\equal{} (y\plus{}4)^{2}$
$ (d) none of these$

[hide]I got an answer different from the correct answer, I got choice b.  Real answer's choice c...please help.[/hide]

[hide]The correct answer is $ \sqrt{x^{2}\plus{}y^{2}}\minus{}1\equal{}y\plus{}3$. I think what you messed up on was the distance formula.[/hide]

[quote="aufha"]How would you apply the distance formula to this problem?[/quote]

[hide]The distance from $ (x,y)$ to $ (0,0)$ is $ \sqrt{x^{2}\plus{}y^{2}}$. Hence, the shortest distance from this point to the circle must be one less than this, or

$ \sqrt{x^{2}\plus{}y^{2}}\minus{}1$

The distance from $ y\equal{}\minus{}3$ to $ (x,y)$ is $ y\plus{}3$.

So:

$ \sqrt{x^{2}\plus{}y^{2}}\minus{}1 \equal{} y\plus{}3$

$ \sqrt{x^{2}\plus{}y^{2}}\equal{} y\plus{}4$

$ x^{2}\plus{}y^{2}\equal{} (y\plus{}4)^{2}$[/hide]


Hi, i need some BIG help with this question, as it's completely got me pulling my hair. First of all, i've got the question and answer, but i'm afraid it makes no sense to me at all.
The question is:

Let A = {p,q,r,s,t} and B = [p,q,r,s,t}. Which of the following relationes on A x B is a function which is neither 1-1 nor onto.

a) {(p,q), (q,s), (r,p), (s,r), (t,p)}
b) {(p,q), (q,s), (r,p), (s,r), (t,t)}
c) {(p,q), (r,s), (r,p), (s,r), (t,t)}
d) none of the above.

Any ideas ?

a) is a function.
It is not onto because t is not reached.
It is not 1-1 because p is the image of both r and t.


Factoring $x^{12}-x^{11}+x^9-x^8+x^6-x^4+x^3-x+1$ that one of factor is 6.degree.

This does not appear to factor; can you check the polynomial?

Did you accidentally miss $x^7$ and $x^5$?  :?

Isn't it related with generating functions?


calculus
integration
inequalities
analytic geometry
geometry
geometric transformation
rotation

Find $ \int{\int_S{(x\plus{}y)}}dS$, where $ S\equal{}\left\{(x,y): x^2\plus{}y^2\leq x\plus{}y \right\}$.

The inequality that defines $ S$ can be rewritten as $ \left(x \minus{} \frac12\right)^2 \plus{} \left(y \minus{} \frac12\right)^2\le\frac 12.$

That makes $ S$ a circular disk centered at $ \left(\frac12,\frac12\right).$

Polar coordinates do look like an attractive option for dealing with that, but my first instinct was just to translate the center to the origin. (Actually I tried a 45 degree rotation - while that works, too, I realized I didn't need it.)

Let $ u \equal{} x \minus{} \frac12$ and $ v \equal{} y \minus{} \frac12.$ Then the Jacobian is simply $ 1$ and we have $ x \plus{} y \equal{} u \plus{} v \plus{} 1.$ So the integral becomes
\[ \iint_{u^2 \plus{} v^2\le \frac12}u \plus{} v \plus{} 1\,du\,dv\]
As for that integral: $ u$ is an odd function in $ u$ and integrates to zero over that set which is symmetric about the vertical axis. For the same reason, $ v$ integrates to zero. That leaves us with the integral of $ 1,$ which is simply the area of the circle.

The value of the integral is $ \frac {\pi}2.$

Here's another thought. The average value of any linear function (that is, any function of the form $ ax\plus{}bx\plus{}c$) over any circular disk is simply the value of that function at the center of the disk. In this case, $ x\plus{}y$ evaluates to $ 1$ at the center $ \left(\frac12,\frac12\right)$ so that's its average value. To get the integral, multiply the average by the area of the circle over which we're integrating.

Thank you, professor Kent. Your answers are always very useful and educational.

I tried using usual substitution $ x\equal{}r\cos{\alpha}, y\equal{}r\sin{\alpha}$, which gives $ r\leq \cos{\alpha}\plus{}\sin{\alpha}$, so that $ r\in [0,\cos{\alpha}\plus{}\sin{\alpha}]$ and $ \alpha \in [\minus{}\frac{\pi}{4},\frac{3\pi}{4}]$ and after a lot of calcultion
we get $ \int^{\frac{3\pi}{4}}_{\minus{}\frac{\pi}{4}}d\alpha{\int^{\cos{\alpha}\plus{}\sin{\alpha}}_{0}{r^2(\cos{\alpha}\plus{}\sin{\alpha})}}dr\equal{}\frac{\pi}{2}$.

I would have rewritten $ (\cos\alpha\plus{}\sin\alpha)$ as $ \sqrt{2}\cos\left(\alpha\minus{}\frac{\pi}4\right).$

Eventually you're going to wind up with the integral of $ \cos^4$ of something over two full quadrants of that something. At that point, I'll note that

$ \int_0^{\frac{\pi}2}\cos^n\theta\,d\theta\equal{}\frac{n\minus{}1}{n}\int_0^{\frac{\pi}2}\cos^{n\minus{}2}\theta\,d\theta$ so that

$ \int_0^{\frac{\pi}2}\cos^4\theta\,d\theta\equal{}\frac34\int_0^{\frac{\pi}2}\cos^2\theta\,d\theta\equal{}\frac34\cdot\frac12\int_0^{\frac{\pi}2}1\,d\theta\equal{}\frac{3\pi}{8}.$

Multiply that by however many full quadrants you have.


The class was asked to add two positive numbers. John subtracted by mistake and got 2. Maria multiplied and got 120. What was the correct answer?

We get $ p_1 \minus{} p_2 \equal{} 2$, and $ p_1p_2 \equal{} 120$. Common sense (because I don't want to plug in) gives $ 10$, and $ 12$. Since we want the sum, so our answer is $ \boxed{22}$.

[hide=Answer] Let x be the larger positive number and y be the smaller positive number that we are looking for.
We have:
x - y = 2

xy = 120
$ y \equal{} \frac{120}{x}$

Plug it in, we get:

$ x \minus{} \frac{120}{x} \equal{} 2$

$ x^2 \minus{} 120 \equal{} 2x$

$ x^2 \minus{} 2x \minus{} 120 \equal{} o$

(x+10)(x-12) = 0

x = {-10, 12} but because x is a positive number, it must be 12

Plug x in to x - y = 2, we get y = 10

Sum of x and y is 22

[quote="quattayangel"]
$ x^2 \minus{} 2x \minus{} 120 \equal{} o$

(x+10)(x-12) = 0[/quote]

don't post a solution unless it's different.

WE WANT $ x \plus{} y$ WE HAVE $ x \minus{} y \equal{} 2$ $ x > y > 0$ AND $ xy\equal{}120$ square this gives $ x^2 \plus{} y^2 \minus{} 2xy \equal{} 4$ and now since we also know \[ xy\equal{}120 \\ 4xy\equal{}480\]
we add this with the previous.
\[ x^2 \plus{} y^2 \minus{} 2xy \plus{} (4xy) \equal{} 4 \plus{} (480)\\
x^2 \plus{} y^2 \plus{} 2xy \equal{} 484\\
(x \plus{} y)^2 \equal{} 484 \equal{} 22^2\\
x \plus{} y \equal{} 22\](it's positive, as implied in problem)
yay.
of course, latex's solution is faster, but mine is generic :)


function
inequalities
logarithms
limit
absolute value
algebra unsolved
algebra

Let $f:Q^+->Q^+$ be function such that: $f(x+y) \leq f(x)+f(y)$;
$f(xy)=f(x)f(y); f(2000)=1$. Prove that: $f(2004) \leq 1$

This seems to be a very difficult problem, at least to me. It's almost as if the conditions don't really generate much at all! I have some trivial observations, hopefully it will help someone else to solve it:

It's plain to see that f(1)=1. Now 1=f(2000)=(f(2)) 4 *(f(5)) 3 . Hence f(2)<=1 or f(5)<=1.

That's all I can come up with!!! What can we get from f(2)<=1? The inequality condition seems absolutely useless (it's too weak to give any restrictions...) I don't know, hope someone could solve this... :)

I am curious, btw, hxtung, where exactly did you get these interesting problems?

f(2000) = f(1) = f(5) * f(4) = 1

f(2004) =< 1 + f(4) is a trap, doens't lead to a result.

f(5) =< 1 + f(4)
f(5) =< 1 + f(4) + 2.f(4)

f(5) =< f(5) + 1 + 2.f(4)
f(4) >= (f(5)-f(5)+1)/2

perhaps a whole pack of these will help out, but it seems a bit much work at all :)

It's very interesting and tricky problem. OK, the key observation in my solution is [tex]1=f(2000)=f(2)^4\cdot f(5)^3 \Rightarrow f(2)\leq 1 \text{ or } f(5)\leq 1[/tex], still that is obvious. If [tex]f(2)\leq 1[/tex], then take the representation of  interger [tex]n\geq 3[/tex]  in base [tex]2[/tex]. Let [tex]n=2^{k_1}+2^{k_2}+\dots+2^{k_m}[/tex], where [tex]k_i > k_{i+1}[/tex] are natural numbers. Therefore [tex]k_1 \leq \log_2{(n)}[/tex]. We can derive [tex]f(x_1+x_2+\dots+x_m) \leq f(x_1)+f(x_2)+\dots+f(x_m)[/tex] and putting [tex]x_i=2^{k_i}[/tex] we obtain
[tex]f(n)\leq f(2^{k_1}) +f(2^{k_2})+\dots+f(2^{k_1})\leq \sum_{i=0}^{k_1} f(2)^i\leq (\log_2{(n)}+1)\leq 2\cdot \log_2{(n)}[/tex], for all integers [tex]n\geq 3[/tex]. Set [tex]n=n^k[/tex] with [tex]k \in \mathbb{N}[/tex] in the last inequality, [tex]f(n)^k\leq 2k\cdot \log_2{(n)}[/tex] or [tex]f(n)\leq (2k\cdot \log_2{(n)})^\frac{1}{k}[/tex]. Since [tex]\lim_{k\rightarrow \infty} k^\frac{1}{k} = \lim_{k\rightarrow \infty} \text{Const}^\frac{1}{k}=1[/tex], [tex]f(n)\leq 1[/tex]. So [tex]f(2004)\leq 1[/tex].

Similarly if [tex]f(5)\leq 1[/tex], then take [tex]M=\max{\{f(1),f(2),f(3),f(4)\}}[/tex]. So [tex]f(n)\leq 2M\cdot \log_5{(n)}[/tex] and using technique as above, we can establish [tex]f(n)\leq 1[/tex] or [tex]f(2004)\leq 1[/tex].

[quote="Peter VDD"][quote="uTOPi_a"]Similarly if [tex]f(5)\leq 1[/tex], then take [tex]M=\max{\{f(1),f(2),f(3),f(4)\}}[/tex]. So [tex]f(n)\leq 2M\cdot \log_5{(n)}[/tex] and using technique as above, we can establish [tex]f(n)\leq 1[/tex] or [tex]f(2004)\leq 1[/tex].[/quote]

Can you explain this step please? I can't follow... :?[/quote]
Take the representation of [tex]n[/tex] in base [tex]5[/tex]. Then [tex]n=\sum_{i=0}^k a_i\cdot 5^i[/tex], so [tex]f(n)=f(\sum_{i=0}^k a_i\cdot 5^i)\leq\sum_{i=0,\@ a_i\neq 0}f(a_i\cdot 5^i)=\sum_{i=0, \@ a_i\neq 0}f(a_i)\cdot f(5)^i \leq \sum_{i=0}^k M\cdot f(5)^i\leq \sum_{i=0}^k M \leq M(\log_5{(n)}+1)<2M\cdot \log_5{(n)}[/tex].
That [tex]+1[/tex] in the last line is for a good measure.

A digression into higher mathematics:

The only solutions of this functional inequality (ignoring the condition on $f(2000)$) are the absolute values on Q; the ordinary absolute value and the p-adic norms. Either $f(x)=x$ for all $x$, $f(x)=1$ for all $x$, or $f(p^n)=a^n$ for some $a<1$ and $f(q)=1$ if neither the numerator nor denominator of $q$ is divisible by $p$. In particular, if $f(m) \le 1$ for any integer $m>1$, $f(n) \le 1$ for all positive integers $n$.
Also, $f(x)=x^r$ is possible for $r$ a negative integer; this is not an absolute value, but the restriction to $Q^+$ allows it.
If $\log_af(a) \neq \log_bf(b)$ and either $f(a)>1$ or $f(b)>1$ with $a$ and $b$ integers, we can pursue a similar chain of inequalities to those described by uTOPIa and find a contradiction; $f(n)>n$ for some integer $n$. The potential solutions $f(x)=x^r,0<r<1$ are ruled out by the fact that they do not take on only rational values.

The specific values 2000 and 2004 are completely irrelevant to the problem- any two integers would work.

Third instance of this problem I see on this forum.
http://www.mathlinks.ro/Forum/viewtopic.php?t=23869&highlight=1996


Jordan built an octagonal aquarium. Each sheet of clear acrylic was 10 inches wide by 45 inches tall. The aquarium was taller than it was wide. What was the perimeter of the aquarium???

[hide]Since it's taller than it's wide, we take 10 inches for each side. 10*8=80 inches. [/hide]


Show that for any coloring of the edges of a complete graph of $ 11$ vertices with three colors, there exists a monochromatic cycle of length $ 4$

Also, show that if $ 11$ is changed to $ 10$, the above doesn't hold true.

The counterexample for $ 10$ is probably based on a Petersen graph.  This topic should have been placed in one of the Combinatorics fora.


modular arithmetic
quadratics
number theory
relatively prime
prime factorization
Divisor Functions

Show that if $m$ and $n$ are relatively prime positive integers, then $\phi( 5^m -1) \neq 5^{n}-1$.

Assume that $ \phi( 5^m \minus{} 1) \equal{} 5^{n} \minus{} 1$ for some relatively prime positive integers $ m$ and $ n$.

We have $ \gcd(5^m \minus{} 1,5^n \minus{} 1) \equal{} 5^{\gcd(m,n)} \minus{} 1 \equal{} 4 \equal{} 2^2$. Therefore, all odd primes in prime factorization of $ 5^m \minus{} 1$ must come in the first degree.

Suppose that $ 5^m \minus{} 1 \equal{} 2^s p_1 p_2 \dots p_k$ where $ s\geq 2$ and all $ p_i$ are distinct odd primes. Then $ \phi( 5^m \minus{} 1) \equal{} 2^{s \minus{} 1} (p_1 \minus{} 1)(p_2 \minus{} 1)\dots(p_k \minus{} 1)$ is divisible by $ 2^{s \plus{} k \minus{} 1}$. Hence, $ \min\{s,s \plus{} k \minus{} 1\}\leq 2$.
Consider two cases:

1) If $ k \equal{} 0$ then $ s \equal{} 3$, implying that $ 5^m \minus{} 1 \equal{} 8$, a contradiction;

2) If $ k > 0$ then $ s \equal{} 2$ [s]and $ k \equal{} 1$, implying that $ 5^m \minus{} 1 \equal{} 4 p$ and $ \phi( 5^m \minus{} 1) \equal{} 5^{n} \minus{} 1 \equal{} 2(p \minus{} 1)$. Hence, $ 4p \equal{} 5^m \minus{} 1 \equal{} 2\cdot 5^n \plus{} 2$, implying that $ \minus{} 1\equiv 2\pmod{5}$, a contradiction.[/s]

The obtained contradiction proves that $ \phi( 5^m \minus{} 1) \ne 5^{n} \minus{} 1$ for all relatively prime positive integers $ m$ and $ n$.

In the case k>0, I understand s=2, but why k=1? I think ur wrong

We show the contrapositive. Suppose $\gcd(m,n)=1$, and write $5^m-1=2^e\prod_{i=1}^{k}p_i^{e_i}$ where $p_i$ are all odd primes. We have \[5^n-1=\varphi(5^m-1)=2^{e-1}\prod_{i=1}^{k}p_i^{e_i-1}\cdot\prod_{i=1}^{k}(p_i-1). \text{    } (*)\] Thus, $2^e|5^n-1$.  Now, it is well known that $\gcd(5^m-1, 5^n-1)=5^{\gcd(m,n)}-1=4$, and it follows that $e_i=1$ for every $1\le i\le k$, and that $e=2$. Moreover, $2^3|5^j-1$ for all $2|j$, so $2\nmid m$. Thus, write $m=2m^{\prime}+1$.

Now, since $p_i|5(5^{m^{\prime}})^2-1$ for $1\le i\le k$, $5$ must be a quadratic residue in $\mathbf{F}_{p^i}$, and it follows that $p_i \equiv\pm 1\pmod{5}$. However, the above work shows that $5\nmid p_i -1$ for any $1\le i\le k$, which means $p_i\equiv -1\pmod{5}$. Now, consider the prime factorization of $5^m-1$. Reducing this $\pmod{5}$ yields $(-1)^k = 1$, so $k$ is even. However, reducing $(*)$ mod $5$ yields $(-2)^{k+1}\equiv 1\pmod{5}$, which means $k\equiv 3\pmod{4}$, which contradicts $2|k$. We are done.


[quote="13375P34K43V312"]Given three positive real numbers $x$, $y$, $z$, the following inequality holds:
$\frac{z}{x+y}<\frac{x}{y+z}<\frac{y}{x+z}$.
Order $x$, $y$, $z$ from least to greatest.

Given three real positive real numbers $a$, $b$, $c$ such that $abc=1$, prove that at least one of $a-\frac{1}{b}$, $b-\frac{1}{c}$, $c-\frac{1}{a}$ does not exceed 1.

Please tell me if my problem statements are wrong.[/quote]

[hide="1"]$\frac{z}{x+y}<\frac{x}{y+z}\Leftrightarrow-y(z-x)>(x+z)(z-x)$

If $(z-x)$ is positive, then that means $-y>(x+z)$ which obviously isn't true.

So $z-x<0 \Rightarrow x>z$

Apply the same concept to the second inequality to get $z<x<y$[/hide]

[hide="2"]$a-\frac{1}{b}, b-\frac{1}{c}, c-\frac{1}{a}=a(1-c), b(1-a), c(1-b)$

Since $abc\ge 1$ at least one of $(1-a),(1-b),(1-c)$ is $\le 0$ so at least one of $a(1-c), b(1-a), c(1-b)$ is $\le 0$ thus not exceeding 1[/hide]

[quote="13375P34K43V312"]Your solution to #2 seems a little iffy...[/quote]

How? unless the question was stated wrong, which is what I think.

[quote="13375P34K43V312"]Your solution to #2 seems a little iffy...[/quote]

Have in mind this: $a(1-c)=a-ac=a-{1\over b}$ etc.

Edit:  Right.  That case doesn't exist.  :oops:

[quote="t0rajir0u"]... both $a$ and $(1-c)$ are negative.[/quote]

Behold:

[quote="13375P34K43V312"]Given three real [b]positive[/b] real numbers $a$, $b$, $c$ such that $abc=1$...[/quote]

This looks like it can be done by AM-HM, but I haven't come up with a way yet.

[hide="2"]
As SplashD also showed, we can even prove that not all are positive. If all were positive, their sum, $a+b+c-\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$, would be. But also, their product would be positive. We have
\begin{eqnarray*}\left( a-\frac{1}{b}\right) \left( b-\frac{1}{c}\right) \left( c-\frac{1}{a}\right) &=&abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}\\ &=&-a-b-c+\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\\ \end{eqnarray*}
Then we can add the sum and the product and get a positive number. But if we do this, we get 0 [b]OH NO!!!!!!!!!!!!!!!![/b]
[/hide]


search
function
number theory proposed
number theory

Find all $ x,k,l,m\in Z^\plus{}$ such that$ :$
$ (1\plus{}x^k)^l\equal{}1\plus{}x^m$

If $ l\equal{}1$, then $ k\equal{}m$ a nd any x solution. $ x\equal{}0$ solution for any $ l,k,m$.
If $ l>1$, then $ l\plus{}C_l^2x^k\plus{}...\equal{}x^{m\minus{}k}.$ If $ x\equal{}1$, then $ l\equal{}1$, else $ x|l$.
If $ l>2$, then $ p|l$ for any $ p|x$ and $ v_p(l)\equal{}v_p(C_l^2)\plus{}v_p(x)\to x\equal{}2$.
From $ (1\plus{}2^k)^l\equal{}1\plus{}2^m$ we get $ k\plus{}v_2(l)\equal{}m$. But $ kl<m$, therefore only $ x\equal{}2,k\equal{}l\equal{}2,m\equal{}3$ is solution.

[quote="DCTPKTCSPHN"]Find all $ x,k,l,m\in Z^ \plus{}$ such that$ :$
$ (1 \plus{} x^k)^l \equal{} 1 \plus{} x^m$[/quote]
I don't understand what Rust posted . I have some hints :
1. If $ x>1$ and $ x^k \plus{} 1|x^m \plus{} 1$ then we must have $ \frac {m}{k}$ is an odd number . Therefore ,you only need to consider this problem in case $ k \equal{} 1$
2. When k=1 ,it is a problem that were posted before by Zetax. Search function is a good idea ! .

[quote="TTsphn"]1. If $ x > 1$ and $ x^k \plus{} 1|x^m \plus{} 1$ then we must have $ \frac {m}{k}$ is an odd number . Therefore ,you only need to consider this problem in case $ k \equal{} 1$
[/quote]
$ x\equal{}2,k\equal{}2,m\equal{}3$ satysfyed $ x^k\plus{}1|x^m\plus{}1$ but $ \frac{m}{k}$ is not integer.

Hello Rust . Could you please explain me the last two lines of your solution

[quote="Rust"][quote="TTsphn"]1. If $ x > 1$ and $ x^k \plus{} 1|x^m \plus{} 1$ then we must have $ \frac {m}{k}$ is an odd number . Therefore ,you only need to consider this problem in case $ k \equal{} 1$
[/quote]
$ x \equal{} 2,k \equal{} 2,m \equal{} 3$ satysfyed $ x^k \plus{} 1|x^m \plus{} 1$ but $ \frac {m}{k}$ is not integer.[/quote]
Are you sure ? For those numbers , $ 2^k\plus{}1\equal{}5$ and $ 2^m\plus{}1\equal{}9$ .


I have a general question

Are all fractions rational numbers? My teacher told me it is so but I'm sure there is bound to be an irrational number amongst fractions. Is my teacher right?

Rational means that it can be made into a fraction, so pretty much if a number is rational, that means it can be a fraction....

How about $\frac{e}{2}?$ Actually, that's still rational. But how about $\frac{e}{\pi}?$ Actually I think that's rational too :P

ask yourself first what is fraction. now ask yourself what is rational. be very specific and reply my post by posting your definition. then we can talk about what is wrong it :P  because there is a difference between a fraction and rational...

Sorry, what I meant was:

Are ANY and ALL fractions rational numbers. By rational I mean non-repeating and non-terminating decimals.

I will be more than happy to clarify anything else you don't understand. I apologize for my poor english, I'm playing four chess games right now :P

[quote="xdl615x"]Sorry, what I meant was:

Are ANY and ALL fractions rational numbers. [b]By rational I mean non-repeating and non-terminating decimals.[/b]

I will be more than happy to clarify anything else you don't understand. I apologize for my poor english, I'm playing four chess games right now :P[/quote]

i see you learned wrong definitions for BOTH of them and you may want to stop playing chess for a minute

fractions represents division of two expressions. this is very general definition and so it can be tricky

rational numbers are numbers that are either finite or with repeating decimals

it has been proven that repeating decimals numbers can be represented by a fraction whose numerator and denominator are BOTH integers.

is that clear? if not, think about it some more before you go back to gaming...

edit: also one more thing, in school usually teachers refer fractions as division of two integers, so can't blame them cuz it is simpler to say that than saying everything that it is a division of two integers

One can define rational in a number of different ways.  Most usually, rational numbers are defined as [i]ratios of integers[/i] (which is where they get their name).  However, this is the class of things that you are calling "fractions," while you are defining "rational numbers" as those with decimal expansions that eventually repeat or terminate.

I claim that those two classes are equal to each other -- that is, every ratio of integers (except when dividing by 0, of course) has a terminating or repeating decimal expansion while every number with a terminating or repeating decimal expansion is a ratio of integers.

So, you appear to already believe that every number with a terminating or repeating decimal expansion is a ratio of integers.  This is easy to verify by writing out the decimal expansion, multiplying by a large power of 10, subtracting and dividing.

We must convince you that every ratio of integers has a terminating or repeating decimal expansion.  There are a number of ways to show this.  Informally, one might notice that if you multiply a non-repeating decimal by an integer, you get another non-repeating decimal (and thus not an integer).  So, if you claim $\frac{a}{b} = x_0.x_1x_2x_3\ldots$, I can simply point out that the number $b\cdot x_0.x_1x_2x_3\ldots$ has a non-repeating decimal and thus cannot be the integer $a$.
Another method is by examining the process of long division.  Say we are dividing an integer $b$ into another integer $a$.  Note that, at any step of long division, the "left over" (before you carry down) is always some integer between $0$ and $b - 1$ (because that's how long division works).  At some point, since $a$ has a decimal expansion of all 0's, you'll only be carrying down a zero.  Thus, after a certain point, the next digit in the decimal expansion depends only on the remainder from the previous step (and not on $a$ at all any more).  But, since there are only finitely many choices for any given remainder, and that remainder determines the next digit and thus next remainder, we see that a single remainder must determine the entire decimal expansion after it, and that this decimal expansion must eventually repeat when it gets back to a remainder it has already covered (in no more than $b$ steps).  Indeed, if you consider all the fractions that you know of that can be written $\frac{a}{b}$, you'll note that the repeating part always has length less than the size of $b$.
(For example, $\frac{3}{7} = .428571428571\ldots$ has a repeating part of length $6 < 7$, $\frac{19}{17} = 1.11764705882352941176\ldots$ has a repeating part of length $16 < 17$ and $\frac{26}{15} = 1.7333\ldots$ has a repeating part of length $1 < 15$.)

Thanks a lot, that helped me understand.


Be
$\left\{ \begin{array}{} \sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}=\frac{3}{2} \\ x+xy+y=9 \\ \end{array} \right .$

find $x-y$

[hide="Solution"]
Squaring the top equation, we have $\frac{x}{y} - 2 + \frac{y}{x} = \frac{9}{4} \implies \frac{x}{y} + \frac{y}{x} = \frac{17}{4}$.  Substitute $q = \frac{x}{y}$, then we have $q^2 - \frac{17q}{4} + 1 = 0$ and $q = 4, \frac{1}{4}$.  So either $x = 4y$ or $y = 4x$.

Rewriting the bottom equation, we have $(x+1)(y+1) = 10$.  We either have $x = 1, y = 4$ or $x = 4, y = 1$, so $x - y = \boxed{\pm 3}$.[/hide]


trigonometry
calculus
calculus computations

[color=darkblue]Let $ y \equal{} e^x.sinx$. Find $ y^{(2008)}(x)$. From that find $ S \equal{} \sum_{i \equal{} 0}^{2008} C_{2008}^{i} \sin^{(i)}{x}$

[u][b]Note[/b][/u]: I have get $ y^{(2008)}(x) \equal{} ( \minus{} 4)^{502}.e^x.sinx$, because it's very easy.[/color]


modular arithmetic
induction
limit
Miscellaneous Problems

For every natural number $n$, denote $Q(n)$ the sum of the digits in the decimal representation of $n$. Prove that there are infinitely many natural numbers $k$ with $Q(3^{k})>Q(3^{k+1})$.

The problem must be $ Q(3^k)\geq Q(3^{k \plus{} 1})$. In that case, the solution is rather well-known; suppose the statement doesn't happen. Then there exists some $ n_0$ such that for all integers $ n\geq n_0$, $ Q(3^n) < Q(3^{n \plus{} 1})$. But we all know that $ Q(k)\equiv k \pmod {9}$, so basically $ Q(3^n ) \equiv Q(3^{n \plus{} 1}) \pmod 9$, therefore $ Q(3^{n \plus{} 1})\geq 9 \plus{} Q(3^n)$, and by induction we get
\[ Q(3^{n \plus{} k}) \geq 9k \plus{} Q(3^n),
\]
for all positive integers $ k$.

Problem is that if $ k\to\infty$, then $ \lim_{k\to\infty} \frac { 3^{n \plus{} k} }{10^k} \equal{} 0$, so at some point $ k_0$ we have $ 3^{n \plus{} k_0} < 10^{k_0}$, which means that $ 3^{n \plus{} k_0}$ has at most $ k_0$ digits, so $ Q(3^{n \plus{} k_0})\leq 9\cdot k_0$, which is a contradiction.


I just noticed that all of my previous classes (and the transcripts, problem sets, etc) are now listed in My Classes once again. Will it stay this way from now on?

Just this week.  A few students needed some of their old stuff - in general, we really would not like to have to do this, so if you're in a class, please collect the material you'd like to hold on to within a few weeks of the class ending.


Solve of equation:
$sinx \sqrt{\frac{1}{sinx}-1}+cosx\sqrt{\frac{1}{cosx}-1}=\frac{\sqrt{2}}{sinx+cosx}$

Really?
There is not a real number solution than a graph.


Asymptote
MATHCOUNTS
ratio
analytic geometry
geometry
similar triangles

I don't have any Asymptote drawing skills, so I'll just display the problem and not the picture (the picture isn't really needed).

"A $ 500$ foot tall pole is $ 1000$ feet from a $ 400$ foot tall pole.  A taut wire runs from the top of each pole to the ground $ 1000$ feet from the opposite pole.  How high is the point where the wires cross?"

The classic two-pole problem, which I can't seem to be able to do.  Yes, I see the right triangles and similar triangles, but I still can't solve it.  All I could get was the fact that the distance between the points where each pole's wire touches the ground is $ x$, and that the answer to the problem is $ \frac{2000x}{9000\plus{}9x}$.

this shouldn't be in intermediate topics, its a mathcounts-style problem.

If you find ratios from similar triangles, you get 1/500+1/400=1/x

[hide="Hint"]
[img]http://myfiles.zapto.org/sim_triangles.PNG[/img]

You want to find $ n$. Similar triangles will do the trick.[/hide]

Uh... You read the problem incorrectly.  If it was that easy, then I'd have solved it.  The distance from where one pole's wire touches the ground to where the other pole touches the ground is 1000 for either pole.  Here's a link to a picture (this picture is the same as the one given in the original problem):
http://s210.photobucket.com/albums/bb214/OwnageOwnage_2007/?action=view&current=Yesuh.jpg

[quote="MathAndKnowledge"]"A $ 500$ foot tall pole is $ 1000$ feet from a $ 400$ foot tall pole..."[/quote]
Your diagram suggests there is more than $ 1000$ feet between the poles. This statement needs to be clarified.

I don't think your diagram makes sense. How can the distance between the two poles be 1000 feet when there is already 2000 feet in between them?

I would like to point out that the distance between the poles doesn't matter.

[quote="MellowMelon"]Assuming the diagram he posted is what the problem intended, it makes quite a bit of difference how far away the poles are. For instance, if they were $ 2000$ apart, the answer would be $ 0$.[/quote]

the lines will never be flat, so the answer will never be 0.  edit: sorry, you're right.  i was assuming that the diagram was somewhat accurate  :D

but on a contest like this that is probably pretty small, the writers probably didn't pay much attention to little cases like that. They're just probably seeing if you know how to use similar triangles.

[quote="MathAndKnowledge"]
The classic two-pole problem, which I can't seem to be able to do.  Yes, I see the right triangles and similar triangles, but I still can't solve it.  All I could get was the fact that the distance between the points where each pole's wire touches the ground is $ x$, and that the answer to the problem is $ \frac {2000x}{9000 \plus{} 9x}$.[/quote]

wait assuming what you did was correct, didn't you solve the problem?

 you have $ x\equal{}\frac {2000x}{9000 \plus{} 9x}$.

[hide]
The two incorrect diagrams confused me, but I think the intention is something like
[asy]size(400);
pair A=(0,0), B=(1000,0), C=(2000,0), D=(3000,0), E=(2000,400), F=(1000,500);
D(MP("$X$",IP(A--E,F--D),N));
D(MP("$Y$",IP(A--E,F--B),NNE));
D(MP("$Z$",IP(E--C,F--D),NE));
D(MP("$A$",D(A),SW)--MP("$B$",D(B),S)--MP("$C$",D(C),S)--MP("$D$",D(D),S)--MP("$F$",D(F),N)--B);
D(A--MP("$E$",D(E),N)--C);
MP("$1000$",midpoint(A--B),S);
MP("$1000$",midpoint(B--C),S);
MP("$1000$",midpoint(C--D),S);
MP("$500$",midpoint(F--B),W);
MP("$400$",midpoint(E--C),W);[/asy]
I can't figure out how to drop a perpendicular from $ X$ to $ AD$(help please?), but let's say I did and call it $ Y$. Let $ BY \equal{} a$, $ YC \equal{} b$, and $ XY \equal{} x$.
By similar triangles,
\[ \frac {b \plus{} 1000}{2000} \equal{} \frac {x}{500}
\]

\[ \frac {a \plus{} 1000}{2000} \equal{} \frac {x}{400}
\]

\[ a \plus{} b \equal{} 1000
\]
Now just add the first two equations to find $ x \equal{} \boxed{\frac {1000}{3}}$
[/hide]

Note that I am the one who posted the problem.  Also note that I am the one who said that there was picture that CAME WITH the problem.  The link that I posted is almost an EXACT REPLICA of the picture that came with the problem.
I understand that the question is not worded very well, but the picture is undeniably correct since it accompanied the problem (if something was wrong with the picture, the contest organizers would have corrected it after hearing complaints).
Also, if you actually read the problem carefully, the picture is logically correct.  Just look at the picture I gave you and (if you can) give me a solution to this problem.

And to answer unimpossible's question, no, the problem isn't solved because I called the distance between the points where the hypotenuses touch the ground as x, not the height.

Please explain the diagram. The problem says the two given poles are 1000 feet apart. However, the diagram labels two distances that are [i]wholly contained[/i] within this distance but are each 1000 feet long.

[quote="MathAndKnowledge"]Note that I am the one who posted the problem.  Also note that I am the one who said that there was picture that CAME WITH the problem.  The link that I posted is almost an EXACT REPLICA of the picture that came with the problem.
I understand that the question is not worded very well, but the picture is undeniably correct since it accompanied the problem (if something was wrong with the picture, the contest organizers would have corrected it after hearing complaints).
Also, if you actually read the problem carefully, the picture is logically correct.  Just look at the picture I gave you and (if you can) give me a solution to this problem.[/quote]

If that is so, then either
[list]
[*]we're all mis-interpreting the meaning of the first sentence
[*]the question is bad/wrong, because the height is not determined (see MellowMelon's post, for example). [/list]

[quote="MathAndKnowledge"]I guess the answer is not able to be determined.  I was given 5 choices for this problem.  Four of them had numerical answers.  The fifth said "none of these."  It turns out that the fifth answer was correct.
Since "none of these" is correct, I don't know if that's supposed to mean that there is an actual answer other than the other four listed, or if it's impossible to determine such an answer.
All I know is that my picture is correct and that I don't know how to determine the answer.[/quote]

Well, if the picture is right, the problem is obviously wrong. The problem describes something along the lines of my picture :wink:

No it doesn't; risingstar's diagram should be correct.

Ok, look at problem #30 and stop doubting the picture that I showed you guys:
http://www.indianamath.org/Math_Contest/2007%20Math%20Contest/Comprehensive2007.pdf

So the picture is right.  Now stop arguing about it.

EDIT: RisingStar, you are correct.

The yards part is not a typo. Your "correction" made the problem contradict with the diagram. 1000 yards = 3000 ft.

[hide="Revised solution"]
$ \frac {x}{400} \equal{} \frac {a}{1000 \plus{} a \plus{} b}$
$ \frac {x}{500} \equal{} \frac {b}{1000 \plus{} a \plus{} b}$
$ a \plus{} b \equal{} 1000$

Answer: $ x \equal{} \boxed{\frac {1000}{9}}$
[/hide]

As RisingStar said, the problem is as is correct.

Then you have the distance from the two points on the bottom is 100 ft. Similar triangles and what not you have the answer.

[quote="123456789"]The only reason why we were "arguing" over this problem was because you originally put feet instead of yards.[/quote]

If you read my last post that has the EDIT on it, I told RisingStar that he/she is right.  That means I already acknowledged the fact that you so deftly pointed out, 123456789.  But, yes, I apologize that my stupidity made many of the posts here pointless, and I realize that the problem could've been solved a lot faster if I had recognized that the problem [i]didn't[/i] have a typo.  The only reason I assumed there was a typo is because the test was filled with typos (and if you look at the answer key, it acknowledges that one of their answers was incorrect and one of their problems had two different multiple choice answers to it).

Let us end this thread and solve other problems.

[quote="MathAndKnowledge"]Well, the problem was taken from a precalculus level Indiana statewide math test.  That's why I put this problem in this section of the forum; however, I do agree that MathCounts problems are similar to this problem.

I'll try getting that ratio right now.[/quote]

wait what statewide contest was this ICTM?

sorry just wanted to ask...


Let $ a,b\in R$ such that $ f(x)\equal{}x^2\plus{}ax\plus{}b\equal{}0$ have no nonnegative real roots. Prove that there exist two polynomials $ g,h$ with nonnegative real coefficients such that $ f(x)\equal{}\frac{g(x)}{h(x)}$

A lemma :
$ h(x)\equal{}(x\plus{}1)^n$
then for n is large enough then all coefficient  of $ g(x)\equal{}f(x)h(x)$ is negative.
Prove by use quadric polynomial with $ \Delta <0$


When I can assume a>=b>=c

Thank you very much .

when the given inequality is symmetric i.e. if you permute the variables the expression remains unchanged. for example
i. $a^{2}+b^{2}+c^{2}\ge a+b+c$
ii. $a^{2}b+b^{2}c+c^{2}a\ge 3abc$
in i. you can assume that $a\ge b\ge c$ but in ii. you cannot. :wink:

were these of any help?

this is a very useful fact. you can solve quite a number of problems using it. here's a problem from an olympiad which can be solved in two lines using the fact.
Solve in reals:$a+b=c^{3}, b+c=a^{3}, a+b=c^{3}$ :D

Yeah - all this 'arranging' business comes under a general heuristic called the Extremal Principle. This looks at extremal points or configurations and solves problems rather easily - especially when you need to prove existence.
  Andy

[quote="nayel"]when the given inequality is symmetric i.e. if you permute the variables the expression remains unchanged. for example
i. $a^{2}+b^{2}+c^{2}\ge a+b+c$
ii. $a^{2}b+b^{2}c+c^{2}a\ge 3abc$
in i. you can assume that $a\ge b\ge c$ but in ii. you cannot. :wink:

[/quote]

also in  ii you can assume $\ a\geq b\geq c\geq $

[quote="goldbach"][quote="nayel"]when the given inequality is symmetric i.e. if you permute the variables the expression remains unchanged. for example
i. $a^{2}+b^{2}+c^{2}\ge a+b+c$
ii. $a^{2}b+b^{2}c+c^{2}a\ge 3abc$
in i. you can assume that $a\ge b\ge c$ but in ii. you cannot. :wink:

[/quote]

also in  ii you can assume $\ a\geq b\geq c\geq$[/quote]

because its true for all a,b,c? i'm not telling you to find mistakes, i was just trying to help out. besides [b]in proving ii [/b]i think you cannot assume that $a\ge b\ge c$ :wink:

[quote="nayel"]Solve in reals:$a+b=c^{3}, b+c=a^{3}, a+c=b^{3}$ :D[/quote]
You probably meant that problem :wink:


abstract algebra
vector
logarithms
superior algebra
superior algebra unsolved

Let $Aut(G)$ be the number of automorphisms of a group $G$. Is it true that the limit of $Aut(G)$ when the cardinal of $G$ tends to infinity is also infinity?

Well I thought of inner automorphisms but that will not work for abelian groups of course
Have you done it yet for small cases, like abelian groups or p groups?

Abelian groups have the automorphism $x\to x^{-1}$ unless they can be written as a vector space over $\mathbb{Z}_2$.  In that case, take a Hamel basis and permute it.  So I think the answer is 'yes'.

This will be my last post before Christmas.

Merry Christmas, everyone!

Yes, rarely the automorphism group will be trivial, but he asked for going to infinity, we still only have two, so we need more, or am I wrong?

The abelian case is simple by the structure theorem:

Write them as product of cyclic groups of prime-power-orders $n_1,n_2,...,n_m$, each of them with $\phi(n_i)$ automorphisms. Let $s$ be the amount of $C_2$-groups in the decomposition and $u$ the product of the rest. Then by permutating the $s$ groups and using the known automorphisms for the others we get at least $s! \sqrt u$ automorphisms, which clearly grows to $\infty$ with $2^s u$.

Does anyone have anymore ideas about this problem (posted a while back)? I tried for some time, but...no final answer.
I can find an infinite $ G$ s.t. $ Aut(G)$ is finite, or I can show that if $ |G|\equal{}n$, then $ |Aut(G)|\leqslant n^{\log_{2}n}$.

A [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163158]related problem[/url]

That same example would have $ c\cdot n^{\log_{2}(n)}$ automorphisms in the large $ n$ limit, where $ c\equal{}\prod_{k\ge 1}1\minus{}2^{\minus{}k}$.

The original question isn't really clear.


The team is officially announced today (we were advised not to disclose the team until now)

[b]Lin Fei[/b], Don Mills Collegiate Institute, Grade 10, Toronto
[b]Elyot Grant[/b], Cameron Heights Collegiate Institute, Grade 12, Kitchener
[b]Yang (Richard) Peng[/b], Vaughan Road Academy, Grade 11, Toronto
[b]Dong Uk (David) Rhee[/b], McNally High School, Grade 11, Edmonton
[b]Peng Shi[/b], Sir John A. MacDonald Collegiate Institute, Grade 11, Toronto
[b]Yufei Zhao[/b], Don Mills Collegiate Institute, Grade 11, Toronto

Congratulations Yufei ! Now you can win gold at the IMO!!

Grats guys!

BTW, Elyot (not to steal your thunder elyot) got a whole page of the life section of our local newpaper today, and on the front page it says "Whiz kid in top 10 of canada math" showing his picture :)

Did any of you guys also get into the paper?

[quote="mrguozheng"]Congratulations to all the IMO team members and good luck![/quote]

Sorry off topic, but its nice to see more people from alberta on the site, hehehe.

My school's principal made an announcement on last Thursday. The announcement included the fact that I'm in IMO team. How did he figure it out? :?  I only told my family and Dr. Liu, so he couldn't possibly figure it out. :?

Hey Elyot congrats, nice of you to come to waterloo next year esp with that nice 25k, are you making the IOI team also?  heard you are going to / have gone to visit Bush and his dad or something?  tell us about it!!

Hahah... yes the Bush "dumb and dumber" t-shirt is a classic.  I didn't intentionally wear it; I didn't know they were gonna come and do the photo shoot that day.

No IOI for me.  There were 5 gold medal winners this year at the CCC, and they picked the other 4.  5th in Canada isn't bad though.  If IOI had teams of 6 like the IMO does, I'd get to go for sure.  Oh well... I think IMO is much more highly regarded internationally than IOI anyway.

[quote="Elyot"]Oh well... I think IMO is much more highly regarded internationally than IOI anyway.[/quote]
True true. But I think the main reason behind it is that IOI is way younger a competition than IMO.

[quote="weileaf"]Congrat to everyone who made the team. I think this year's Canadian team is stronger than last year. Maybe not strong enough to make top 10, but it will be close  ;)[/quote]

I'd like to see someone on the team tie Jacob's perfect score, lol.

[quote="TheDreamer"]Did someone other than Jacob win a gold medal last year?[/quote]
No, we got 1 gold and 3 bronze last year.


AMC USA(J)MO USAMO probability trigonometry

Would the identity tan(A) + tan(B) + tan(C) = tan(A) * tan(B) * tan(C), given A, B, and C are angles of a triangle, have to be proved if used on the USAMO? If so, can somebody please provide a proof for me? (I think it will involve sum-to-product rules, which I am not especially familiar with) Speaking of which, are sum-to-products (and vice versa) necessary tools for the USAMO?

Sum-to-product rules can be useful, but they are generally easy to prove/derive.  As for formulas, if you can name it, I believe they accept it.  There is a 99.9% chance that they'll know it anyway if you just call it a "well-known theorem."

Although I am not familiar with that I believe that it is well known.

Wow, am I like the most oblivious USAMO qualifier ever?  I had no idea, but thanks!

(I'll just call it a well-known thm if it's actually useful.)

I'm absolutely certain that theorem would not need to be proved to be used on the USAMO.

Thanks for the replies then.

Btw, how would one prove this?

I don't actually know, or at least not off the top of my head, but I think there is a high probability that you would substitute C = 180 - A - B.

Yeah substitute $C$ for $\pi - A - B$ and then use $\tan(A + B) = \frac{\tan{A} + \tan{B}}{1 -\tan{A}\tan{B}}$.  That should do the trick.

I believe you wanted to do tan(A+B) but yeah that does indeed work.


let[tex] x,y, z\succ0[/tex] and [tex] x+y+z=3[/tex]
Prove that:
[tex]\surd x +\surd y +\surd z \ge xy+yz+zx[/tex]

[quote="grobber"]This has been posted before (twice, I think :)). Try searching for it.[/quote]

Indeed. It's here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169

must be a very popular problem, that so many people are posting it :D


function
algebra
domain
floor function
trigonometry
set theory
real analysis

Domain of $ log_{[x\minus{}1]}sin^2(\pi x)$ is $ (3,\infty )$ ~ {$ p,p\plus{}1,p\plus{}2,...n$},where n is multiple of 4 and 25
Find all possible values of p+(p+1)+(p+2)+....+n

Haven't you studied set theory ?

~ means all elements in set (3,infinity) except (p,p+1,p+2,...n)

It is basically called difference of sets

I've usually seen it written with a minus sign or the $ \setminus$ symbol. The tilde isn't standard notation, at least where I'm from. In context, I can at least tell what you meant; nothing else would fit.
Also, did you mean $ \lfloor$ and $ \rfloor$ by those square brackets?

Then again, your question doesn't make sense. The points we know can't be in the domain are those at which $ \sin^2(\pi x)\equal{}0$ - every integer, including those greater than $ n$. All possible values? None, because the domain of a function with that formula can't be of the form specified.

[.] represents greatest integer function

find the domain of the function it will be (3,inf) - {4,5,6,...,n}

So, p=4  and n=100k  as it is a multiple of 4 and 25

So p + p+1 + p+2 + ... + n = (100k-3)(4+100k)/2 = (100k-3)(2+50k) (using the sum of an AP)

[x-1] > 0 => x-1 belongs to [2,infinity)
further [x-1] cannot be 1 so x cannot be [2,3)

further sin^2(pix) cannot be zero => pix is not npi => x is not an integer.

So, the domain is [3,infinity) - {3,4,5,6,...,n}

However I think you have mistyped the question a bit. I had seen this question some days back in a test paper and the question was "the domain obtained is  [3" and not (3 .. So that changes the complete answer

in that case p=3 and you can further proceed

So the required sum = (100k-


geometry
circumcircle
perimeter
inradius
inequalities
trigonometry
Pythagorean Theorem

Two right angled triangles are given, such that the incircle of the first one is equal to the circumcircle of the second one. Let $S$ (respectively $S'$) be the area of the first triangle (respectively of the second triangle).

Prove that $\frac{S}{S'}\geq 3+2\sqrt{2}$.

My solution:
[hide]
It's easy to see that $S' \leq r^2$, that $S=pr$ where $p$ is the half perimeter of the first triangle and that $r=\frac{b+c-a}{2}$.
Hence $\frac{S}{S'}\geq \frac{p}{r} = \frac{a+b+c}{b+c-a}$.
Fix $a$. Then $\frac{a+b+c}{b+c-a}=\frac{a+\sqrt{a^2-c^2}+c}{\sqrt{a^2-c^2}+c-a}$. Now define $f(x)=\frac{a+\sqrt{a^2-x^2}+x}{\sqrt{a^2-x^2}+x-a} (0 < x < a)$. Then $f'(x)$ is of the sign of $x-\sqrt{a^2-x^2}$. Thus $f$ admits a minimum at $x=\frac{a}{\sqrt{2}}$. Then $b=c=\frac{a}{\sqrt{2}}$.
And finally: $\frac{a+b+c}{b+c-a} \geq \frac{a+\sqrt{2}a}{\sqrt{2}a-a}= 3+2\sqrt{2}.$
[/hide]

[quote="Igor"]Two right angled triangles are given, such that the incircle of the first one is equal to the circumcircle of the second one.
Let $S$ (respectively $S'$) be the area of the first triangle (respectively of the second triangle).
Prove that $\displaystyle \frac{S}{S'}\geq 3+2\sqrt{2}$.[/quote]

It seems to be difficult to give a non-bash solution of this problem, but let me try:

Let ABC be the first right-angled triangle, with the right angle at C, and with the sides a = BC, b = CA, c = AB and the angles A = < CAB, B = < ABC, C = < BCA. Let A'B'C' be the second right-angled triangle, with the right angle at C', and with the sides a' = B'C', b' = C'A', c' = A'B' and the angles A' = < C'A'B', B' = < A'B'C', C' = < B'C'A'. According to the conditions of the problem, the inradius r of triangle ABC equals to the circumradius of triangle A'B'C'.

The circumradius of a right-angled triangle equals $\frac12$ of its hypotenuse. Since the triangle A'B'C' is right-angled with the hypotenuse c', its circumradius thus equals $\frac{c^{\prime}}{2}$; on the other hand, we know that its circumradius is r. Thus, $r=\frac{c^{\prime}}{2}$, so that 2r = c'. But by the Pythagorean theorem, applied to the right-angled triangle A'B'C', we have $c^{\prime}\ ^2=a^{\prime}\ ^2+b^{\prime}\ ^2$. On the other hand, the area S' of the right-angled triangle A'B'C' equals $\frac12$ of the product of its catets a' and b'; in other words, $S^{\prime}=\frac12 a^{\prime}b^{\prime}$. By the inequality $2xy\leq x^2+y^2$ for any two real numbers x and y, we thus have

$4S^{\prime}=4\cdot\frac12 a^{\prime}b^{\prime}=2a^{\prime}b^{\prime}\leq a^{\prime}\ ^2+b^{\prime}\ ^2=c^{\prime}\ ^2=\left(2r\right)^2=4r^2$,

so that $S^{\prime}\leq r^2$.

On the other hand, by Lemma 1 from http://www.mathlinks.ro/Forum/viewtopic.php?t=4972 post #19, we have

$S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$,

since S is the area and r is the inradius of triangle ABC. Since C is a right angle, we have $\frac{C}{2}=\frac{90^{\circ}}{2}=45^{\circ}$, so that $\cot\frac{C}{2}=\cot 45^{\circ}=1$, and this equation becomes

$S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}$.

The angles A and B are the two acute angles of the right-angled triangle ABC; thus, they satisfy 0° < A < 90° and 0° < B < 90°. Consequently, $0^{\circ}<\frac{A}{2}<45^{\circ}$ and $0^{\circ}<\frac{B}{2}<45^{\circ}$. Consequently, by the inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=21384 , we have

$\sqrt{\tan\frac{A}{2}\tan\frac{B}{2}}\leq\tan\frac{\frac{A}{2}+\frac{B}{2}}{2}$.

Since the triangle ABC is right-angled with the right angle at C, we have A + B = 90°, so that $\tan\frac{\frac{A}{2}+\frac{B}{2}}{2}=\tan\frac{A+B}{4}=\tan\frac{90^{\circ}}{4}=\tan 22,5^{\circ}=\sqrt2-1$, and thus this becomes

$\sqrt{\tan\frac{A}{2}\tan\frac{B}{2}}\leq \sqrt2-1$.

Since both sides of this inequality are nonnegative, we can square it:

$\tan\frac{A}{2}\tan\frac{B}{2}\leq\left(\sqrt2-1\right)^2$.

Hence,

$S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}=\frac{r^2}{\tan\frac{A}{2}\tan\frac{B}{2}}\geq \frac{r^2}{\left(\sqrt2-1\right)^2}=r^2\cdot\left(3+2\sqrt2\right)$.

Combining this with the inequality $S^{\prime}\leq r^2$ obtained before, we get

$\frac{S}{S^{\prime}}\geq\frac{r^2\cdot\left(3+2\sqrt2\right)}{r^2}=3+2\sqrt2$,

and the problem is solved.

  Darij

This one seems a bit easier:

Let c, and c' be the hypotenuses of the two triangles.
We have $a+b=c+c'$.

We shall prove that:
$p\geq{h'}\cdot(3+2\sqrt{2})$ or
$p^2\geq{{h'}^2}(17+12\sqrt{2})$ (1).
 Since ${h'}^2\leq{{c'}^2}\frac{1}{4}$  (1) becomes:
$(2c+c')^2\geq{(c')^2(17+12\sqrt{2})}$. This reduces(after opening the paranthesis and factorising) to:
$c\geq{c'(1+\sqrt{2})}$ which is true because if not then:
$a+b=c+c'>c\cdot{\sqrt{2}}$ or $2ab>c^2=a^2+b^2$ contradiction!

Hence the inequality is proved!

Let $R'=r=t$, hence $c'=2t$ and $4t^2=a'^2+b'^2\geq 2a'b'=4S' \implies  S'\leq t^2$
$ab=2S=2rs=t(a+b+\sqrt{a^2+b^2})\geq t(2\sqrt{ab}+\sqrt{2ab})$ , we used AM-GM and AM-RMS.
So $\sqrt{ab}\geq t(2+\sqrt{2}) \implies S\geq t^2(1+\sqrt2)^2$
Combining both we have $\displaystyle\frac{S}{S'}\geq (1+\sqrt2)^2=3+2\sqrt{2}$
Equality iff both triangles are isoceles.


In the following triangular lattice distance of two vertices is length of the shortest path between them. Let $ A_{1},A_{2},\dots,A_{n}$ be constant vertices of the lattice. We want to find a vertex in the lattice whose sum of distances from vertices is minimum. We start from an arbitrary vertex. At each step we check all six neighbors and if sum of distances from vertices of one of the neighbors is less than sum of distances from vertices at the moment we go to that neighbor. If we have more than one choice we choose arbitrarily, as seen in the attached picture.

Obviusly the algorithm finishes
a) Prove that when we can not make any move we have reached to the problem's answer.
b) Does this algorithm reach to answer for each connected graph?

(a) Let's assign to each lattice point $P$ the number $f(P)$ which is the sum of the distances from $P$ to the $n$ constant vertices.

We will use a few lemmas to solve this problem.

[b] Lemma 1. [/b] If $ABCD$ is a rhombus of side length $1$, where $A, B, C, D$ are lattice points so that $\angle BAC = \angle BDC = 60$, then we have $f(A) + f(C) \ge f(B) + f(D).$

[b] Proof [/b]. It suffices to show this when there is only one $A_i$, then sum over all $A_i$'s. This is not hard to show by casework on where the $A_i$ is.

$\blacksquare$

[b] Lemma 2. [/b] If $A, B, C$ are three collinear lattice points so that $AB = BC = 1$, then $f(A) + f(C) \ge 2 f(B).$

[b] Proof. [/b] This is proven in the same way as Lemma $1.$

$\blacksquare$

[b] Lemma 3. [/b] If $ABCD$ is a parallelogram so that $A, B, C, D$ are lattice points, its sides are parallel to the gridlines of the triangular lattice, and $\angle BAC = 60$, then we have $f(A) + f(C) \ge f(B) + f(D).$

[b] Proof. [/b] Sum Lemma $1$ over all "sub-rhombi" of $ABCD$ which have parallel sides to $ABCD$ and are of length $1.$All interior points and lattice points on the perimeter of $ABCD$ appear an equal number of times on either side of the $\ge$ sign, except for $A, C$ which appear once each to the left of the $\ge$ sign, and $B, D$ which appear once each to the right of the $\ge $sign. This implies the lemma.

$\blacksquare$

[b] Lemma 4. [/b] If $A, B, C, D$ are four lattice points so that $AD = t, AB = 1, BC = t-1, CD = t-1$ for some $t \ge 3$ and $AD$ lies entirely on some gridline, then $f(A) + f(D) \ge f(B) + f(C).$

[b] Proof. [/b] Our strategy will be to use Lemma $2$ repeatedly. Let's label the lattice points on $AD$ $P_0, P_1, \cdots, P_t$ in the obvious manner, with $A = P_0$ and $D = P_t.$ Notice that by Lemma $2$ we have:

$$f(P_0) - f(P_1) \ge f(P_1) - f(P_2) \ge f(P_2) - f(P_3) \ge \cdots \ge f(P_{t-1}) - f(P_t).$$

This implies the lemma.

$\blacksquare$

From here, we're ready to solve the problem.

Suppose that our process ended at $A$, where $f(A)$ is less than or equal to $f$ of all its neighbors. Assume that $A$ is not the point where $f(A)$ is minimized, for contradiction. Let $B$ be the point closest to $A$ such that $f(B)$ (where "closeness" is measured in any sensible manner).

We consider two cases.

[b] Case 1. [/b] $AB$ lies on a gridline.

Then, since $AB$ is clearly not $1$, we can consider points $C, D$ on segment $AB$ so that $AC = DB = 1.$ Then Lemma $4$ implies that $2 f(A) > f(A) + f(B) \ge f(C) + f(D)$, and so one of $f(C), f(D) < f(A).$ But $C, D$ are both closer to $A$ than $B$, contradiction.

[b] Case 2. [/b] $AB$ doesn't lie on a gridline.

Then, there are points $C, D$ so that $ACBD$ is a parallelogram with $\angle CAD = \angle CBD = 60.$ Lemma $3$ then implies that $f(C) + f(D) \le f(A) + f(B) < 2f(A)$, so one of $f(C), f(D) < f(A).$ But $C, D$ are both closer to $A$ than $B$, contradiction.

As we've derived a contradiction in all cases, the problem is finished.

$\square$

(b) No, we can construct a counterexample as follows.

Our vertices will be $P_1, P_2, A_1, A_2, A_3, A_4, A_5, B_1, B_2, B_3, B_4, B_5.$

Start by connecting $P_2$ to all the $B_i$'s and $P_1$ to all the $A_i$'s. Now, for each $1 \le i \le 5$, connect $A_i$ to $B_i.$

Let $B_1, B_2, \cdots, B_5$ be the five constant points. Define $f$ similarly as in part (a). Then we have that $f(P_1) = 10$, $f(P_2) = 5$, $f(A_i) = 13$, and $f(B_i) = 8$ for all $1 \le i \le 5.$ If we begin our algorithm at $P_1$, then since all neighbors of $P_1$ have larger $f$ value (they all have $13$), our algorithm will tell us that $A$ is the global minimum of $f.$ However, that's actually $B$, which attains $5.$

$\square$


Could someone please order the following nuclear emission types in order from least penetrating to most penetrating?  Thanks in advance!

Alpha, Beta, Gamma, Neutron, Proton, Positron

First of all we all know that $ \gamma > \mbox{n} > \beta > \alpha$.

This is what I believe to be correct:

-Gamma = highest
-Neutron
-Beta
-Positron (Not completely sure)
-Proton
-Alpha = lowest

(I am not sure if my placement of a proton is correct; proton emission from radioisotopes is decidedly, rare. My guess is that the proton will interact with matter in a way similar to that of an alpha particle.)

I think beta and positron would be equal, and I think they are ahead of neutron and proton which are relatively equal.  This is probably wrong though.


1. Let $ a_1,a_2...a_n$ be positive with a sum of $ 1$. Show that $ \sum_{i\equal{}1}^n a_i^2 \ge \frac{1}{n}$

2. Let $ x,y,z > 0$ with $ xyz \equal{} 1$. Prove that $ x\plus{}y\plus{}z \le x^2\plus{}y^2\plus{}z^2$

[quote="Kono"]1. Let $ a_1,a_2...a_n$ be positive with a sum of $ 1$. Show that $ \sum_{i \equal{} 1}^n a_i^2 \ge \frac {1}{n}$

2. Let $ x,y,z > 0$ with $ xyz \equal{} 1$. Prove that $ x \plus{} y \plus{} z \le x^2 \plus{} y^2 \plus{} z^2$[/quote]

1. By Cauchy-Schwarz inequality , $ (\sum_{i \equal{} 1}^n 1)(\sum_{i \equal{} 1}^n a_i^2) \ge (\sum_{i \equal{} 1}^n a_i)^2\equal{}1$
$ \Rightarrow \sum_{i \equal{} 1}^n a_i^2 \ge \frac{1}{n}$

2. $ (x\plus{}y\plus{}z\minus{}3)^2 \ge 0$
$ \Rightarrow 6(x\plus{}y\plus{}z) \le (x^2\plus{}y^2\plus{}z^2)\plus{}2(xy\plus{}yz\plus{}zx)\plus{}9(xyz)^{\frac{2}{3}} \le 6(x^2\plus{}y^2\plus{}z^2)$.
So $ x\plus{}y\plus{}z \le x^2\plus{}y^2\plus{}z^2$

Thanks for that!

For Q 2 how did you think of using $ (x \plus{} y \plus{} z \minus{} 3)^2 \ge 0$? (That's like a needle in a haystack for me  :rotfl: )

Are there any other ways?

equality holds when $ x \equal{} y \equal{} z \equal{} 1$ , then think of an expression to put $ x^2, y^2, z^2$ and $ x,y,z$ together.

Another way:

$ (x \minus{} 1)^2 \plus{} (y \minus{} 1)^2 \plus{} (z \minus{} 1)^2 \ge 0 \Rightarrow x^2 \plus{} y^2 \plus{} z^2 \minus{} x \minus{} y \minus{} z \ge x \plus{} y \plus{} z \minus{} 3 \ge 0$

[quote="Kono"]

2. Let $ x,y,z > 0$ with $ xyz \equal{} 1$. Prove that $ x \plus{} y \plus{} z \le x^2 \plus{} y^2 \plus{} z^2$[/quote]

We have that

$ x^2\plus{}y^2\plus{}z^2\geq \frac{1}{3}\left(x\plus{}y\plus{}z\right)^{2}\equal{}\frac{1}{3}(x\plus{}y\plus{}z)(x\plus{}y\plus{}z)\geq \frac{1}{3}\cdot 3\sqrt[3]{xyz}(x\plus{}y\plus{}z)\equal{}x\plus{}y\plus{}z$


linear algebra
matrix
linear algebra theorems

Let $ A$ be a matrix.Study how the inverse of A ($ A^{\minus{}1}$) changes if we make the following changes on $ A$:

a)we interchange two rows
b)we multiply all elements of a row with a nonzero constant $ c$
c)we add at the elements of a row the corresponfing elements of another row,each of these element multiplied by a constant c.

I am intersted not in the answers(they are almost obvious),but the simplest proofs of them.

That depends on what you mean by "simplest proof."  I would use $ (AB)^{\minus{}1} \equal{} B^{\minus{}1} A^{\minus{}1}$ and the explicit form of the elementary matrices, but some might prefer a less explicitly matrical proof.


function
number theory unsolved
number theory

By considering the $ n$ fractions $ \frac{1}{n}, \frac{2}{n}, \ldots, \frac{n}{n}$, or otherwise, prove that $ n \equal{} \sum_{d|n} \phi (d)$.

Any hints to point me in the right direction?

Reduce the fractions $ \frac{k}{n}$ that you have described. How often does $ d$ show up as a denominator?

It shows up exactly $ m$ times, where the number $ n$ has $ m$ factors? But I can't link that to the phi functions of the various $ d$.  :maybe:

Edit: The above is a mistake, checked only small values of $ n$. Will get back to that soon  :blush:

For a divisor of n like d, define $ S_{d}$ to be the set of all positive integeres $ a\in[1,n]$ such that gcd(a,n)=n/d. Then $ S_{d}$ consists of all elements of the form b.n/d where b is a positive integer $ \in[1,d]$ and gcd(b,d)=1 So $ S_{d}$ contains $ \phi(d)$ elements. Every integer belongs to a unique set and so the identity follows.

You wanted a hint so how about:
$ gcd(n,d)\equal{}1\Leftrightarrow gcd(n,n\minus{}d)\equal{}1$.


function
algebra
functional equation
algebra solved

Provided the functional equation f: R->R which satisfies x^2 * f(x) + f(1-x) = 2x-4. Determine f(x).

[i]German IMO Selection Test 1982, problem 3.[/i]

Change x in 1-x we have a system of two equations with
unknown f(x) and f(1-x)

x^2 * f(x) + f(1-x) = 2x-4
(1-x)^2*f(1-x) + f(x) = -2-2x

That's wrong. Please plug in your function in the condition. So you will see that your answer is wrong. Have another try !  :)

I read it :

x^2*f(x) + f(1 - x) = 2x - x^4

like in http://65.36.154.75/Forum/viewtopic.php?t=2015

but here it's x^2*f(x) + f(1 - x) = 2x - 4.

I find f(x) = 2(x^3-4x+6x-1)/(x^4-2x^3+x-1) but I don't have the courage to check it  :(

I prefer x^2*f(x) + f(1 - x) = 2x - x^4   :D

[quote="belenos"]I find f(x) = 2(x^3-4x+6x-1)/(x^4-2x^3+x-1)
I prefer x^2*f(x) + f(1 - x) = 2x - x^4   :D[/quote]

I guess your formula is the solution to the system Moubinool proposed.
If you look downstairs you will see that
(x^4-2x^3+x-1)=(x 2- x+1)(x 2 -x-1). This function has real 0s, so the problem is not over yet.

Math and generally life have a funny way of forcing you to face up with problems you wouldn't want to.

I don't know (who would like to check that !) if 2(x-4x+6x-1)/(x^4-2x^3+x-1) is a solution to the equation .... and I don't want to know  :D

I think your solution is almost right: f(x) = 2 - (x^3-4x^2+6x-1)/(x^4-2x^3+x^2-1)


Given a polynomial with integer coefficient f(x), for natural numbers a,b,if f(a) and f(b) is coprime,then a and b is coprime.Prove that f(0)=0,or exists positive integer d>1,d|f(n) for all positive integer n.

In mech-mat forum MSU I proved:
$Let \ P(x) \in Q[x] \ satisfyed \ conditions:$
$1) \ for \ big \ primes \ p \ and \ q \ (P(p),P(q))=1,$
$2) \ exist \ two \ rational \ numbers \ r1 \ and \ r2: \ (r1,r2)=1, \ P(r1) \ and \ P(r2) \ are \ whole,$
$then \ P(x)=\pm x^n.$

Your problem's, because your conditions stronger.


I got a 113 :(  :(  Maybe I'll be in the top 1%! :?

I got a 150, unless I get disqualified. I'm worried that I forgot to sign that it was my own work. That said, it seemed easier than in the past. did anybody else get that impression

I did seem a little easier - I got a 74 last year.

If you don't mind, I'm going to delete this as there's another thread on the same topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25569]here[/url].

This one [i]was[/i] the first one, but if you could just have MCrawford's I think rhat's fine.

what do you all think the qualifying score will be? I got a 112 cause of 4!!! silly mistakes!!! :mad:

I got a 133.5 due to a stupid mistake on #22 (LCM (4,6) = 24 (i thought)) :blush: . Oh well, I am generally happy with what I got.


[quote="FactorMagic"]This problem should be easy, but for some reason in not solving it correctly
Problem:
3x^3-9x^2-27x=15
my solution
3x^3-9x^2-27x-15=0
x=5
3x^2+6x+3=0
(3x+3)(x+1)
im getting x=-1 x=-1

Where did i go wrong? my factoring seems ok, my sythetic division seems ok, im new at algebra, and i am teaching myself. Not the easiest way to learn algebra i tell you.
Thanks.[/quote]
It doesn't look like you did anything wrong.  :wink: I looked through your steps and didn't find any errors.
By the way, you should try to type math symbols in $\LaTeX$. When you use $\LaTeX$ you can produce nice math expressions like this:
$3x^{3}-9x^{2}-27x-15=0$
For a tutorial on how to type in $\LaTeX$ you can look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261 .
There's a .pdf tutorial file on the first post.

[quote="nebula42"][quote="FactorMagic"]I will look into that, the answer key states thats
x=-1 x=5
I guess the book may be wrong?[/quote]
But $x=5$, $x=-1$ is exactly what you got.  :wink:  :huh:[/quote]

ok, please explain, how is that the same as my answer?
i get it, nevermind. thanks!!!

Are you worried about the double root at $-1$? Even though you get 2 of the same root, it would still be listed as only one answer. I'm not very good at explaining things.  :roll:

well thats one half of it, it confused me getting the double root, and i also forgot about the 5 that i solved for at the beginning of the equation.

The double $-1$ just means that $(x+1)^{2}$ is a factor of the original polynomial.


search
function
absolute value
complex numbers

Let $f(x) = |x - p| + |x - 15| + |x - p - 15|$, where $0 < p < 15$.  Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \le x \le 15$.

This has already been posted before, in a marathon, I believe. :)

[hide="Solution"]It is best to get rid of the absolute value first.

Under the given circumstances, we notice that $|x-p|=x-p$, $|x-15|=15-x$, and $|x-p-15|=15+p-x$.

Adding these together, we find that the sum is equal to $30-x$, of which the minimum value is attained when $x=15$.

The answer is thus $15$. [/hide]

Ok. I can't use the search function right now because my school computer is slow.

Can you confirm that for me? Thanks.

I posted it in the "Speed" topic in this forum a few days ago. Your answer is correct.

Ok, since it has already been posted, this topic is locked.  ;)


modular arithmetic
number theory unsolved
number theory

Determine all pair of positive integers $ (m,n)$ such that $ m$ is divisible by every integer from $ 1$ to $ n$(inclusive) but not divisible by $ n\plus{}1,n\plus{}2,n\plus{}3$

[quote="math10"]Determine all pair of positive integers $ (m,n)$ such that $ m$ is divisible by every integer from $ 1$ to $ n$(inclusive) but not divisible by $ n \plus{} 1,n \plus{} 2,n \plus{} 3$[/quote]

For $ i \in \{1,2,3 \}$ and $ n \ge 3$,

if $ n\plus{}i$ is not a power of prime, we have $ n\plus{}i\equal{}pq$ , where $ p,q > 1$ are coprime .
Then $ p,q \in \{1,2,...,n \} \Rightarrow n!|pq\equal{}n\plus{}i$

So $ n\plus{}i$ are all powers of primes.

[b]Case 1:[/b] $ n$ is odd, then $ n\plus{}1\equal{}2^x, n\plus{}3\equal{}2^y$
$ \Rightarrow 1\plus{}\frac{2}{n\plus{}1}\equal{}\frac{n\plus{}3}{n\plus{}1}\equal{}2^{y\minus{}x}$ is integer
$ \Rightarrow n\equal{}1$, contradiction.

[b]Case 2:[/b] $ n$ even, then $ n\plus{}2\equal{}2^x$.

For $ x\equal{}2k$ is even, $ n\plus{}1\equal{}2^{2k}\minus{}1\equal{}(2^k\plus{}1)(2^k\minus{}1)$ is power of prime
$ \Rightarrow k\equal{}1 \Rightarrow n\equal{}2$, contradiction.

For $ x\equal{}2k\plus{}1$ is odd, $ n\plus{}3\equal{}2^{2k\plus{}1}\plus{}1 \equiv 0 \pmod{3}$

$ \Rightarrow 2^x\plus{}1\equal{}3^y$ for some $ y$. mod 4 we get $ (x,y)\equal{}(1,1),(3,2)$ is the only solution for it

$ \Rightarrow (m,n)\equal{}(60k,6)$, where $ k$ is not divisible by $ 7,8,9$ as the only solution for this case.

For $ n\equal{}1$, $ (m,n)\equal{}(6k\plus{}1,1),(6k\minus{}1,1)$ , where $ k$ is integer, are the solutions.

For $ n\equal{}2$, $ (m,n)\equal{}(2k,1)$, where $ k$ is not divisible by $ 3,4,5$, are the solutions.


[quote="mr.danh"]Caculus:
$ P \equal{} \prod_{i \equal{} 1}^{m \minus{} 1}sin\frac {i\pi}{2m}$[/quote]
Posted before many times.

Could you give me solution of this problem ? If you do, thanks

One of methods is: Let \[ f(x)\equal{}1\plus{}x\plus{}...\plus{}x^{2m\minus{}1}\equal{}\frac{x^{2m}\minus{}1}{x\minus{}1}\equal{}\prod_{k\equal{}1}^{2m\minus{}1}(x\minus{}e_m^k),\] were $ e_m\equal{}exp(\frac{\pi i}{m})$. Then \[ f(1)\equal{}2\prod_{k\equal{}1}^{m\minus{}1}|1\minus{}e_m^k|^2\], because $ |1\minus{}e_m^k|\equal{}|1\minus{}cos\frac{\pi k}{m}\minus{}isin\frac{\pi k}{m}|\equal{}2sin\frac{k\pi }{2m}, 1\le k<m$ we get $ 2m\equal{}f(1)\equal{}2^{2m\minus{}1}P^2$ or $ P\equal{}\sqrt m 2^{\minus{}(m\minus{}1)}$.

Thanks for your help. It is dificult to express my gratitude.


[url=http://www.theonion.com/news/index.php?issue=4124&n=1]a funny article[/url]

(: I find this website to be highly amusing.

Seriously, though, it's true that we (Americans) do buy/use/waste too much stuff.

That has bad language. You should put a warnning there.

I found the fake article kind of disgusting because it really is the truth. There is so much junk.

It was not funny to me. Just because something uses the s-word or just sounds stupid, does not make it funny.

[quote="CLee"][quote="IntrepidMath"]That has bad language. You should put a warnning there.[/quote]

At the bottom of the page, it says "The Onion is not intended for readers under 18 years of age."  That explains the bad language...  It does need to have a warning, though.[/quote]

I didn't think it was funny because of the s-words (does that really offend you??) I thought it was funny because it was making fun of American waste, etc. the curse words aren't meant to be taken seriously.

[quote="Drunken_Math"]I found it really funny since I'm Chinese!  :lol:[/quote]

me too :D

[quote="The Onion®"]The Onion® is not intended for readers under 18 years of age.[/quote] putting it at the bottom is just like saying "The previous may contain expicit content and should not have been veiwed by those under 18"
 :rotfl:

How many times will I use a taco shell holder?
I'm vegetarian. How many times will I eat a taco?

Wow. They make alot of junk.

[quote="Palytoxin"]right...so, are you guys going to buy more s***? i liked the article, whether it's fake or not, it's believable, and i think this s*** is sort of epidemic considering that all this could have helped out a lot to a poor children with just $\textdollar$1 instead of buying a crap. it's going to help Earth too because there will be less s*** trash :)[/quote]

I loved this article. because I"m chinese, and I hate wasting. :D

[quote="juicybooty911"]haha, that's how my dad talks :rotfl:[/quote]

[quote="the onion"]

How many times will you use a taco-shell holder? 'Oh, I really need this silverware-drawer sorter or I will have fits.' Sh-t up, stup-d American.

[/quote]

ehh, I had to dash the 'curse words' so people wouldn't get offended. but I LOVE THIS QUOTE. your dad seems really cool :D


geometry
circumcircle
geometric transformation
reflection
ratio
conics
ellipse

The tangents at $ B$ and $ A$ to the circumcircle of acute triangle $ ABC$ meet the tangent at $ C$ at $ T$ and $ U$ respectively. Let $ P\equal{}AT\cap BC,R\equal{}BU\cap CA$. Let $ Q,S$ be the midpoints of $ AP,BR$ respectively. Prove that $ \angle ABQ\equal{}\angle BAS$.

This is [b]IMO shortlist 2000. G5 problem[/b] Solution can be found from IMO Compendium book.

Notation is highly inconvenient, changing it: Let $ U \equiv t_A \cap t_B,$ $ V \equiv t_C \cap t_A,$ $ W \equiv t_A \cap t_B$ be pairwise intersections of the circumcircle tangents at $ A, B, C$ and $ P \in BC, Q \in CA, R \in AB$ feet of the symmedians $ AU, BV, CW$ concurrent at the symmedian point $ K.$ Let $ M, N$ be midpoints of $ AP, BQ.$ Show that $ \angle ABM \equal{} \angle BAN.$
______________________________

Let $ C_1, K_1, P_1$ be reflections of the $ C, K, P$ in the perpendicular bisector of the segment $ AB.$ By a central projection, project the triangle circumcircle $ (O)$ to a circle $ (O')$ and a line $ w \parallel AB$ through $ W$ to infinity. Symmedians and the symmedian point $ K$ of the $ \triangle ABC$ go to the symmedians and symmedian point $ K'$ of the projected $ \triangle A'B'C'.$ All parallels to $ w$ remain parallel without changing any segment ratios on these lines, in particular, $ C_1', K_1', P_1'$ remain reflections of $ C', K', P'$ in the perpendicular bisector of $ A'B'.$ Tangents $ t_A', t_B'$ at $ A', B'$ become parallel to each other $ \Longrightarrow$ the projected triangle is right at the vertex $ C'$ and the C-symmedian identical with its C'-altitude. The symmedian point $ K'$ divides the C'-symmedian $ C'R'$ in the ratio

$ \frac {C'K'}{K'R'} \equal{} \frac {a'^2 \plus{} b'^2}{c'^2} \equal{} 1$

(by Pythagorean theorem), hence $ K'$ is the midpoint of the C'-altitude. Let a perpendicular to $ A'B'$ through $ P'$ cut $ A'B', A'C', (O')$ at $ X', Y', Z'.$ From similarity of the right $ \triangle A'X'Y' \sim \triangle P'X'B',$

$ X'Y' \cdot X'P' \equal{} XA' \cdot X'B' \equal{} X'Z'^2$

But $ \frac {P'X'}{Y'X'} \equal{} \frac {K'R'}{C'R'} \equal{} \frac {1}{2},$ hence $ \frac {P'X'}{Z'X'} \equal{} \sqrt {\frac {P'X'}{Y'X'}} \equal{} \frac {1}{\sqrt {2}}.$ Thus the A'-symmedian foot $ P'$ is on an ellipse $ \mathcal E$ with the main axis $ A'B',$ pedal circle $ (O')$ and minor/major axis ratio $ \frac {1}{\sqrt {2}}$ and similarly, the B'-symmedian foot $ Q' \in \mathcal E.$

Let $ K'K_1' \parallel A'B'$ cut $ t_A'$ at $ D'$ and let $ C'C_1' \parallel A'B'$ cut $ t_B'$ at $ E'.$ Using parallel projection, project the ellipse $ \mathcal E$ to the circle $ (O').$ The line $ A'B'$ remains in place, the symmedian feet images $ P^*, P_1^*, Q^*$ go the circle $ (O')$ and the vertex image $ C^*$ leaves it. In effect, the $ \triangle A'B'C^*$ is no longer right at $ C^*,$ the symmedian images $ A'P^*, B'Q^*$ are its altitudes, and the symmedian point image $ K^*$ its orthocenter. The ratio $ \frac{K^*R'}{C^*R'} \equal{} \frac{K'R'}{C'R'} \equal{} \frac{1}{2}$ is preserved.  Let $ D^*K^*K_1^*, C^*C_1^*E^* \parallel A'B'$ be images of $ D'K'K_1', C'C_1'E' \parallel A'B'$ in this parallel projection.

The quadrilateral $ A'K_1^*P_1^*D^*$ is cyclic on account of the right angles $ \angle A'D^*K_1^*, \angle A'P_1^*K_1^*$ and similarly, the quadrilateral $ A'K^*Q^*D^*$ is cyclic on account of the right angles $ \angle A'D^*K^*, \angle A'Q^*K^*.$ From their circumcircles and from the isosceles trapezoid $ A'B'K_1^*K^*$ then follows

$ \angle D^*Q^*A' \equal{} \angle D^*K^*A' \equal{} \angle P_1^*K_1^*D^* \equal{} \angle P_1^*A'D^*$

$ \angle D^*A'Q^* \equal{} \angle D^*K^*Q^* \equal{} \angle A'K_1^*D^* \equal{} \angle A'P_1^*D^*$

which means that the $ \triangle D^*A'Q^* \sim \triangle D^*P_1^*A'$ are oppositely similar and the points $ D^*, Q^*, P_1^*$ collinear. Similarly, the quadrilaterals $ B'P_1^*C_1^*E^*, B'Q^*C^*E^*$ are also cyclic on account of right angles. In exactly the same way, from their circumcircles and from the isosceles trapezoid $ A'B'C_1^*C^*$ then follows that the $ \triangle E^*B'Q^* \sim \triangle E^*P_1^*B'$ are oppositely similar and the points $ E^*, Q^*, P_1^*$ collinear. If the line $ E^*Q^*P_1^*D^*$ cuts the line $ A'B'$ at a point $ F',$ then $ K^*K_1^*D^* \parallel A'B', A'D^* \perp A'B'$ are midlines of the right $ \triangle B'E^*F'$ on account of $ K^*$ being the midpoint of $ C^*R'$ and $ F'A' \equal{} A'B'.$

Returning to the original configuration before the two projections, the line $ QP_1$ cuts $ AB$ at $ F,$ such that $ FA \equal{} AB.$ If $ M_1$ is reflection of the symmedian $ AP$ midpoint $ M$ in the perpendicular bisector of $ AB,$ then $ NM_1 \parallel QP_1$ cuts $ FB$ at its midpoint $ A.$ As a result, $ \angle ABM \equal{} \angle BAM_1 \equiv \angle BAN.$

Solving by calculation: With the changed notation, let $ a = BC, b = CA, c = AB$ and let $ A_1, B_1$ be midpoints of $ BC, CA.$ $ m_a = AA_1, m_b = BB_1$ are medians and $ k_a = AP, k_b = BQ$ the corresponding symmedians. Using their isogonality WRT the angles $ \angle A, \angle B,$

$ \frac {k_a}{m_a} = \frac {AP}{AA_1} = \sqrt {\frac {PB \cdot PC}{A_1B \cdot A_1C}} = \frac {2bc}{b^2 + c^2}$

$ \frac {k_b}{m_b} = \frac {BQ}{BB_1} = \sqrt {\frac {QC \cdot QA}{B_1C \cdot B_1A}} = \frac {2ca}{c^2 + a^2}$

$ M, N$ are midpoints of $ AP, BQ$ $ \Longrightarrow$ $ BM, AN$ are medians of the $ \triangle ABP, \triangle ABQ,$

$ \frac {BM}{AN} = \sqrt {\frac {2BP^2 + 2AB^2 - AP^2}{2AQ^2 + 2AB^2 - BQ^2}}$

Substituting $ BP = \frac {ac^2}{b^2 + c^2},$ $ AQ = \frac {bc^2}{c^2 + a^2},$ ${ 4m_a^2 = 2b^2 + 2c^2 - a^2},$ ${ 4m_b^2 = 2c^2 + 2a^2 - b^2},$

$ \frac {BM}{AN} = \frac {c^2 + a^2}{b^2 + c^2} \sqrt {\frac {2c^2a^2 + 2(b^2 + c^2)^2 - b^2(2b^2 + 2c^2 - a^2)}{2b^2c^2 + 2(c^2 + a^2)^2 - a^2(2c^2 + 2a^2 - b^2)}} = \frac {c^2 + a^2}{b^2 + c^2}$

By repeated use of the sine theorem for the $ \triangle ABM, \triangle ABN, \triangle ABP, \triangle ABQ, \triangle ABC,$

$ \frac {\sin \widehat{ABM}}{\sin \widehat{BAN}} = \frac {AM}{BM} \cdot \frac {AN}{BN} \cdot \frac {\sin \widehat{BAP}}{\sin \widehat{ABQ}} = \frac {AP}{BM} \cdot \frac {AN}{BQ} \cdot \frac {BP}{AP} \cdot \frac {BQ}{AQ} \cdot \frac {\sin B}{\sin A} =$

$ = \frac {AN}{BM} \cdot \frac {BP}{AQ} \cdot \frac {CA}{BC} = \frac {b^2 + c^2}{c^2 + a^2} \cdot \frac {ac^2}{b^2 + c^2} \cdot \frac {c^2 + a^2}{bc^2} \cdot \frac {b}{a} = 1$

$ \Longrightarrow$ $ \angle ABM = \angle BAN$ (both acute).

See also [url]http://www.mathlinks.ro/viewtopic.php?t=19806[/url].

It is easy to prove this nice problem, as a particular case of the below well known and very simple [b][size=100]Lemma[/size][/b].

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup DEF,$ the Cevian triangle of an arbitrary point $ P$ inwardly to it. We draw two lines through the point $ D$ and parallel to $ BE,\ CF,$ which intersect the side-segments $ AC,\ AB$ respectively, at points $ E',\ F'.$ Prove that $ E'F'\parallel EF.$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – We denote the points $ K\equiv EF\cap DE'$ and $ L\equiv EF\cap DF'.$

Because of the points $ Q,\ B,\ D,\ C,$ where $ Q\equiv BC\cap EF,$ are in harmonic conjugation as well, we have that the pencil $ E.QBDC,$ is in harmonic conjugation too.

So, because of $ DK\parallel EB,$ we conclude that $ DE' \equal{} E'K$ $ ,(1)$

Similarly, because of the pencil $ F.QBDC,$ is also in harmonic conjugation and $ DL\parallel FC,$ we conclude that $ DF' \equal{} F'L$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ E'F'\parallel EF$ and the proof is completed.

[b][size=100]REMARKS.[/size][/b]

[b][size=100](1).[/size][/b] – It is easy to show that the line segments $ EF',\ FE',$ pass through the midpoints $ M,\ N,$ of the segments $ FX,\ EZ$ respectively, where $ X\equiv DE\cap CF$ and $ Z\equiv DF\cap BE.$

[b][size=100](2).[/size][/b] – Also, it is easy to show that the line segment $ AR,$ where $ R\equiv EM\cap FN,$ bisects the segment $ EF$ ( from the trapezium $ FEE'F'$ ).

$ \bullet$ In our problem now, let us to change the notations as follows:

[b][size=100][color=DarkBlue]PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ D,\ E,\ F,$ the tangency points of its incircle $ (I),$ with the side-segments $ BC,\ AC,\ AB,$ respectively. We denote the points $ X\equiv DE\cap CF,\ Z\equiv DF\cap BE$ and let $ M,\ N$ be, the midpoints of the segments $ FX,\ EZ,$ respectively. Prove that $ \angle EFN \equal{} \angle FEM.$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – Let be the points $ R\equiv EM\cap FN$ and $ E'\equiv AC\cap FN$ and $ F'\equiv AB\cap EM.$

The line segments $ AD,\ BE,\ CF,$ are concurrent at point $ G_{e}$ as well, which is called the [b][size=100]Gergone point[/size][/b] of $ \bigtriangleup ABC.$

Because of now $ AE \equal{} AF,$ it is enough to prove that the point $ R,$ lies on the angle bisector of the angle $ \angle A.$

But, it is clear based on the above [b][size=100]Lemma[/size][/b], because of $ E'F'\parallel EF$ and so, the line segment $ AR,$ bisects the segment $ EF.$

Hence, we conclude that $ \angle EFN \equal{} \angle FEM$ and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Christos Patilas[/size][/b].

Kostas Vittas.

[quote][color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ ABC$ be a triangle which is touched by its incircle in $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ .

Denote $ K\in BE\cap DF$ , $ L\in CF\cap DE$ and the midpoints $ M$ , $ N$ of $ [KE]$ , $ [LF]$ respectively. Prove that $ \widehat {EFM}\equiv\widehat {FEN}$ . [/color][/quote]

[color=darkblue][b][u]Proof (metric).[/u][/b]  ${ \frac {FD}{FE} = \frac {\sin\widehat {FED}}{\sin\widehat {FDE}} = \frac {\sin(90^{\circ} - \frac B2)}{\sin(90^{\circ} - \frac A2)} = \frac {\cos\frac B2}{\cos\frac A2}} = \sqrt {\frac {b(p - b)}{a(p - a)}}$ $ \implies$ $ \boxed {\ \frac {FD}{FE} = \sqrt {\frac {b(p - b)}{a(p - a)}}\ }$ . Denote $ S = [ABC]$ .

$ \frac {KF}{KD}\stackrel{1^*}{\ = \ } \frac {EA}{EC}\cdot\frac {BF}{BD}\cdot\frac {BC}{BA} = \frac {p - a}{p - c}\cdot\frac ac$ $ \implies$ $ \frac {KF}{a(p - a)} = \frac {KD}{c(p - c)} = \frac {FD}{a(p - a) + c(p - c)}$ $ \implies$ $ \boxed {\ KF = \frac {a(p - a)}{a(p - a) + c(p - c)}\cdot FD\ }$ .

Denote $ x = m(\widehat {EFM})$. Observe that $ m(\widehat {KFM}) = 90^{\circ} - \frac C2 - x$ . Therefore, $ MK = ME$ $ \stackrel{2^*}{\implies}$ $ KF\cdot\sin\widehat {KFM} = FE\cdot\sin\widehat {EFM}$ $ \implies$

$ KF\cdot\cos(\frac C2 + x) =$ $ FE\cdot\sin x$ $ \implies$ $ \frac {\sin x}{\cos(\frac C2 + x)} =$ $ \frac {KF}{FE} =$ $ \frac {a(p - a)}{a(p - a) + c(p - c)}\cdot\frac {FD}{FE} =$ $ \frac {a(p - a)}{a(p - a) + c(p - c)}\cdot\sqrt {\frac {b(p - b)}{a(p - a)}} =$

$ \frac {\sqrt {ab(p - a)(p - b)}}{a(p - a) + c(p - c)}$ $ \implies$ $ \frac {\tan x}{\cos\frac C2 - \sin\frac C2\cdot\tan x} =$ $ \frac {\sqrt {ab(p - a)(p - b)}}{a(p - a) + c(p - c)}$ $ \implies$ $ \tan x = \frac {S}{a(p - a) + c(p - c) + (p - a)(p - b)}$ $ \implies$

 $ \tan x = \frac {4S}{4a(b + c) - 3a^2 - (b - c)^2}$ , which is a symmetrical form w.r.t. the pair $ (b,c)\ \implies\ \boxed {\ \tan x = \tan y = \frac {4S}{4a(b + c) - 3a^2 - (b - c)^2}\ }$ .[/color]

[hide="Lemma 1*."][size=100][color=darkred][b] Let $ ABC$ be a triangle. For the points $ \|\begin{array}{c} D\in (BC) \\
 \\
E\in (CA) \\
 \\
F\in (AB)\end{array}$ denote $ P\in EF\cap AD$ . Then $ \boxed {\ \frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}\ }$ .[/b][/color][/size]

[color=darkblue]Indeed, $ \|\begin{array}{c} \frac {PF}{PE}\stackrel{2^*}{\ = \ } \frac {AF}{AE}\cdot\frac {\sin \widehat {DAB}}{\sin \widehat {DAE}} \\
 \\
\frac {DB}{DC}\stackrel{2^*}{\ = \ }\frac {AB}{AC}\cdot\frac {\sin \widehat {DAB}}{\sin \widehat {DAE}}\end{array}\|\ \implies\ \frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}$ . [b]Remark.[/b] If $ AE = AF$ , then $ \frac {PF}{PE} = \frac {DB}{DC}\cdot\frac {AC}{AB}$ .[/color] [/hide]
[hide="Lemma 2*."] [color=darkred][b]Let $ ABC$ be a triangle and let $ M\in BC$ be a point. Then $ \frac {MB}{MC} = \frac {AB}{AC}\cdot\frac {\sin\widehat {MAB}}{\sin\widehat {MAC}}$[/b] .[/color]

[color=darkblue][b]Remark.[/b] In the particular case $ MB = MC$ obtain $ AB\sin\widehat {MAB} = AC\sin\widehat {MAC}$ .[/color][/hide]

i have a solution using harmonic ranges and pencils and poles and polars whih i will be posting shortly

The tangents at $ B$ and $ A$ to the circumcircle of acute triangle $ ABC$ meet the tangent at $ C$ at $ T$ and $ U$ respectively. Let $ P \equal{} AT\cap BC,R \equal{} BU\cap CA$. Let $ Q,S$ be the midpoints of $ AP,BR$ respectively. Prove that $ \angle ABQ \equal{} \angle BAS$.
[size=200]DEFINATIONS[/size]
let $ A'$,$ B'$ and $ C'$ be midpoints of sides $ BC$, $ AC$ and $ AB$ respectively
let tangent at $ A$ and $ B$ intersect in $ V$
 we define $ Q \equal{} B'C' \cap AT$ and $ S \equal{} C'A' \cap BU$
we also construct lines $ DE$, $ EF$ and $ DF$ such that de is parallel to $ AB$, $ BC$ and $ AC$ such that $ D$, $ E$ and $ U$ are collinear and similarly for $ EF$ and $ DF$
let $ BQ \cap AS \equal{} K$
let $ CA \cap TV \equal{} Z$
 let $ BC \cap UV \equal{} X$
let $ UV \cap EZ \equal{} M$
let $ DX \cap TV \equal{} N$
let $ NA \cap DV \equal{} U$
[size=200]SOLUTION[/size]
all poles and polars are WRT circumcircle of $ \Delta ABC$
observe that $ TUV$ is the orthic triangle of triangle $ DEF$
$ Q$ lies on $ B'C'$ and $ AT$
$ Q$ lies on the polar of $ U$ and $ X$ therefore the polar of $ Q$ is $ UX$
$ S$ lies on $ C'A'$ and $ UB$
$ S$ lies on the polar of $ E$ and $ X$ therefore the polar of $ S$ is $ EX$
therefore pole of $ BQ$ is $ N$ and pole of $ AS$ is $ M$
the polar of $ K$ is $ MN$
$ (U,V: A,X)$ is a harmonic range
therefore $ D,(U,V: A,X)$ is a harmonic pencil
let $ NA \cap DU \equal{} F"$
therefore $ (N,A: R,F")$ is a harmonic range
as $ TUV$ is the orthic triangle of triangle $ DEF$
$ UV$ bisects angle $ NVA$
therefore $ F"V$ is perpendicular to $ RV$
this implies $ F" \equal{} F$
similarly we get that $ NBF$ are collinear
now we jus apply the cevas theorem to triangle $ ABF$ and prove that $ MN$ is parallel
this implies their poles are concurrent with the centre of the circle
ie $ K$ lies on the perpendicular bisector of $ AB$
hence done

Let $ D$, $ E$ be the midpoints of the sides $ BC$, $ CA$, respectively. And $ M$, $ N$ the intersections of $ AD$, $ BE$ with the circumcircle of triangle $ ABC$. We have $ \triangle ABP\sim\triangle AIC$, it implies that $ \measuredangle ABQ\equal{}\measuredangle AME$. Similarly, $ \measuredangle BAS\equal{}\measuredangle BND$. Hence, we have to prove that $ \measuredangle AME\equal{}\measuredangle BND$ $ (1)$

Indeed, we have $ ABMN$ is a cyclic quadrilateral, and $ DE\parallel AB$, so we conclude that $ EDMN$ is a cyclic quadrilateral. It follows $ \measuredangle DME\equal{}\measuredangle END$ $ (2)$

Combining $ (1)$ and $ (2)$, the result follows.

$ \bullet$ This solution is dedicated to [b]Kostas Vittas[/b], who always try to give us the most elementary and nicely solutions :lol:

[quote="April"]Let $ D$, $ E$ be the midpoints of the sides $ BC$, $ CA$, respectively. And $ M$, $ N$ the intersections of $ AD$, $ BE$ with the circumcircle of triangle $ ABC$. We have $ \triangle ABP\sim\triangle AMC$, it implies that $ \measuredangle ABQ = \measuredangle AME$. Similarly, $ \measuredangle BAS = \measuredangle BND$. Hence, we have to prove that $ \measuredangle AME = \measuredangle BND$ $ (1)$

Indeed, we have $ ABMN$ is a cyclic quadrilateral, and $ DE\parallel AB$, so we conclude that $ EDMN$ is a cyclic quadrilateral. It follows $ \measuredangle DME = \measuredangle END$ $ (2)$[/quote]
Very beautiful solution dear [b][size=100]Son[/size][/b] and thank you a lot for the dedication.

Let me please to say some more explanatory details, about the relations $ (1),\ (2),$ for the beginners in the symmedians and antiparallel lines theory.

So, the triangles $ \bigtriangleup ABP,\ \bigtriangleup AMC$ are similar, because of they have $ \angle ABP = \angle AMC$ and $ \angle BAP = \angle CAM,$ from the line segment $ AP,$ as the $ A$-symmedian  of $ \bigtriangleup ABC$ as well $ ($ The symmedian through a vertex of a given triangle, passes through the intersection point of the tangent lines of its circumcircle, through the other vertices. Some proofs of this result, have already been posted lately, in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=214829]Symmedian[/url], of [b][size=100]77ant[/size][/b] $ ).$

$ \bullet$ In the triangle now, $ %Error. "bigtrinaleup" is a bad command.
KMN,$ where $ K\equiv AM\cap BN,$ we can say that every circle taken as cord the segment $ MN,$ intersects its sidelines $ KM,\ KN,$ at two points lie on a line with constant direction, as the antiparallel line to $ MN,$ with respect to the angle $ \angle K$ $ ($ which means, with respect to the line segments $ KM,\ KN$ of the angle $ \angle K$ $ ).$

So, because of the circle $ (O)$ is one of these circles $ ($ taken as cord the segment $ MN$ $ )$ and because of $ DE\parallel AB,$ we conclude that the line segment $ DE$ is antiparallel to $ MN$ wrt $ \angle K$ and hence, the conclusion of the cyclic quadrilateral $ EDMN,$ it follows.

Thank you again and best regards, Kostas Vittas.

PS. Although it is difficult to use the argument of antiparallel lines in trying to solve a problem, however it is a useful geometric tool, for elegant synthetic solutions.

[quote="LawOfSigns"]How can you prove that AP (extended, AT) is the A-symmedian?  Does it have to do with knowing that TB = TC (as tangents) and then using the fact that TBC is isosceles?[/quote]
Take a look please in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=214829]Symmedian[/url].

kostas Vittas.

Nice one!!!(Although my solution is ugly)
[hide=Solution]Let $BQ\cap AS=V$
I will prove that $AV=BV$.The tangent to circumcircle at $A$ has equation $b^2z+c^2y=0$.Using this it's easy to find that $U=(a^2,-b^2,c^2)$, $T=(-a^2,b^2,c^2)$.
Intersecting $AT$ and $BC$ we get $P=(0,b^2,c^2)$ and similarly $R=(a^2,0,c^2)$.Normalizing $A$ and $P$ we get $Q=(b^2+c^2,b^2,c^2)$ and similarly $S=(a^2,a^2+c^2,c^2)$. Intersecting $AS$ and $BQ$ we get $V=(\frac{b^2+c^2}{K},\frac{a^2+c^2}{K},\frac{c^2}{K})$ where $K=a^2+b^2+2c^2$.
Then using Distance formula $AV^2\cdot K^2=2c^4(a^2+c^2)+b^2c^2(a^2+2c^2)$ and $BV^2\cdot K^2=2c^4(b^2+c^2)+a^2c^2(b^2+2c^2)$. So we get $AV=BV$ $\implies$ $\angle ABS=\angle BAQ$.


$ a\in\mathbb{N},a>1$ aya lozuman adade aval va farde $ p$ vojud darad ke $ a$ yek risheye avaliye be peymaneye $ p$ bashad?

[quote="khashi70"]Fekr konam in soal ye hadse va hanooz sabet ya rad nashode . intor nist ?[/quote]
age baman budi,nemidunam:D
vali aslan masaleye sadeie be nazar nemiad...

in soali ke to gofti javabesh na hastesh chon har adade $ a \equal{} x^{2}$ nemitoone rishe avvalie be peymaneye ye adade avval bashe . dar vaghe martabash hadeaksar $ \frac {p \minus{} 1}{2}$ hast .
vali ye hads hast ke mige :
agar $ a > 1$ va $ a$ morabae kamel nabashe dar oon soorat $ a$ rishe avvalie be peymaneye binahayat $ p$ e avval hast .
in hadsam be hadse Artin maroofe age search koni mitooni darbarash matlab peyda koni .


ratio
LaTeX
geometric sequence
geometric series

Hi!

I have a progression:

Term    Value
----------------
0          4
1          4y
2          4y^3
3          4y^4

The problem is how to retrieve the sum of the first 156 elements.  The answer should look something like

((y^155)-1)/y-1

Or something!  Can anyone help?  Much appreciated.

Thanks

[quote="jb05"]Gullanian: since you asked, I assume you don't know this formula (?). I suggest that you learn it. The sum of a geometric sequence with original amount a, ratio r, and t terms is (sorry, I can't use Latex):

a(r^(t+1)-1)/(r-1)

As a simple exercise, you may want to find the sum of an infinite geometric series using this formula (when the ratio is less than 1--for instance .5+.25+.125+...). The answer is given below:

[hide]a/(1-r)[/hide][/quote]
Jb, your formula is a bit messed up, its actually $r^{t}$, not $r^{t+1}$.  So it would be:
$\frac{a(r^{t}-1)}{r-1}$
[hide]$\frac{a}{1-r}$ is the formula for infinte geometric sequcne.[/hide]


This is a game to see how much you think like other people.

Since I do not have all day to be on AoPS, I will only allow, hmm, let's say 6 players per game. Or if someone is willing to help me with the scoring, it'd be OK to have more people  :D . If you want to join, post on this thread.

I will post a word. Then each player will pm me (or the scorer) the first 3 words they think of from that word. Your score is the sum of the 3 words' frequencies. For example:

WORD:POPCORN
[b]Alice:[/b]
butter
yummy
movie
[b]Bob:[/b]
butter
microwave
corn
[b]Carl:[/b]
yummy
microwave
butter
[b]Scoring:[/b]
Alice = 3+2+1 = 6
Bob: = 3+2+1 = 6
Carl: = 2+2+3 = 7

Carl wins round.

[hide="details"]
After each round, the players will be scored, and then the next 6 players for the next round will be the next 6 people who post on this thread saying that they will join. In other words, each time, you must re-register. Every, say, 5 rounds, a winner will be announced.

Pm me if you want to be a scorer. If you are a scorer, you do not get to play, but you get 3(#of people)/2 points  :lol: .

No pm-ing or teaming up with each other. If someone pms you, then you can expose them and win 3(#of players) points. The person who attempted to cheat will lose that same amount of points.

Examples of words that will be considered the same:
write-writes-writing-writer-writings
pencil-pencils
soda-pop-coke

Examples of words that will not be considered the same:
computer-keyboard-monitor-mouse
teach-learn
rainbow-rain-bow
[/hide]

First word: [b]PLUTO[/b]

Wait, wouldn't Alice get 3+2+1=6 points?

Anyways, I'll play.

Oops, yeah, you're right andersonw.  :oops:

Round 1 Info (so far):
Wickedestjr: submissions received.
andersonw: submissions received.

Round 1 Info (so far):
Wickedestjr: submissions received.
andersonw: submissions received.
erabel: submissions received.

No new players after 6-6-08 3pm AoPS time.

Wickedestjr: submissions received.
andersonw: submissions received.
erabel: submissions received.
zapi2007: submissions received.

YAY! People are actually joining! And the words being associated are actually different (I was worried everyone would put the same words)!

Wickedestjr: submissions received.
andersonw: submissions received.
erabel: submissions received.
zapi2007: submissions received.
chenhsi: submissions received. Next time post you're joining on the thread.
BOGTRO: have not received submissions.

I'll grade if you want. Not like I have anything worthwhile to do anyway.... *looks at homework*

So, when BOGTRO gives his words, then can we begin?

Ok, lesee....

The words were
planet x6
dwarf x3 (someone put dwarf planet, which I counted as dwarf)
cold x2
dog x1
dark x1
universe x1
Disney x1
small x1
debate x1
eight x1

Which means the scores are....
Wickedestjr: 9
andersonw: 9
erabel: 10
zapi2007: 10
chenhsi: 10
BOGTRO: 8

Close game. So the tied winners are erabel and zapi2007! Congratulations to you both.

Sign-ups for next round begin NOW. Thanks Brut3Forc3 for being willing to grade. How many players are you willing to take? Say Brute3Forc3 is willing to grade x players. Hmm.... and I'll do 6. So 6+x players max.

Now, how is this gonna work? Hmm.... idk. For now, just pm both to me and Brut3Forc3.

Round 2 Word: [b]STAPLER[/b]

Round 2 so far:
Wickedestjr: submissions received.

I'm.... very bored. I could probably take up to 20 people. :P

This sounds like the TV show, "Password."

Round 2 so far:
Wickedestjr: submissions received by me.
zapi2007: submissions received by me.
crazychinesemaniac: submissions received by me.

I dunno about Brut3Forc3.

Nevermind about having to post on this thread to enter each round. That was a stupid policy.

Deadline for entrees for Round 2 is June 8, 6pm AoPS time.

Round 2 so far...
Wickedestjr: submissions received by me.
zapi2007: submissions received by me.
crazychinesemaniac: submissions received by me.
erabel: submissions received by me.

Round 2 Results

The words were
staple x4
office x3
desk x3
paper x2
staple remover x1
metal x1
black x1

Which means the scores are....
erabel: 10
zapi2007: 6
BOGTRO: 10
crazychinesemaniac: 6
Wickedestjr: 9

The tied winners are erabel and BOGTRO! Congratulations to you both.

Sign-ups for next round begin NOW.
Pm words both to me and Brut3Forc3.

Round 3 Word: [b]JUICE[/b]

And total scores are:

erabel: 20
Wickedestjr: 18
BOGTRO: 18
zapi2007: 16
chenhsi: 10
andersonw: 9
crazychinesemaniac: 6

Round 3 so far:
erabel: received.
zapi2007: received.

Once again, zapi is the only one sending to me...

Oh well, it doesn't matter unless there are more than 6 people...

Round 3 so far:

erabel: received.
zapi2007: received.
BOGTRO: received.

I guess this means Game Over so....

FINAL SCORE
1) erabel: 20

2) Wickedestjr: 18

2) BOGTRO: 18

4) zapi2007: 16

5) chenhsi: 10

6) andersonw: 9

7) crazychinesemaniac: 6

Wow, that's a wierd set of scores, because I only played one round. Other than that, I didn't do anything. Why is my score so high?

uhh wickedstjr. plz don't revive old threads

I was just wondering how I got so many points when I barely even did anything.