ARML
rotation
geometry
geometric transformation
algebra
polynomial
complex numbers

is this an old arml problem?

[hide]

we have a regular dodecagon and its asking for the product of $(x_n+iy_n)$, so we should be thinking complex numbers.[/hide]

i dont have an exact solution, but is the idea that you have
let (a,b) be the center of the dodecagon) and r is the radius (considering that the first point is rotated) in polar notation
$x =  a + bi + ^{12}\sqrt{r}$
then you do to the 12 power and find the constant term?

well, i got an answer like $\approx 3.55 + -10.29i$, is the answer a clean answer or not?

i found a way to write this through complex numbers but when i did the product on my calculator, it came out badly

$\prod^{11}_{k=0} 4* \sqrt{ 2 - \sqrt{3} } * e ^{ i (\frac{11 \pi}{12} + \frac{ \pi}{6}*k) } - 1 + i (-3 + 2\sqrt{3})$
the thing is that using $k= 0,7$, you get the 2 points that are integers, so i think it is right, and they are 6 apart like the $P_1$ to $P_6$, just going from $0,12,11,10,9,8,7$, but that should not matter, the thing is that the center of the dodecagon could be either slightly above or below the line connecting the 2 points, so i am not sure what to do, i did above

well, after looking for an easy answer, are you sure that the points are $P_1$ and $P_6$ because, there would be an integer answer if you were given $P_1$ and $P_7$ because that way the center would be $(-1,1)$
therefore think of the equation
$ x = \sqrt[12]{2^{12}} - 1 + i$
the verticies of the dodecagon would be given by that
(considering the 12 complex roots of $2^{12}$)
so the constant term will be the product of the roots of the equation, and to find the constant term,
$( x + (1-i )= \sqrt[12]{2^{12}})^{12}$
doing the binomial expansion is not necessary since all we need is the constant term which would be
$(1-i)^{12} - 2^{12}$
$(\sqrt{2}e^{i* \frac{5 \pi}{4}})^{12} - 2^{12}$
$64*e^{15 \pi i} - 2^{12}$
$-64 + -4096 = -4160$

It says P_6 on my sheet, but I'd believe it could be a typo.  However I don't get what you did when you turned it into an equation and found the constant term.

well, if you have the nth roots of a number, you get a regular polygon (if it is graphed on a complex plane)
so this is a polygon just it is translated over left 1 and up 1, so you add $i$ and subtract $1$ (since left and right is the real axis and up and down is the imaginary axis)
since the 12th root represents all the verticies of the dodecagon (there are 12 complex roots for the $\sqrt[12]{2^{12}}$, and with the translation, they are the points of the dodecagon given in the problem, so in any polynomial where the lead coeffieient is 1, the product on the roots is the constant term (or negative of the constant term when the lead coefficient is odd)
the reason that i needed to rearrange the equation is because the complex roots could be 1 of the roots of $\sqrt[12]{2^{12}}$
if i did do this, i would not know which root to use for some of the middle terms
$(\sqrt[12]{2^{12}} - 1 + i)^{12}$
but we do know that any of the roots to the 12th power = $2^{12}$
and rearranging removes the confusion for the different roots so it works
and i was not sure of that step myself, but after, i checked using a calculator and it was correct the other way to do it would be like:
$\prod^{11}_{k=0}(2*e^{\frac{\pi * k * i}{6}} -1 + i)$
and this does infact yield the same answer, i hope this helped explain a little better


geometry
integration
LaTeX
trigonometry
function
absolute value
calculus

This is equivalent to $\frac{1}{2}\int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}sec(2x) dx$
Let u=2x
 $\frac{1}{4}\int^{\pi}_{-\pi} sec(u) du=\frac{1}{4}ln(|sec(u)+tan(u)|)]^{\pi}_{-\pi}=0-0=0$
But of course you wanted area which is impossible because of the undefined portion of the graph at $x=\frac{\pi}{4}$

I downloaded TeXnicCenter Setup, Version 1 Beta 6.21 but don't have the MiKTeX 250mb engine. And I guess I need Go2PDF as well. I should have typed (1/2) sorry about that. What do I need to do to integrate the absolute value? |(1/2)cos2x| ? I tried pi/2 to 0 plus 0 to -pi/2 but that didn't work.

You don't need anything to post $\LaTeX$ on this forum- that's handled by the forum software. Just put the code inside dollar signs.

[quote="cole"]I downloaded TeXnicCenter Setup, Version 1 Beta 6.21 but don't have the MiKTeX 250mb engine.[/quote] I can recommend you to install MikTeX small installation package (about 20MB). I think, it will cover all you needs.

[quote="cole"] What do I need to do to integrate the absolute value? |(1/2)cos2x| ? I tried pi/2 to 0 plus 0 to -pi/2 but that didn't work.[/quote] First, figure out where the function is positive and where it is negative.

[quote="fedja"][quote="cole"]find area bounded by y = 1/2cos2x, y=0, x = -pi/2, x= pi/2

answer is 1

I can't figure out the anitderivative. would it be 1/4sin2x ??[/quote]

That's why $\LaTeX$ is useful: I guess cole meant $\frac 12 \cos 2x$, not $\frac 1{2\cos 2x}$ (at least, such interpretation gives the right answer). As to the question itself, yes, the antiderivative of $\frac12\cos 2x$ is $\frac 14\sin 2x$. Diophantus made one more common mistake though: the function changes sign, so to find the area, one needs to integrate not the function itself but its absolute value...[/quote]
If you read my post more carefully, I mentioned that the area cannot exist because of the asymptopes in the equation that I interpreted.


please rewrite it more clearly.I'm sure if you write in this way you won't know meaning what you wrote in some years :D


quadratics
algebra
polynomial
binomial theorem
number theory
modular arithmetic
absolute value

Find the smallest value of $n$ for which $n^{2}+n+41$ is not a prime number.

[hide="Incomplete Solution"]The expression is not prime when $n^{2}+n$ is divisible by $41,$ so the smallest $n$ is $40.$ (Which was the answer)

However, I didn't show that $n=40$ was the smallest value, so how does that work?[/hide]

[quote="srulikbd"][quote="pkerichang"][quote="10000th User"]I heard that a European mathematician (Matiyasevich?) solved a very general problem (a Hilbert problem) about polynomials generating primes. I'm not 100% sure what it was about but I think it is similar to this, perhaps....[/quote]

It is mentioned in ACoPS.  Basically, they want to see if there's ever a polynomial that only generates prime.  The end result is never (the argument is kind of simple), and you have an interesting property: for a polynomial with integer coefficient, $f(a+f(a))$ will always be a multiple of $f(a)$.[/quote]
for $a$ integer? what is the proof? thanks.[/quote]

well...obviously $a$ is integer (or else you won't get integers as output)
[hide="proof"]
consider the polynomial $f(x) = a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{0}$, then we wish to prove that $f(x+f(x))$ will always be a multiple of the above expression.
So if we use binomial theorem and expand, we get:
$f(x+f(x)) = a_{n}(x^{n}+nx^{n-1}f(x)+\dots)+\dots+a_{0}$
if we actually expand everything out, you can see that most of the terms contains $f(x)$ in it.  If it didn't contain $f(x)$, it must be the first term in the binomial expansion, and you can group them all together, and guess what, they add up to be $f(x)$, therefore $f(x+f(x))$ must be a multiple of $f(x)$.
This also prove that prime-generating polynomials are not possible.  Because suppose $f(x)$ is prime $p$, then $f(x+f(x))$ will be a multiple of this prime, for this polynomial to still generate primes, $f(x+f(x))$ must equal to $p$, continue this logic we see that this polynomial must be linear, a contradiction. (because a polynomial of degree $n$ must have at most $n$ identical values, but you can go on this process indefinitely)
[/hide]

Using modular arithmetic gives another, perhaps simpler, proof.

[hide="Another Proof"]Let's start with the equation \[a+f(a) \equiv a \mod{f(a)}\, .\] Now, let's apply $f$ to both sides.  We get \[f(a+f(a)) \equiv f(a) \equiv 0 \mod{f(a)}\, .\] [/hide]

I've found a pattern!
[hide="You just need to"]add consecutive odd numbers. with n=0, add $2+4+...+2x-2+2x$ to get to where $n=x$[/hide]
So the answer [hide="is...."]I'm too lazy to go through the first 40 even numbers. [hide=" :lol: "]Really! I am![/hide][/hide]

[quote="Altheman"]you pretty much have to brute force[/quote]

There is a way to shorten it (wheter or not that way is appropriate for intermediate we may discuss later on ;) ):

Assume that $p=n^{2}+n+41$ is not prime for some $n<40$.
Let $q$ be the smallest prime divisor of $p$. Then $2<q<41$. We also have that $0 \equiv 4(n^{2}+n+41) = (2n+1)^{2}+163 \mod q$.
Thus $-163$ is a square $\mod q$. Now with the Law of Quadratic Reciprocity, we can easily check this to be wrong for all primes $<41$.

@nutz: ¿?

[quote="ZetaX"]Now with the Law of Quadratic Reciprocity, we can easily check this to be wrong for all primes $<41$.

@nutz: ¿?[/quote]

could you explain how to apply that, first, you need to make $n\cdot q-163$ a prime (which of course is possible since 163 is prime), then you can switch $q$ and $nq-163$, what additional insight does that give?  you still have to list the quadratic residues of many primes either way

I don't need to make $-163$ prime by shifting it (when making it prime, changing the same is at least shorter ;) ).
The trick is to reduce $\mod$ the "denominator" and to use simple factoring (the last one is clearly possible since we speak of number with at most two digits) or the Jacobi symbol, for which still the same rules hold (especially the Law of Quadratic Reciprocity) for all odd numbers. See http://www.artofproblemsolving.com/Wiki/index.php/Quadratic_residues for some stuff.

So e.g. for the primes $31,37$ (the two biggest ones to check, the others are even faster):
factoring:
$\left( \frac{-163}{37}\right) = \left( \frac{-15}{37}\right) = \left( \frac{-1}{37}\right) \left( \frac{3}{37}\right) \left( \frac{5}{37}\right) = \left( \frac{37}{3}\right) \left( \frac{37}{5}\right) = 1 \cdot (-1) =-1$
Jacobi:
$\left( \frac{-163}{31}\right) = \left( \frac{23}{31}\right) =-\left( \frac{31}{23}\right) =-\left( \frac{8}{23}\right) =-\left( \frac{2}{23}\right) =-1$
(note that $\left( \frac{-163}{31}\right) = \left( \frac{-8}{31}\right)$ would make it shorter, but I wanted to show the Jacobian way ;) )

Just take the representative of the residue class that has smallest absolute value and factor then, us the Law and do so again.

Edit: error corrected.


solve the equation :
$|y|\sqrt{x^3+4}-x^2=2-\sqrt{y^2-x-1}$  :D


m and  n are two naturel numbers such that
m<n
n divides m^2
show that there is a perfect square number that divides n.

n divides m^2.  So any factor which is in the prime factorization of n is also in that of m^2.  Therefore, every prime which divides n also divides m.  So, if n>m, then in their prime factorizations, one of the factors of n must be raised to higher power than it is in the prime factorization of m (because n doesn't have any factors that m doesn't have, yet n>m).  Therefore, at least one of the exponents in the prime factorization of n must be >1, and therefore, n is divisible by a perfect square.

If n is squarefree, it is the product of primes p_1 p_2 p_3...p_k Because m^2 is a multiple of n (given), m has a factor of p_1 p_2 ... p_k, implying m >= n.  Contradiction.  So if m<n and n|m^2, then there exists a square that divides n.


Euler
function
floor function
logarithms
number theory

[b]1.[/b] A number is created by writing the numbers [b]3[/b] to [b]333[/b] in order after the decimal point, that is, 0.3456789101112.....333.  What is the 700th digit to the right of the decimal point?

[b]2.[/b]  How many digits are there in the expanded form of 2 to the 40th power?

[b]3.[/b]  What is the remainder when 5 to the 999,999th power is divided by 7?

Looking for an easier way..... :)

[quote="NapoleonXIV"]Third one can easily be done by using Euler Totient function and Euler theorem.[/quote]

Is that the only way to solve it?? :|  (just asking..)

Probably not, but I believe its the easiest, and it doesn't take long.

since 7 is prime, it becomes Fermat's Little Theorem

for the second one, we have

$ 2^{10}>10^3\Rightarrow 2^{40}>10^{12}$, and $ 2^{13}<10^4\Rightarrow 2^{40}<2\cdot 10^{12}$, so $ 2^{40}$ has $ \boxed{13}$ digits

#1 is just cruel

Heh, for number 2, I have memorized that log 2=0.3010.

Number of digits in $ x^{n}$ is $ \lfloor x\log n\rfloor \plus{}1$.

Useful Things to Memorize

$ \log e\approx .434294481903$
$ \log 2\approx .301029995664$
$ \log 3\approx .47712125472$
$ \log 5\approx .698970004336$
$ \log 7\approx .845098040014$
$ \log p$, for as many prime $ p$ as you can

It isn't necessary to memorize all that, 4 or 5 digits is sufficient for most problems

[quote="alanchou"]Number of digits in $ x^{n}$ is $ \lfloor x\log n\rfloor \plus{} 1$.
[/quote]

I think its  $ \lfloor n\log x\rfloor \plus{} 1$


I would.  It makes the solution look more organized, in my opinion.

[quote="jtw"]Should I put the parts of the solution to a problem with two parts on separate pages? Or what?[/quote]It makes no difference.  Just make sure that each numbered problem starts on its own page, and that the solutions are in numeric order (1 through 5).

I just put them on the same page because I want to use the same lemmas for both.

[quote="timwu"]By the way, in LaTeX, how can I start a new page even if I don't finish the current page?[/quote]

I think this also works, but a little differently:
[code]\pagebreak[/code]


Solve these equation with $ x\in R$:

[u][b]equation 1[/b][/u]

$ \sqrt {2 \minus{} x^3} \equal{} \sqrt {x^2 \minus{} 2}$

[u][b]equation 2[/b][/u]

$ x \equal{} \sqrt {(3 \minus{} x)(4 \minus{} x)} \plus{} \sqrt {(4 \minus{} x)(5 \minus{} x)} \plus{} \sqrt {(5 \minus{} x)(3 \minus{} x)}$

[u][b]equation 3[/b][/u]

$ \sqrt [3]{x^2} \minus{} 2\sqrt [3]{x} \minus{} (x \minus{} 4)\sqrt {x \minus{} 7} \minus{} 3x \plus{} 28 \equal{} 0$

[u][b]equation 4[/b][/u]

$ \sqrt {5x \minus{} 1} \plus{} \sqrt [3]{9 \minus{} x} \equal{} 2x^2 \plus{} 3x \minus{} 1$

[color=red][u][size=150][b]Note:[/b][/size][/u][/color] [b][color=darkblue]Please post full solution. Thank you very much.[/color][/b]

[quote="thanhnam2902"]Solve these equation with $ x\in R$:

[u][b]equation 1[/b][/u]

$ \sqrt {2 \minus{} x^3} \equal{} \sqrt {x^2 \minus{} 2}$

[u][b]equation 2[/b][/u]

$ x \equal{} \sqrt {(3 \minus{} x)(4 \minus{} x)} \plus{} \sqrt {(4 \minus{} x)(5 \minus{} x)} \plus{} \sqrt {(5 \minus{} x)(3 \minus{} x)}$

[color=red][u][size=150][b]Note:[/b][/size][/u][/color] [b][color=darkblue]Please post full solution. Thank you very much.[/color][/b][/quote]
[b]Equation 1:[/b]
$ \begin{cases} 2 \ge x^3 \\ x^2 \ge 2 \end{cases} \Leftrightarrow x \le \minus{}\sqrt{2}$

So $ 2\minus{}x^3\equal{}x^2\minus{}2 \Leftrightarrow x^2(x\plus{}1)\equal{}4$ . But $ x^2(x\plus{}1)<0$ , contradiction. So no real solution for equation 1 .

[b]Equation 2[/b]

Put $ x\equal{}4\minus{}y$ , the equation becomes $ 4\minus{}y\equal{}\sqrt{y(y\minus{}1)}\plus{}\sqrt{y(y\plus{}1)}\plus{}\sqrt{y^2\minus{}1}$
$ \Rightarrow [(4\minus{}y)\minus{}\sqrt{y^2\minus{}1}]^2\equal{}(\sqrt{y(y\minus{}1)}\plus{}\sqrt{y(y\plus{}1)})^2$
$ \Rightarrow \minus{}8y\plus{}15\equal{}8\sqrt{y^2\minus{}1}$
$ \Rightarrow y\equal{}\sqrt{289}{240}$
$ \Rightarrow x\equal{}\frac{671}{240}$ , which is a solution after checking.


Let $ A\equal{}\left(
\begin{array}{cc}
a & b \\
0 & a
\end{array}
\right)$, by judging from $ A^2,\ A^3,\ \cdots$ to expect $ A^{n}$ for every positive integer $ n$, then prove that the expectation is true by induction.

$ A^2 \equal{} \left(\begin{array}{cc}a^2 & 2ab \\
0 & a^2\end{array}\right)$, $ A^3 \equal{} \left(\begin{array}{cc}a^3 & 3a^2b \\
0 & a^3\end{array}\right)$. Assume that $ A^n \equal{} \left(\begin{array}{cc}a^n & na^{n \minus{} 1} b \\
0 & a^n\end{array}\right)$. It holds true for $ n \equal{} 1$, assume that it is true for all $ k$ s.t. $ k < n$, now
$ A^n \equal{} A^{n \minus{} 1}A \equal{} \left(\begin{array}{cc}a^{n \minus{} 1} & (n \minus{} 1)a^{n \minus{} 2} b \\
0 & a^{n \minus{} 1}\end{array}\right)\left(\begin{array}{cc}a & b \\
0 & a\end{array}\right) \equal{} \left(\begin{array}{cc}a^n & na^{n \minus{} 1} b \\
0 & a^n\end{array}\right)$, and $ A^n \equal{} \left(\begin{array}{cc}a^n & na^{n \minus{} 1} b \\
0 & a^n\end{array}\right)$ is true for every positive integer $ n$, by induction.

Here's what I thought when I saw it:

$ \begin{bmatrix}a&b\\0&a\end{bmatrix}\equal{}a\begin{bmatrix}1&a^{\minus{}1}b\\0&1\end{bmatrix}$

I'm pretty familiar with $ \begin{bmatrix}1&x\\0&1\end{bmatrix}^n\equal{}\begin{bmatrix}1&nx\\0&1\end{bmatrix}.$

So the rest follows quickly from that.

Of course, proving the thing I said I was already familiar with is only barely easier than directly proving the result here - it's the same ideas. It's just that that familiarity tells me up front what the answer has to be.

[quote="Kent Merryfield"]Here's what I thought when I saw it:

$ \begin{bmatrix}a & b \\
0 & a\end{bmatrix} \equal{} a\begin{bmatrix}1 & a^{ \minus{} 1}b \\
0 & 1\end{bmatrix}$[/quote]

Why is $ a\neq 0?$

Ah, yes - I thought $ a\ne0$ was a condition and it isn't. Of course if $ a\equal{}0,$ the matrix is obviously nilpotent, and that's also easy to account for in the result.


Call $ (s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)....(s\minus{}z)$ "M". We have $ (\frac{M}{M})*M$ which is just M.

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl: It is a joke.
$ M\equal{}(s\minus{}a)(s\minus{}b)...(s\minus{}z)\equal{}(s\minus{}a)(s\minus{}b)(s\minus{}c)...(s\minus{}r)(s\minus{}s)(s\minus{}t)...(s\minus{}z)\equal{}$
$ (s\minus{}a)(s\minus{}b)(s\minus{}c)...(s\minus{}r)0(s\minus{}t)...(s\minus{}z)\equal{}0$
$ \frac{0}{0}.0\equal{}\frac{0}{0}\equal{}$[b]Undefined.[/b]

Yes i agree with Akashnil, the answer is undefined

Thanks. I treated "s" as a seperate variable than every other letter in the alphabet.:cleaning:  :cleaning: Why you should never let me into your house.  :rotfl:

Well, it could be......but that will make it confusing.


Putnam
geometry
rectangle
calculus
calculus computations

I found this question in my math book:

Inscribe a rectangle of base b and height h and an isosceles triangle of base b in a circle of radius one as shown. For what value of h do the rectangle and triangle have the same area?

rectangle meets circle at four points (x,y)

$ \left(\pm \frac{h}{2}, \pm \frac{b}{2}\right)$

$ h^2\plus{}b^2\equal{}4$

triangle has height $ 1\minus{}\frac{h}{2}$, base b

$ A_R\equal{}bh\equal{}\frac{b}{2}(1\minus{}\frac{h}{2})\equal{}A_T$

$ 4bh\equal{}b(2\minus{}h) \to h\equal{}\frac{2}{5} \to b\equal{}\frac{\sqrt {21}}{5}$

[quote="JRav"]This seems sort of easy to be a Putnam problem.[/quote]
Sometimes they are. If you're taking the Putnam, don't overlook that possibility - and you really need to get one like this correct.

I re-solved the problem myself and while my answer for $ h$ agrees, my answer for $ b$ is $ \frac{4\sqrt{6}}{5}$, which disagrees with kenn4000.  If you solve with my values, the areas of the two fiqures are equal.

hey hey yeah i definitely mis solved b, i think i used 1 instead of 4.


[quote="Ignite168"]:oops: <- This counts as 2 smilies instead of 1 for some reason.[/quote]

I think that is because it has both a  :o and  :oops:

[code]:o and :oops[/code]


This question arises from the problem being considered [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87224]here[/url].

a) If $P(x)$ and $Q(x)$ are polynomials, what are all possible degrees of $P(Q(x))$?
b) If $P(x)$ and $Q(x)$ are polynomials, and $Q(x) = P'(x)$, what are all possible degrees of $P(Q(x))$?  Of $Q(P(x))$?
c) If $P(x)$ is a polynomial and $Q(x) = P'(x)$, is it possible for $\deg(P(Q(x)) - Q(P(x))) \not= \deg(P(x))\cdot\deg(Q(x))$?
d) If $P(x)$ is a polynomial and $Q(x) = P'(x)$, is it possible for $\deg(P(Q(x)) - Q(P(x))) = \deg(P(x))\cdot\deg(Q(x)) - 1$?

a) $\deg P(Q(x))= \deg P(x) \times \deg Q(x)$

c)  Yes.

Considering only the first term, let $P(x) = ax^n, Q(x) = nax^{n-1}$.  Then

$P(Q(x)) = a^{n+1} n^n x^{n(n-1)}$
$Q(P(x)) = a^n n x^{n(n-1)}$

If the two coefficients are equal, then $\deg(P(Q(x)) - Q(P(x)) < \deg(P(x)) \cdot \deg(Q(x))$.  This occurs iff

$a^{n+1} n^n = a^n n$, or
$a = n^{1-n}$.

d)  I think so.  For example, let $P(x) = \frac{x^2}{2} + 3x$.

Huh, I wasn't expecting that to happen on part (d).  What if I replace the minus signs in parts (c) and (d) with plus signs?  Are the answers the same?

Then $a = - n^{1-n}$ would also work analogously, I think.

Yes, it would for part (c).  Part (d) will (hopefully) be more interesting.


algebra
polynomial
factorial
superior algebra
superior algebra unsolved

Is it true that for every cyclotomic polynomials $ \Phi_m, \Phi_n, m\neq n$ they share no common
factor when we consider them in $ \mathbb{Z}/p\mathbb{Z}$, $ p$ a prime s.t. $ p\not|n,m$
(i.e. if we take coefficients modulo $ p$) ?

edit: if $ p|n,m$ this is not true, pick $ \Phi_4, \Phi_8$ and $ p\equal{}2$

It is true.
If $ f(X)$ not constant is a common factor of $ \Phi_m(X),\Phi_n(X)$ over $ \mathbb{F}_p\equal{}\mathbb{Z}/p\mathbb{Z}$.
Then the formula $ X^{mn}\minus{}1\equal{}\prod_{d|mn}\Phi_d(X)$ gives
$ f(X)^2|\Phi_m(X)\Phi_n(X)|X^{mn}\minus{}1$, i.e. $ X^{mn}\minus{}1\equal{}f(X)g(X)$ for some $ f(X)|g(X)\in\mathbb{F}_p[X]$.

Deriving this equation formally gives
$ f(X)|f'(X)g(X)\plus{}f(X)g'(X)\equal{}mnX^{mn\minus{}1}\in\mathbb{F}_p[X]$.

But since $ p\nmid mn$ and $ f(X)$ not constant and $ \mathbb{F}_p[X]$ [url=http://en.wikipedia.org/wiki/Factorial_ring]factorial[/url] with $ X$ irreducible
we get $ X|f(X)|X^{mn}\minus{}1$. Contradiction!


let $ x,y,z\ge 0$ prove that
$ \frac{1}{x^3\plus{}y^3}\plus{}\frac{1}{y^3\plus{}z^3}\plus{}\frac{1}{z^3\plus{}x^3}\ge \frac{20}{(a\plus{}b\plus{}c)^3}$

is there any relation between x,y,z and a,b,c

[quote="bunhiacovski"]let $ x,y,z\ge 0$ prove that
$ \frac {1}{x^3 \plus{} y^3} \plus{} \frac {1}{y^3 \plus{} z^3} \plus{} \frac {1}{z^3 \plus{} x^3}\ge \frac {20}{(a \plus{} b \plus{} c)^3}$[/quote]
I think, it can be $ \frac {1}{x^3 \plus{} y^3} \plus{} \frac {1}{y^3 \plus{} z^3} \plus{} \frac {1}{z^3 \plus{} x^3}\ge \frac {20}{(x\plus{} y \plus{} z)^3}.$
The expanding kills it easy enough.

can anyone tell me how to use muirhead to solve this? i am not good at using muirhead


vector
linear algebra
matrix
Ring Theory

Let $M_{22}$ be the vector space of 2×2 matrices with real entries, and with
addition and scalar multiplication defined as the standard operations on matrices. Let W be the subset of $M_{22}$ that consists of all matrices of the form
|a b|
|0 d|.
Determine whether W is a subspace of $M_{22}$.

Find a basis for W.

Masoud Zargar

$W=\text{span}\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}.$

Since the span of any particular set of vectors is a subspace and since this set is independent, this answers both of your questions.

Note that $W$ is also a subring of $M_{2,2}$ since it is closed under matrix multiplication.


probability
number theory
least common multiple

Here are some questions for warm-up. Please don't solve them, unless they are new to you.

1. What is the probability that if you take a marble out of a bag containing 3 blue marbles, 6 green marbles, and 7 yellow marbles, that it will be green?

2. How many odd numbers between 200, and 350 are there?

3. What is the smallest number that is divisible by 12, 56, and 35?

I'll just hide these:

[hide="#1"]
There are 3+6+7=16 total marbles, and 6 are green, so the probability is $ \frac{6}{16}\equal{}\frac{3}{8}$.[/hide]

[hide="#2"]
350-200=150. 150/2=75. There are 75 odd numbers.[/hide]

[hide="#3"]
Each can be prime factorized like this:
$ 12\equal{}2^2\times3$
$ 56\equal{}2^3\times7$
$ 35\equal{}5\times7$

Using this to take the LCM, we get $ 2^3\times3\times5\times7\equal{}840$[/hide]

P.S. Ever noticed that people wil still post solutions, even if they are too easy? :wink:  :P  :rotfl:

[quote="isabella2296"]Please see post #4 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23215]here[/url].[/quote]

Oops, sorry about that.  :oops:


find all functions $ f: R \rightarrow R$ , $ f(f(x) \plus{} y) \equal{} f(x \plus{} y) \plus{} f(0)$

[hide="Solution"]Since $ f$ maps from $ \mathbb{R} \to \mathbb{R}$, we can study the case:

$ (f(x),y) \equal{} (u,0) \implies f(u) \equal{} u \plus{} f(0)$

Hence, $ f$ must satisfy $ f(x) \equal{} x \plus{} C$ for arbitrary C.

Then, $ f(f(x) \plus{} y) \equal{} f(x \plus{} y) \plus{} C \implies x \plus{} y \plus{} 2C \equal{} x \plus{} y \plus{} 2C$, so there is no contradiction or further restriction.[/hide]

[b]@ whoever called this spam:[/b] Please post a better/more correct/more relevant solution  :)

Zuton Force, you have $ f(u) \equal{} u \plus{} f(0)$, but this equation only holds if $ u$ is in the image of $ f$, since you have $ f(x) \equal{} u$. One solution you've missed is $ f(x) \equiv 0$.

Right, I see I missed the $ f\equal{}0$ solution because I took an inverse. But if $ f$ maps all reals to all reals, wouldn't $ u$ have to be the image of $ f$?  :huh:

$ f: \bf R \to R$ means the image of $ f$ is [i]in[/i] $ \bf R$, not that the image of $ f$ [i]is[/i] $ \bf R$.


Albatross has 25 monkeys. He feeds 20 monkeys a day. What is the probabilitity that one of his monkeys will have to go hungry for 3 days in a row.

Albatross has 25 monkeys. He dicides to to start a monkey show. He must choose to go to 8 towns out of 12 towns. In how many ways is this possible. Order does NOT Matter.

[quote="IntrepidMath"]Albatross has 25 monkeys. He feeds 20 monkeys a day. What is the probabilitity that one of his monkeys will have to go hungry for 3 days in a row.

Albatross has 25 monkeys. He dicides to to start a monkey show. He must choose to go to 8 towns out of 12 towns. In how many ways is this possible.[/quote]

  [hide]  1. $\frac{5}{25}^3 = \frac{1}{125} $

2.  Im not sure if the order of the towns they visit matters.... If it does the answer is  $P ( 12, 8 ) = \frac{12!}{4!}=19958400$  if order does not matter its  $C ( 12, 8) = \frac{12!}{8!*4!}=495 $[/hide]

[quote="ajai"]Someone please tell me if I did something wrong.

[hide]
I got the same for the first problem, but I'm not sure if combinations or permutations should apply to the second problem. So it could either be:

$\frac{12!}{8!*4!}=495$

or:

$12P8=19958400$[/hide][/quote]

 you've basically contradicted yourslef


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove the subgroup of $S_{7}$ generated by $(1234567),(26)(34)$ has order 168.

uhm.. just a question: does this problem come from automorphisms of a riemann surface of genus 3?

Hi.

Some ideas:

I will call the two elements a and b. The 7-cycle a generates a 7-sylow of $S_{7}$ and so it generates a 7-sylow of our subgroup U. If you conjugate a with b you get a 7-cycle that doesn't lie in $\langle a\rangle$, so we have at least 2 7-sylows in U. Sylows theorem tells us, that in fact there are 8 or 15 of them (these are the only numbers that are equivalent to 1 mod 7 and are divisors of 120, which is the number of 7-sylows n $S_{7}$).

$bab^{-1}a=(172)(365)$, which is an element of order 3. So the order of U is a multiple of $lcm(7,8,3)=168$ or of $lcm(7,15,3)=105$.

Maybe this helps...

uhm..
just another approach, suggested by a friend.
let's call $\sigma$ the $7$-cycle, $\tau$ the other generating permutation, and $H$ the subgroup they generate
$168$ is the order of $G = GL_{3}(\mathbb{F}_{2})$, so let's try to find an isomorphism between our group $H<S_{7}$ and $G$.
after a couple of attempts, one gives the following correspondence:
$1\mapsto \left(\begin{array}{c}1\\0\\0\end{array}\right), 2\mapsto \left(\begin{array}{c}1\\0\\1\end{array}\right), 3\mapsto \left(\begin{array}{c}1\\1\\1\end{array}\right), 4\mapsto \left(\begin{array}{c}0\\1\\1\end{array}\right)$, $5\mapsto \left(\begin{array}{c}1\\1\\0\end{array}\right), 6\mapsto \left(\begin{array}{c}0\\0\\1\end{array}\right), 7\mapsto \left(\begin{array}{c}0\\1\\0\end{array}\right)$.
and send $\sigma\mapsto\left(\begin{array}{ccc}1 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{array}\right), \tau \mapsto\left(\begin{array}{ccc}1 & 0 & 1\\0 & 1 & 0\\0 & 0 & 1\end{array}\right)$.
so you have injected $H$ in $G$.
but if you show that $G$ contains an element of order $3$ (as gockel did) and that $H$ contains a copy of dihedral $D_{4}$, you win.

uhm... :D ,ma_go,This problem is given for an example of simple group of order $168$.

Thank you for your nice solution.

My approach is like Gockel's,Because the subgroup contains an element of order 3,and a copy of $D_{4}$,thus $|H| \ge 168$,and obviously $H \subset A_{7}$,So we just have to should there is no element in $H$ has order 5,But I don't know how to show it?

Thus the another question is how to show $H$ is simple,or equivalently,how to show $GL_{3}(\mathbb F_{2})$ is simple?


Let $ a,b$ are coprime positive integers. Prove that there are infinitely many coprime positive integers $ m,n$ such that $ m \mid a^{n\minus{}1} \minus{} b^{n\minus{}1}$ and $ n \mid a^{m\plus{}1} \minus{} b^{m\plus{}1}$

bwoc, list all possible value for $ n$ and $ m$.  Then look at the equations in modular form and multiply them together.


quadratics
vector
linear algebra
matrix
function
algebra
polynomial

Hi everyone,

Here is my question let's $ Q_1$ and $ Q_2$ be two quadratic forms specified as follows :

$ Q_1(x_1,...,x_n)\equal{}a_1\plus{}B_1^t.Vect(x_1,...,x_n)\plus{}Vect(x_1,...,x_n)^t.C_1.Vect(x_1,...,x_n)$
$ Q_2(x_1,...,x_n)\equal{}a_2\plus{}B_2^t.Vect(x_1,...,x_n)\plus{}Vect(x_1,...,x_n)^t.C_2.Vect(x_1,...,x_n)$
where $ B_i$ is a vector and $ C_i$ a square matrix( and no assumptions on definitiveness of $ Q_i$).

Is it possible to express (if possible explicitly) $ Q_1$ as a function of any kind of $ Q_2,Q_3$ where $ Q_3$ is, in some sense to make precise, "independent" of $ Q_2$ ?

Thank's for any help or reference

PS :
I have a very partial result if if $ Q_1$ is positive definite and $ a_i$ , $ B_i$ are null (then $ Q_3$ is null), but was wondering if a general result was available.

$ Q_1,Q_2$ are not quadratic forms but polynomials of degree 2.
If $ n\equal{}1$,  $ Q_1(x)\equal{}a_1\plus{}b_1x\plus{}c_1x^2,Q_2(x)\equal{}a_2\plus{}b_2x\plus{}c_2x^2$. $ Q_1$ is a function of $ Q_2$ iff the 2 associated  parabolas have same axis of symmetry that is iff $ c_2b_1\minus{}b_2c_1\equal{}0$.
Now if you want $ Q_1$ as a function of the independant variables $ Q_2,Q_3$, then it is very easy:
you take $ Q_3\equal{}Q_1$.
Have you a clear idea about what you want ?

Salut vieux loup blanc  :lol:

Thanks for your interest, first you are right about $ Q_1$ and $ Q_2$ these are not a quadratic forms but a polynomial forms of degree 2.

By a change of variable I think I can replace $ Q_1$ by a real quadratic form (assuming it has degree 2 or 0 to on all its components).

Then if in this basis if $ Q_2$ is a quadratic form (i.e. the parabolas have the same symetry axis as you noted) everthing goes smoothly

The problem I really wanted to address was in fact motivated by a probablistic problem where I have two polynomial forms of (independent ) gaussian variables and I wanted to get conditionnal expectation of one form with respect to other one.

For the sympthetic case I mentionned earlier ( same symetry axis or independent coordinate) that's ok, so only the "linear" case remains I think (am I wrong ?)

So I wanted to get a representation in some way of $ X$ with respect to $ X^2$ but the more I think about it the more I find this task quite hard as I am really dumb in all algebra matters.

I have tried a series of even polynomials but after a moment of reflection this doesn't make sense, so  if you have any other ideas "je suis preneur"

thx again and see ya

In the end, I think that the simplest way to go is to write $ X$ as the difference of two polynoms of degree 2.
For example we have $ X\equal{}(X\minus{}\frac{1}{4})^2\minus{}(X\plus{}\frac{1}{4})^2$

this should do the trick but is only a small step before I can solve my global issue

see ya


[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that:
1)
\[ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\geq\frac {3}{2}\]
2)
\[ \frac {1}{2 \plus{} 3a^3} \plus{} \frac {1}{2 \plus{} 3b^3} \plus{} \frac {1}{2 \plus{} 3c^3}\geq\frac {3}{5}\]
3)
\[ \frac {1}{3 \plus{} 5a^4} \plus{} \frac {1}{3 \plus{} 5b^4} \plus{} \frac {1}{3 \plus{} 5c^4}\geq\frac {3}{8}\]
[/quote]
$ 1$ is simplified using the $ uvw$ technique. Upon expanding it becomes $ Aw^6 \plus{} Bw^3 \plus{} C \le 0$, where $ A \equal{} \frac{3}{2} > 0$, so we should prove it when $ a\equal{}b$ and when $ c\equal{}0$.

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that:
1)
\[ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\geq\frac {3}{2}\]

[/quote]

It is equivalent to :

$ \sum (a\minus{}b)^2 [abc(a\plus{}b\plus{}c)\plus{}a^2b^2\plus{}2abc^2] \geq 0$

Done !

:)


geometry
3D geometry
prism
rhombus
trigonometry
MATHCOUNTS
arithmetic series

Here is a PDF file I made with lots of useful [b]formulas[/b].

All rights are reserved. No reproduction of this material is allowed without written permission from the author.

I think you got the diagonals in a figure wrong. It's n(n-3)/2. I'll prove that to you if you want...

For example... if you have a triangle, which obviously has no diagonals, your formula (meant for adding consec. numbers starting at one) would give you three, while n(n-3)/2=0.

Your rectangular prism formula is wrong.  It looks like you copied over one of the cricle formulas by accident.

In the formula for the sum of an arithmetic series, you are missing a subscript n.

Yakko

Sorry guys...

I did this really quickly

this isn't anything wrong, but here are some stranger formulas:

(a+b)2 = a2 + 2ab + b2

a^3 + b^3 = (a+b)(a2 - ab + b2).

a^3 - b^3  = (a+b)(a2 + ab + b2).

Please check my last two formulas, they might be wrong

Yup.. you messed up the last one...
I think you should have put in (a-b) instead

The area formula for the rhombus applies to any quadrilateral with perpendicular diagonals. There are many more triangle area formulas.

$\text{Area}=\frac{ab\sin C}{2}=rs=\frac{abc}{4R}$

The sum of a sequence is called a series. The formula for arithmetic series is ambiguous. It can look like $\frac{n}{2\left(a_1+a\right)}$, when it's $\frac{n\left(a_1+a\right)}{2}$.
In the formula for geometric series, $\left|r\right|< 1$; otherwise you might apply it incorrectly and get negative solutions.

Include the formulas for sums of squares ($1^2+2^2+\cdots+n^2$, not $a^2+b^2$) and maybe cubes. These are not likely to come up in competition, though.

For Pascal's triangle, explain how you generate it, and mention binomial coefficients and maybe some formulas.

I prefer lowercase pi's ($\pi$), radical signs ($\sqrt$), $\times$ over $\text{x}$ and \frac over the division bar.

thank you hockey, the last one should be (a-b).

High school stuff is not for my brain right now, lets keep this a more middle school related thing, no log 9 or sin 65 or anything like that.

[quote="hockeyadi23"]I would appreciate it if you keep this to a mathcounts level aka no trig pls...

What's the formula for the sum of the first n cubes by the way?[/quote]

The sum of the first $n$ cubes is just $(1+2+3+4+5...+n)^2$

Look at the graphics thingies on the sidebars on the lefthand side, occasionally an explanation pops up for this...


everytime i try to put an avatar, there is always something like: this must be a jpg., gif., etc., or it will say only one avatar may be specified, or process failed. the picture is a jpg., and i know most people will rate this spam, and it doesn't have to do with classroom math, but still... :lol:

A mod should move this to the Questions, Suggestions, and Annoucements forum.

well, anyway, while it is in this forum, i'll just answer this question. maybe it is too big space.

It didn't have that much KB, but I don't know how to find the length or width, though.

[quote="AstroPhys"]Put the picture in some program that shows you the measurements, like Paint.[/quote]Or in Windows you can always right click and look at the Properties ;)

EDIT: Wait, this is an old topic? Oops.... Heheh. Maybe I should follow my own advice and check next time.

@ pingala - Please don't revive old topics.


analytic geometry
graphing lines
slope
calculus
calculus computations

$ x \equal{} (37 \plus{} 10y \minus{} 2y^{2})^{\frac {1}{2}}$.

Find $ \frac {dx}{dy}$ in terms of $ y$.

Hence find the equation of the tangent to the curve at the point $ (7, 3)$.

[I make this $ y\equal{}7x\minus{}46$.]

[b]I obtained a different answer for the tangent one[/b]

you get $ \frac{dx}{dy} \equal{} \frac{5 \minus{} 2y}{(37 \plus{} 10y \minus{} 2y^2)^{1/2}}$

At the point (7, 3), slope of tangent = $ \frac{dy}{dx} \equal{} \minus{}7$

So, the tangent should be $ y \equal{} \minus{} 7 x \plus{} 52$ .

Nora.91, Thanks for your reply.

When I substitute $ y\equal{}3$ into $ \frac{dx}{dy} \equal{} \frac{5\minus{}2y}{(37\plus{}10y\minus{}2y^{2})^\frac{1}{2}}$,
I get $ \frac{5\minus{}6}{7} \equal{} \frac{\minus{}1}{7}$.

Yes exactly,

so the slope is dy/dx = - 7

 :)

Do take note of  $ \frac{dy}{dx}$ and $ \frac{dx}{dy}$ :)

Thanks Nora.91 and IndoChina. I've made yet another silly mistake: I must have been thinking of the product of the slopes of two perpendicular lines being $ \minus{}1$.


Where can I find these competition question bookelts?
Can anyone email or help me find these?
My email is pokewiz88@yahoo.com

2004 National
2003 School
2002 School
2001 School
2000 School
1999 School
1998 School, Chapter, State, National
1997 School, Chapter, State, National
1996 School, Chapter, State, National
1995 School, Chapter, State, National
1994 School, Chapter, State, National
1993 School, Chapter, State, National
1992 School, Chapter, State, National
1991 School, Chapter, State, National
1990 School, Chapter, State, National
1989 School, Chapter, State, National

how can you purchase the 1989 contests on mathcounts.org ?

[quote="shinwoo"][quote]Purchase them on mathcounts.org[/quote]

but don't you have to pay for it?[/quote]

Hence the word "purchase"  ;)

I know we all don't want to spend money, but the truth is, Mathcounts makes money off of their old contests, and it would be illegal to copy and send them to others.

ooooooooooo............

don't want to get caught.....

[quote="Silverfalcon"]I thought it was available.[/quote]

I called the vendor. They only have 1999 and later

They used to sell them all back to 1989.  I think I might have bought the last copy though as even later that same year I heard others couldn't get them :(

Do you have a coach that would have a collection of old contest?  Unless you school is new to MC they should have the old school tests lying around.

Good Luck

This is a good site that has '00-'04 states and chapters.

http://goodlymaths.0catch.com

My first post was this same question too!(almost) :cool:


Can someone please show me how to do this problem?  It seemed impossible for me to count all of the regions.


Prove that for every positive integer $ k$ there exists a sequence of $ k$ consecutive positive integers none of which can be represented as the sum of two squares.

Let $ \{ q_i \}$ be an enumeration of the primes congruent to $ 3 \bmod 4$ (recall that there are infinitely many such primes).  The system of congruences

$ x \equiv q_1 \bmod q_1^2$
$ x \plus{} 1 \equiv q_2 \bmod q_2^2$
...
$ x \plus{} (k\minus{}1) \equiv q_k \bmod q_k^2$

has a unique solution $ \bmod q_1^2 q_2^2 ... q_k^2$, and the integers $ x, x\plus{}1, ... x\plus{}(k\minus{}1)$ are all divisible by a square congruent to $ 3 \bmod 4$ with multiplicity $ 1$.  It's well-known that no such number is the sum of two squares.


geometry
3D geometry
tetrahedron
geometry unsolved

Let $T$ be an arbitrary tetrahedron satisfying the following conditions:

[b]I.[/b] Each its side has length not greater than 1,
[b]II.[/b] Each of its faces is a right triangle.

Let  $s(T) = S^2_{ABC} + S^2_{BCD} + S^2_{CDA} + S^2_{DAB}$. Find the maximal possible value of $s(T)$.


AMC
AIME
geometry
3D geometry
octahedron
ratio
rotation

Given that $ O$ is a regular octahedron, that $ C$ is the cube whose vertices are the centers of the faces of $ O,$ and that the ratio of the volume of $ O$ to that of $ C$ is $ \frac mn,$ where $ m$ and $ n$ are relatively prime integers, find $ m \plus{} n.$

http://www.artofproblemsolving.com/Wiki/index.php/2005_AIME_II_Problems/Problem_10
In the first solution of the above link, why does $ AF \equal{} s\sqrt2$? In what ways are a octahedron symmetrical?

Supposing I have a uniformly white octahedron, can I pick it up, rotate it, and put it down to get an identical figure relative to me where up and down matters? ... I know I'm not phrasing this right...

Hes talking about the symmetry of the octahedron, in which it is two regular square pyramids built on top of eachother where the base acts as the plane of reflection, thus everythingis symetrical.

where did you get tetrahedron from?

Once you get height of one regular square pyramid to be $ \frac {\sqrt {2}s}{2}$, the height of the octahdedron must be $ \sqrt {2}s$.

Once you take the cross section of the triangle with two of the midpoints as vertices, you know it is isoscles since it is symmetric.

This problem is actually very simple once you realize this and the fact that the median is 2:1 so use similar triangles and you're done

[quote="aufha"]Whoops. I meant octahedron.

"Once you take the cross section of the triangle with two of the midpoints as vertices, you know it is isoscles since it is symmetric."

Which two midpoints as vertices? Why do we want to know about midpoints?

Also, what figure did you draw when you were doing this problem?[/quote]

Ahh centroids, the cubes vertices are cetnroids (sorry midpoints, centrods gets confusing somethimes), you want to compare the edge length of the cube with something of the octahedron.  And its going to be 1/3 the height of the entire octahedron.

Well the triangle I drew had two of the vertices as centroids of adjacent faces and the other vertex as the midpoints of the edge joining the two faces (this is hard to explain)


dandu-se un patrulater (cu unghiuri < 180 grade), cum pot afla dreptunghiul orizontal (adica cu laturile paralele cu Ox si Oy) cu arie maxima "inscris" in el (de fapt, nu este obligatoriu sa aiba toate cele 4 colturi pe laturile patrulaterului, nici nu cred ca este posibil in toate cazurile) ?

[quote="Valentin Vornicu"]asta suna mai mult a problema de informatica decat de matematica :? :?[/quote]
Într-adevăr, mie îmi trebuie soluţia într-un program; dar totuşi, tot o problemă de geometrie rămâne.

Oricum, am găsit soluţia; pentru cei interesaţi:
http://cgm.cs.mcgill.ca/~athens/cs507/Projects/2003/DanielSud/complete.html


[quote="tamal banerjee"]here is some options for answers :
$ (a)\dfrac{1}{2}(\dfrac{10000}{10001} \plus{} \dfrac{5050}{10202})$
$ (b)\dfrac{1}{2}(\dfrac{10000}{10001} \minus{} \dfrac{5050}{10202})$
$ (c)\dfrac{1}{4}(\dfrac{10000}{10001} \plus{} \dfrac{5050}{10202})$
$ (d)\dfrac{1}{4}(\dfrac{10000}{10001} \minus{} \dfrac{5050}{10202})$ :maybe:[/quote]

So none of these four choices.

Obviously, there is a mistake in your problem statement :  $ \frac{k}{10k\plus{}8}>\frac 1{20}$  and so $ \sum_{k\equal{}1}^{100}\frac{k}{10k\plus{}8}>\frac{100}{20}\equal{}5$

And the four choices you proposed all are less than $ 1$


Please help me. I need a proof of that theorem:
Weisstein, Eric W. "Lukács Theorem." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/LukacsTheorem.html

Sorry to disappoint you but this problem solves automatically one problem from Polish Olympiad which is still ongoing, so topic blocked till 4th December 2006.


inequalities
search
national olympiad
algebra

Let $ a,b,c>0$ be real numbers. Prove that
\[ \frac 1{a(1+b)}+\frac 1{b(1+c)}+\frac 1{c(1+a)}\geq \frac 3{ 1+abc}. \]
[i]Greece[/i]

[b]EDIT BY DARIJ:[/b]

Please use this topic to discuss the [b]sources and history[/b] of the above inequality,
and use http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 to discuss the inequality itself (i. e. different proofs and generalizations).

See this http://www.mathlinks.ro/Forum/viewtopic.php?t=161059

and I have no comments . How some people give a problem very-very-well-known in a big contest  as BMO

yeah, like functional eqaution which was twice on BMO?

Now I know, where I saw this nice method...
What a pity, it's just the same problem ;/

[Moderator edit: The "nice method" referred to here is Mzdanovicz's solution in http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 .]

[quote="Arne"]I can't believe they give this problem at a BMO. It's just so incredibly well known  :huh:[/quote]It was problem 1. I assume they knew it was kind of known but they put it there to be the easy problem in the competition ...  :?

That can be said of 30% of the IMC problems...

[quote="Arne"]What makes it even worse is the fact that this is VERY hard for people who've never seen it before.

It's just unfair![/quote]Yeah, that can also be said about usage of ISL problems in TSTs around the world, given that some of the students might know some of the problems ahead ...  :?

I belive we got very close to that point where almost all problems are known, and those which aren't are sent to the IMO :D

[quote="Arne"]What makes it even worse is the fact that this is VERY hard for people who've never seen it before.

It's just unfair![/quote]

I can NOT believe that. Why don't they hire some inequality experts from Mathlinks?  :lol:  ;)

I guess all team leaders knew this problem. Why didn't they stop it from being selected? :(

That can tell us something about Balkan Olympiad. In the IMO, this way is forbidden.

Is not this problem from [i]Old and new inequalities[/i]?

[quote="Arne"]Might be, although the problem is quite old.

I think the BMO is making a joke of itself this way...[/quote]
I got feeling long time ago, that competition among only a few countries, like BMO ,has been a joke ever since.
However,reasons like making a friends and visiting beutiful parts of Mediteranian   make it sufficient to keep it going every year. :)

I think when the same question in the BMO appeared twice in different years, eyebrows must have been raised.

This question appeared last year on the UK Correspondence Course and I assume it therefore a well publicised inequality.

Still I guess these regional Olympiads should also be about making friends and meeting new people. If it were just about the questions, then why isn't the whole thing done in two days?

Of course, but I just can't believe the problem committee cannot find any better or new questions.

Maybe... *starts writing the plot of a new detective novell*

this question just got proposed

all team leaders knew that their students knew the question

but none knew it was well-known to the other countries too

so they let is pass  :diablo: :D

[quote="Valentin Vornicu"]Let $a,b,c>0$ be real numbers. Prove that \[ \frac 1{a(1+b)} + \frac 1{b(1+c)} + \frac 1{c(1+a)} \geq \frac 3 { 1+abc}.  \][/quote]

This problem appeared some years ago in Crux Mathematicorum. It was proposed by Mohammed Aassila.

Might be, it appeared like everywhere. I'm sure you can find at least 10 threads on this forum about this problem.

Guys if you want to know,this problem even was proposed on Armenian National Olimpiad 1996 :lol: ,I think they have found most pupular ineq. in the world(my be after AM-GM :D ) and proposed it for BMO :oops: .

[quote="stergiu"]I have this inequality in my book on Inequalities(written by Skombris and me, in Greek). As source we give there  << Komal 1993 >> , so it was known before Aassila sent it to CRUX , vol 1998. This nice solution, sent by Silouan , is to be found there, too(by a chinese).Of course the solution was discused in ML very wide.

   My opinion is that the problem should not be proposed in the BMO 2006 , although very good. problem. Competition problems must be original( this is very difficult) or clever and nice variations of nice problems.

 Babis[/quote]

Dear Babis Stergiou,

I have your book, and it unfortunately contains several imprecisions. The inequality appeared later on in Komal (but not in 1993). This inequality was first proposed by the Russian mathematician D. P. Mavlo in the Bulgarian journal Mathematica (number 4, 1987). A little later, it was published in Kvant (problem M1136, number 11-12, 1988). It was also a problem given at the training of the USA IMO team in 1993, and also was used in the training of the South African IMO team in 1994,....etc..etc

actually the problem appears also in the book "mathematical miniatures" by T.Andreescu

The rules were broken since the moment when the leaders allowed two problems from the same country!

(If you want more check the 4th problem)

To: Albanian Eagle

The Leaders do not know the origin of the problems. This is a rule of IMO too. The proposing countries are revealed after the final selsction. In this BMO there was a real danger of having three problems proposed by the same country since the only Combinatorics problem was proposed by Greece! The Problem finally was eliminated with the intervention of the Selection Committee for exactly this reason. I got all this on the grapevine. I was not there!

Dear albanian eagle
1 congratulations
2 which problem is written in this book that you mention cause i have it ?
3 could you please see and think what i've written above and answer to me as soon as possible?
thank you

I'm quite sure that this inequality was proposed by Sotiris Louridas. What I do not understand is the fact that why inequalities are proposed by someone at BMO and JBMO and by someone at IMO??? :?

Wrong. The real author told me that he started from another known inequality and then transformed it by a to 1/a, b to 1/b and c to 1/c and arrived to this (known) inequality! The Problem Selection Committee and the Jury detected it after the Competition! (the problem was being used in training at MIT).

[b] Michel [/b],

  :)  thank you for your nice information.You re better informed. I will use the history of the problem in the future.It is sure now that my source was not the older. I think ,  I have seen Bulgarian Olympiads 1987 and the name Mavlo is known to me.He must have proposed an ieq. problem in Bulgarian olympiad. Any way , probably - as Dulac said -  the greek proposer created the same problem trying to create a new problem. This hapens often. If Louridas is the proposer(Lupu wrote this) , then I'm sure that he gave it as a new problem - in his opinion. He never likes - is a friend -  to propose discussed proplems in competitions.
   [u]Babis[/u]

Louridas is [b]definitely[/b] not the author of the problem! The perpetrator is well-known to the Greek Math Competitions scene.

[quote="DuLac"] Source: Deep throat[/quote] lol @ hidden message in post #33 :D

Also by eding 3/1+abc to both parts of the inequality and regrouping by AM-GM you can found the solution

HELLO GUYS HAS ANYBODY SEEN MY SOLUTION THAT I VE WRITTEN ABOVE? PLEASE SEE IF IT IS CORRECT AND SEN ME A PERSONAL MESSAGE OK? THANK YOU FOR YOUR UNDERSTANDING

Please do not shout.

Your proves looks not entirely rigorous to me, although it might be easy to fix (no time to investigate).

You define y as the best constant, and you conclude $y=y=6\sqrt{abc}$, which is not a constant.

MY FRIEND
FISRTLY I DO NO SHOUT I WRITE THIS WAY  BECAUSE I AM USED TO
SECONDLY WHAT DOES RIGOROUS MEAN?
THIRDLY I DID NOT SAY THAT WE TRY TO FIND Y AND AFTERWARDS TO FIND THE LEAST POSSIBLE VALUE FOR X WHICH IS GREATER THAN THE NUMBER ASKED IN THE PROBLEM OK?
THANK YOU PLEASE ANSWER TO ME WHEN YOU SEE THAT AND FIND IF THIS IS CORRECT OR WRONG OK?
THANK YOU

If you ever make another post using capitals only, it will be deleted. :mad:

Sorry my friend you do not need to bwe so angry  :blush:  now could you please see my solution and tellme if it is correct ? i have a kind of battle with someone and i have to know from people like you if my solution is correct  :D
Thank you

[quote="ambros"]Guys i wish to see if this solution is correct and tell me as soon as possible please
thank you
Let  $\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq x$  where x is the best real .

Let  $a(b+1)+b(c+1)+c(a+1)\geq y$  where  y is the best contsnt .

By multipliing these we have  and using B-C-S
$[\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}][a(b+1)+b(c+1)+c(a+1)]\geq xy\geq 9$

From Cauchy -AM-GM we have $a(b+1)+b(c+1)+c(a+1)\geq 6\sqrt{abc}$

so  $y=6\sqrt{abc}$  and so  $6x\sqrt{abc}\geq 9$  and so

$x\geq\frac{3}{2\sqrt{abc}}\geq\frac{3}{abc+1}$[/quote]

I don't think it works. Let $A = \frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}$ and $B = a(b+1)+b(c+1)+c(a+1)$.

Just because $AB \ge xy$ and $AB \ge 9$ does not imply that $xy \ge 9$. Even if you specify that $x,y$ are the "best" constants, you have $y = 6 \sqrt{abc}$, which is not a constant, so I don't think it holds.

[quote="Peter VDD"]So now the test for paladin, what if we want to know it for all numbers for which $abc=K$ for a certain constant $K$? ;)

[ because if that's true for all constants K, we have proven the original problem ][/quote]

Hmm... well we still have to show that $y = 6\sqrt{K}$ would be the best constant, right? Which is equivalent to showing that there is always an equality case on the AM-GM for any $K$. But I don't think this is true, because the AM-GM has equality only when $a = b = c = 1$ so it only works for $K = 1$. Correct me if I'm wrong, but the proof doesn't seem to work.

Excuse me guys but i want to clarify two things
frist we do not imply xy>=9 we claim  it and we search for which x,y is in charge that xy>=9
secondsice y is got form cauchy we know that for each a,b,c  $y=6\sqrt{abc}$ so we get that for wach a,b,c $6x\sqrt{abc}\geq 9$ and so on
now that i was clarified could you please resee that and tell me ? thank you for all the time that you spend for this i am obliged

Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive real numbers . Prove that $$\frac{1}{a_1(1+a_2)}+\frac{1}{a_2(1+a_3)}+\cdots+\frac{1}{a_{n-1}(1+ a_n)}+\frac{1}{a_n( 1+a_1)}\geq\frac{n}{1+a_1a_2\cdots  a_n}.$$
[url=https://artofproblemsolving.com/community/c6h1829758p13013172]h[/url]

[quote=sqing]Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive real numbers . Prove that $$\frac{1}{a_1(1+a_2)}+\frac{1}{a_2(1+a_3)}+\cdots+\frac{1}{a_{n-1}(1+ a_n)}+\frac{1}{a_n( 1+a_1)}\geq\frac{n}{1+a_1a_2\cdots  a_n}.$$[/quote]
https://artofproblemsolving.com/community/c6h161059p10340314

Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive real numbers . Prove that $$\frac{1}{a_1(n-2+a_2)}+\frac{1}{a_2(n-2+a_3)}+\cdots+\frac{1}{a_{n-1}(n-2+ a_n)}+\frac{1}{a_n( n-2+a_1)}\geq\frac{n}{n-2+a_1a_2\cdots  a_n}.$$


inequalities
geometry
3D geometry
algebra
linear equation

Imagine a box with dimensions measured in units. All dimensions are integers.
The box is hollow on the inside but has a thickness of 1 unit on all sides.
Given dimensions a,b,c you can find the total volume of thick material: abc - (a-2)(b-2)(c-2)

Find all such boxes that satisfy the property that the thickness volume (material used to make the box) is equal to the hollow volume (air inside the box).

Hmm... I haven't solved it completely, but here it is:
We want to find such $a,b,c$ for which $abc=2(a-2)(b-2)(c-2)$.
If $a,b,c>9$ then we can prove that $\sqrt[3]{2}(a-2)>a$ and so for $b,c$. Product of these three inequalitis give us $2(a-2)(b-2)(c-2)>abc$. So at least one of $a,b,c$ must be smaller than $10$. We can now try $a=2,3,\ldots,9$ and using similar approach (inequalities). For example for $a=5$ we have $5bc=6(b-2)(c-2)$. Now for $b,c>22$ we can prove $\sqrt{\frac{6}{5}}(b-2)>b$ and $\sqrt{\frac{6}{5}}(c-2)>c$. Product of them give us $6(b-2)(c-2)>5bc$. So at least one of $b,c$ is smaller than $23$ (let $b$ be that one). So we have finite number of cases for $b$, now we have linear equation for $c$.

Computer can easily compute it all, but it's hard to do without it.
Have you got nicer solution?

Well, the cube with a,b,c=(2^(4/3))/((2^(1/3))-1) works...(sorry about the way i wrote it...unable to put in mathtype stuff)

[quote="Hexaditidom"]All dimensions are integers.
[/quote]

Maybe to make the question easier, just give the total number of boxes that work. It's less than 100.

abc=2(a-2)(b-2)(c-2)

An 8x10x12 box works... :lol:
You know that all of the numbers have to be greater than 4.
5x22x24 works too because (22)(24)=(5/6)(20)(22)
There's gotta be an elegant solution
For instance, for a=5, you know that (b-2)(c-2)(5/6)=bc.  Then you could find, with proof, all integer solutions for that


inequalities
algebra
polynomial
Cauchy Inequality

Let $x,y,z$ be non-negative numbers.
$x+y+z=1$
Prove:
$\frac{1}{x+y^2}+\frac{1}{y+z^2}+\frac{1}{z+x^2}>\frac{13}{2}$

Would substitution work, such as $x=\frac{1}{a}$

Wait, where does $f(0,0.35,0.65)$ come from??

[quote="luimichael"]So, are we proposed to show that $f(x,y,z)\geq f(\frac 13, \frac 13,\frac 13)$ ?[/quote]

and why do we have to prove that :?:  :blush:

[quote="Bourne"]Let $ x,y,z$ be non-negative numbers with $ x \plus{} y \plus{} z \equal{} 1$ and $ yz \plus{} zx \plus{} xy > 0.$ Prove that

$ \frac {1}{x \plus{} y^2} \plus{} \frac {1}{y \plus{} z^2} \plus{} \frac {1}{z \plus{} x^2} > \frac {13}{2}.$[/quote]Using the Cauchy inequality,

$ \left(\sum_{cyc}{\frac {1}{x \plus{} y^2}}\right)\left[\sum_{cyc}(2x \plus{} 3y \plus{} 6z)^2(x \plus{} y^2)\right]\geq \left[\sum_{cyc}(2x \plus{} 3y \plus{} 6z)\right]^2.$

Id est, it remains to prove that

$ 242\left(x \plus{} y \plus{} z\right)^4 > 13\sum_{cyc}(2x \plus{} 3y \plus{} 6z)^2\left[x(x \plus{} y \plus{} z) \plus{} y^2\right].$

But $ 242\left(x \plus{} y \plus{} z\right)^4 \minus{} 13\sum_{cyc}(2x \plus{} 3y \plus{} 6z)^2\left(x^2 \plus{} xy \plus{} xz \plus{} y^2\right)$

$ \equal{} 73\sum_{sym}{x^4} \minus{} 176\sum_{cyc}{x^3y} \plus{} 331\sum_{cyc}{xy^3} \minus{} 121\sum_{sym}{y^2z^2} \plus{} 135\sum_{sym}{x^2yz}$

$ \equal{} \frac {3}{4}\sum_{cyc}{\left(2x^2 \minus{} 6y^2 \plus{} 11yz \minus{} zx\right)^2} \plus{} \sum_{cyc}{\frac {x^2}{10}\left[107(4y \plus{} z \minus{} 2x)^2 \plus{} 2x^2 \plus{} 239yz\right]}$

$ \plus{} \sum_{cyc}{\frac {yz}{5}\left[471(y \minus{} 2z)^2 \plus{} 2yz\right]} > 0.$[quote="Bourne"]$ f\left(\frac {1}{3},\frac {1}{3},\frac {1}{3}\right) \equal{} 6.75 > f(0,0.65,0.35) \equal{} 6.518505\cdots > \frac {13}{2}.$[/quote]The minimum of $ \frac {1}{x^2 \plus{} y} \plus{} \frac {1}{y^2 \plus{} z} \plus{} \frac {1}{z^2 \plus{} x}$ is $ 6.5175776615\cdots,$

the unique real root of the following irreducible polynomial over $ \mathbb{Q}:$

$ 12k^7 \minus{} 136k^6 \plus{} 436k^5 \minus{} 436k^4 \plus{} 612k^3 \minus{} 1760k^2 \minus{} 900k \minus{} 125;$

which is attained when $ x \equal{} 0,y \equal{} 0.35491847725\cdots$ is the unique real root of the following irreducible polynomial

$ y^7 \minus{} y^6 \plus{} y^5 \minus{} 3y^4 \plus{} 9y^3 \minus{} 9y^2 \plus{} 5y \minus{} 1,$

$ z \equal{} 0.64508152274\cdots$ is the unique real root of the following irreducible polynomial

$ z^7 \minus{} 6z^6 \plus{} 16z^5 \minus{} 22z^4 \plus{} 22z^3 \minus{} 16z^2 \plus{} 8z \minus{} 2.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=160017


Euler
limit
advanced fields
advanced fields theorems

Does anyone have more information about this constant? I don't even know if the spelling of topic is correct... :oops:

$(lim_{n\rightarrow \infty}1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n})-ln\ n$

It's the
[url=http://en.wikipedia.org/wiki/Euler-Mascheroni_constant]Euler[/url]-
[url=http://mathworld.wolfram.com/Euler-MascheroniConstant.html]Mascheroni[/url]
[url=http://planetmath.org/encyclopedia/MascheroniConstant.html]Constant[/url].

By the way, you should write that limit as
$\lim_{n \to \infty}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}-\log(n)\right)$.
As you wrote it, it looks like you only take the limit of the sum, not the log.

thank you very much :) , but is that $log(n)$ or $ln(n)$ ?

Generally, in higher mathematics, $\log(n)$ is used to mean $\ln(n)$ and if you want any base other than the natural one you have to specify it.


Err... $\frac{1}{r^{n}}\cdot n$. Is there a typo in the problem statement? :P

EDIT: The problem is still pretty trivial... should be in HSB or sth.

I don't know, but this is an awesome problem.

So yeah, that is the right answer.

[quote="13375P34K43V312"]No, it's a reciprocal. If |r|<1, $r^{n}$ goes to 0 and the reciprocal tends to infinity

pwned[/quote]

Oh right, I was being dumb.

uhh...direct application of geometric series formula...this belongs in HSB

i dont think we are allowed to use that formula..if I havent misinterpreted the question..

Yes but if you know the proof for the formula of a geometric series the question becomes VERY simple... I don't know enough to agree or disagree with whether or not it should be HSB though.

Take a look at the other topics in this forum...

True it is more HSB once you think about it...

[quote="nutz_for2.718281828"]First point: You may NOT use any formula!!!

Second point: Assume that r is greater than 1.

Third point: Just because I make these points doesn't mean that anyone is WRONG, but this statement doesn't mean that you are RIGHT, either.[/quote]

but that doesn't change the fact that the formula, and consequently, the proof for that theorem, are common knowledge

[quote="nutz_for2.718281828"]Evaluate

$\sum_{n=1}^{\infty}\frac{1}{r^{n}}$

in terms of r, without using any formula whatsoever. :evilgrin:[/quote]
[hide="lemma"]$1+r+r^{2}+r^{3}+...=x$
$1+r(1+r+r^{2}+r^{3}+...)=x$
$1+rx=x$
$1=x(1-r)$
$\frac{1}{1-r}=x$[/hide]
You just got owned, buddy.

or...
[hide]We can prove the formula by simply doing the long division $\frac{1}{1-r}$, and get $1+r+r^{2}+\cdots$.[/hide]


Prove that:$\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+d)}+\frac{1}{d(1+a)}\geq 2$
where $a;b;c;d$ are positive real numbers satisfying $abcd=1$

Ban post phan nhieu la bai da co trong sach cua Hung, hoac da co ngay trong dien dan mathlinks. Vi vay nen doc ky truoc ky post nha.


Dear MATHCOUNTS Students and Teachers:

For several years mathleague.org has worked to build at the high school level a program parallel to what MATHCOUNTS offers at the junior high level.  mathleague.org organizes local and state level math contests in several states using a format that incorporates events former MATHCOUNTS students are familiar with.

After a successful pilot last year where we released a set of tests at several contests in January ([b]these tests are available for free download now on our website![/b]), this year we are expanding our offerings and plan to release four sets of tests identical to MATHCOUNTS format – one each in December, January, March, and April.  Students participating in contests at host schools will have an opportunity to take MATHCOUNTS-style tests in a competition setting prior to and during the official MATHCOUNTS season.  We are currently soliciting schools to host contests, and we will post host schools on our website as they become confirmed.  Non-hosting schools that are interested in participating in one of our contests can then contact the contest director of the nearest contest.

To get more information about our junior high program, you may visit our website at [url]http://mathleague.org/jh.php[/url].  There you will find information on how to host a contest, how to participate in one of the contests, or how to purchase a copy of the tests after each contest is over.  Students, be sure to pass this information along to teachers who may be interested in hosting a contest or purchasing practice materials; Please do not hesitate to contact me if I can answer any questions for you.

[Note:  mathleague.org is not affiliated with MATHCOUNTS; tests are written by mathleague.org staff]

Tim Sanders
President, mathleague.org
tim@mathleague.org

This is legal, right?

Because I've seen some that we were not suppose to look at them yet..

DonkeyKong -
yeah, everything we are doing is legal; we have separate testwriters writing our tests, and we have no affiliation with MATHCOUNTS.  As for these tests you're not supposed to have seen, they aren't ours - we are still in the process of writing our tests this year.  please email me (tim at mathleague d0t org) and tell me more about the tests you've seen; i'd like to help you figure out whether this should be a cause for concern for you..


ratio
geometry
parallelogram
perpendicular bisector
geometry proposed

Yetti told me that it was a typo . Now I fixed it .Is it right now yetti  ??
Try to solve it.

You did not fixed the problem.

[quote="silouan"]Let a circle $C_1$ and its chord $BC$. Let another circle $C_2$  tangent internally to the  circle $C_1$  at $L$  and tangent to  $BC$  at the point  $K$. Let  $CC'\bot KL$  with  $C'\in\ KL$ and $C'M//\ LC$ where  the point $M$  belongs to the arc $LB$  of the circle $C_1$. Prove that  the ratio  $\frac{CC'}{C'M}$  is constant as the the circle $C_2$ varies in the arc $BC$[/quote]
Let A be the midpoint of the arc BC of the circle $\mathcal C_1$ opposite to the point L and Z the midpoint of the arc BC opposite to the point A. Let $t_1, t_2$ be tangents from A to the circle $\mathcal C_2.$ Invert the figure in a circle (A) centered at A and with radius AB = AC. The circle $\mathcal C_1$ passing through the inversion center A is carried into the line BC, the line BC into the circle $\mathcal C_1,$ and the tangents $t_1, t_2$ passing through the inversion center A are carried into themselves. It follows that the circle $\mathcal C_2$ is carried into itself. Hence, the tangency point $L \equiv \mathcal C_1 \cap \mathcal C_2$ is carried into the tangency point $K \equiv BC \cap \mathcal C_2$ and the other way around. Thus the points A, K. L are collinear and consequently, the angle $\angle C'LC \equiv \angle KLC \equiv ALC = \angle AZC$ is fixed. Since the angle $\angle CC'L = 90^\circ$ is right, the triangles $\triangle CC'L$ remain similar for all circles $\mathcal C_2.$ Let the parallel to CL through C' meet the parallel to CC' through L at a point N. CC'NL is a parallelogram and the triangles $\triangle NLC' \cong \triangle CC'L$ are congruent. Thus the parallelogram CC'NL also remains similar for all circles $\mathcal C_2$ and consequently, the ratio $\frac{CC'}{C'N}$ is constant. Since the point C is fixed and the point L moves on the circle $\mathcal C_1,$ the locus of points N is also some circle $\mathcal C_3.$ When the circle $\mathcal C_2$ degenerates into the point C, the parallelogram CC'NL also degenerates into the point C. Hence the locus circle $\mathcal C_3$ passes through the point C. Let $P_0$ the midpoint of $CZ.$ Let $AP_0$ meet the line BC at $K_0$ and the circle $\mathcal C_1$ at $L_0$. Let $CC_0 \perp K_0L_0$ with $C_0 \in K_0L_0.$ Since $AZ$ is a dimeter of the circle $\mathcal C_1$, $K_0L_0 \equiv AL_0 \perp L_0Z,$ so that $CC_0 \parallel L_0Z.$ Since in addition, the diagonal $C_0L_0$ cuts the diagonal $CZ$ in half, $CC_0ZL_0$ is a parallelogram, with the angles $\angle CC_0L_0 = 90^\circ,\ \angle C_0L_0C \equiv AL_0C = \angle AZC.$ This means that the point Z lies on the locus circle $\mathcal C_3.$ Thus the circles $\mathcal C_1, \mathcal C_3$ have at least 2 common points C, Z. Hence, they are either identical or they intersect at these 2 points. Returning to the general parallelogram CC'NL, its diagonals CN, C'L cut each other in half at a point P. The normal to CN at P passes through the center of the locus circle $\mathcal C_3.$ When L coincides with Z, the lines BC, CC' coincide, the points K, C' both coincide with the midpoint Q of the chord BC, and the parallelogram diagonal intersection P coincides with the midpoint R of the segment QZ. Since BC is a chord of $\mathcal C_1,$ R is not the center of this circle. Since the normal to CR at R, passing through the center of the locus circle $\mathcal C_3,$ is different from the perpendicular bisector of the chord BC, it does not pass through the center of the circle $\mathcal C_1.$ It follows that the circles $\mathcal C_1,\ \mathcal C_3$ are different. Since $M \in \mathcal C_1,\ N \in \mathcal C_3$ and since the ratio $\frac{CC'}{C'N}$ is constant, the ratio $\frac{CC'}{C'M} = \frac{CC'}{C'N} \cdot \frac{C'N}{C'M}$ could be constant only if the ratio $\frac{C'N}{C'M}$ was constant, in other words, if C' was a similarity center of the circles $\mathcal C_1,\ \mathcal C_3.$ Since AC is fixed and $CC' \perp KL \equiv AC',$ the locus of points C' is a circle $\mathcal C_4$ with diameter AC. Let the line $C'N \equiv C'M$ meet the circle $\mathcal C_4$ again at a point S. Since the angle

$\angle CC'S = \angle C'CL = 90^\circ - \angle C'LC = 90^\circ - \angle AZC = \angle CAZ$

is fixed, the point S is a fixed point of the line $C'M \equiv C'N,$ hence the point C' is not fixed and consequently, it cannot be identical with a fixed similarity center of the fixed circles $\mathcal C_1,\ \mathcal C_3.$ As a result, the ratio $\frac{CC'}{C'M}$ [color=red]is not constant[/color].

Sorry from all mathlinkers who try the problem :blush:    :blush:
My english are awful . :mad:
I wrote <<arc>>  except for  <<chord>>

Now try the problem

[quote="silouan"]Let a circle $C_1$ and its chord   $BC$ .Let another circle $C_2$  tangent internally to the  circle $C_1$  at $L$  and tangent to  $BC$  at the point  $K$ . Let  $CC'\bot KL$  with  $C'\in\ KL$  and  $C'M//\ LC$  where  the point $M$  belongs to the chord $LB$  of the circle $C_1$. Prove that  the ratio  $\frac{CC'}{C'M}$  is constant as the the circle $C_2$  varies in the arc $BC$[/quote]
Now the problem is correct :)
I think you know the fact that $\angle{BLK}=\angle{CLK}=\alpha$ for every circle $C_2$,as we have that $MC'||CL$=>$\angle{MC'L}=\angle{MLC'}=\alpha$=>
\[ MC'=\frac{LC'}{2cos\alpha}  \]
=>\[ \frac{CC'}{C'M}=2cos\alpha\frac{CC'}{LC'}=2cos\alpha*tg\angle{CLK}=2cos\alpha*tg\alpha=2sin\alpha  \]=>
 \[ \frac{CC'}{C'M}=const  \]
 ;)

[quote="silouan"][quote="Tiks"]I think you know the fact that $\angle{BLK}=\angle{CLK}=\alpha$ for every circle $C_2$,as we have that $MK||CL$=>$\angle{MCL}=\angle{MLC}=\alpha$ [/quote]

I don't know the first fact you have mentioned. Could you prove it please  ??
Also in the second line you have some typos I think .
We have $C'M//\ CL$  and not  $MK//\ CL$ . Please correct the post and give me the proof of the first theorem you used.[/quote]
Sorry for typos,I have edited my post :) .
For the fact let that the point O and O' is respectively centers of the cirecles $C$ and $C'$,and $F\in{LK}\cap{C}$,then we have that the points $O,O'$ and $L$ lie in the same line =>$\angle{OLF}=\angle{O'LK}$=>$\angle{OFL}=\angle{O'KL}$=>$O'K||OF$=>$OF\bot{BC}$=>$F$ is the centre of the arc $BC$=>$\angle{BLK}=\angle{CLK}$ ;) .
I hope you will get this solution. ;)

Thank you .In fact I found a solution for this .
Here it is .Let $BL$ cuts the circle $C_2$ at $P$  and $CL$  cuts the $C_2$ at $S$

Then by the power-point theorem and the similarity of the triangles $LSP$ and $LCB$ we have that  $\frac{KC}{KB}=\frac{LC}{BK}$   so $LK$ is bisector . ;)

But your solution has also her beauty  . :D


algebra
polynomial
superior algebra
superior algebra unsolved

Let $\alpha=\sqrt{2}+\sqrt[3]{2}$.  Find a polynomial $p(x)\in \mathbb{Q}[x]$ such that $p(\alpha)=0$.  Prove that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{2})=\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$.

I think I know how to prove the second part, but is there an 'easy' approach to finding $p(x)$?

Very interesting.  I don't recall ever seeing this approach before.  I understand the concept of $k$th roots of unity, but could you explain the purpose superscripts $i$ and $j$?  Thank you so much!

perhaps this makes it more clear: the conjugats are $u \sqrt{2}+v \sqrt[3]{2}$ where $u$ is a second root of unity and $v$ is a third root of unity.

I guess I'm having a hard time understanding why/the logic behind using the second and third roots of unity.

another possible approach: $\alpha-\sqrt2 = \sqrt[3]2$, then $\alpha^{3}+6\alpha-2=(3\alpha^{2}+2)\sqrt2$, so finally, squaring both sides you should get the same polynomial.
to check it's minimal, you should either show it's irreducible, or show that the extension is of degree 6, which can be done showing it contains $\sqrt2$ or in some other messy way..
well, you could also say that the group of automorphism of $\mathbb{Q}(\sqrt2,\sqrt[3]2)$ over $\mathbb{Q}$ acts freely (? is it correct way to say that images are all different?) on $\sqrt2+\sqrt[3]2$.

-oo-I've always been fascinated by this beautiful approach.  What is the most general thing that can be said about it? Would the trick work for something huge like

$\sqrt{17}+\sqrt{29}+5^{\frac{1/3}}+14^{\frac{1}{5}}$

I guess so?

yes. I've learned it from this thesis:

http://www.icm.tu-bs.de/ag_algebra/software/distler/Diplom.pdf


If two distinct numbers are selected at random from the first seven prime numbers, what is the probability that their sum is an even number? Express your answer as a common fraction.

Prime numbers are all odd except 2.

There are 6c2 ways to choose both odd primes, and 7c2 overall. 6*5/7*6=5/7


induction
combinatorics proposed
combinatorics

Some checkers placed on an $n \times n$ checkerboard satisfy the following conditions:

(a) every square that does not contain a checker shares a side with one that does;

(b) given any pair of squares that contain checkers, there is a sequence of squares containing checkers, starting and ending with the given squares, such that every two consecutive squares of the sequence share a side.

Prove that at least $(n^{2}-2)/3$ checkers have been placed on the board.

We'll cover the board with checkers and ghost-checkers.

For each checker we'll have two ghost-checkers available. So we'll have $3k$ units available, where $k$ is the number of checkers. For each checker X with just two neighbors A, B: place the two ghost-checkers for X on the two squares adjacent to X different from A and B. For each checker with three neighbors, place one of its ghosts on the fourth adjacent square and keep the other. For each checker with four neighbors, keep both of its ghosts.

For each checker X with just one neighbor A, place the two ghost-checkers for X on the squares adjacent to X different from A and its opposite. (Of course there are no lonely checkers since the graph is connected). When we're done there may be squares covered more than once, and there may be ghosts outside the board; it doesn't matter. The only squares that may be left uncovered are those squares A with just one checkered neighbor X, where X has just one checkered neighbor itself (call it Y), and A, X, Y are collinear.

Now note that, for the connected graph formed by the checkers, the number of "endpoints" of the graph (that is, checkers with just one neighbor) is at most 2+the number of checkers with 3 neighbors+twice the number of checkers with 4 neighbors (the event "a checker with 4 neighbors" is equivalent to the event "a checker with 3 neighbors" ocurring two different times), which is easy to prove by induction (take an endpoint, go along the graph till you find a point where the path has two or more branches, and remove the string connecting the taken endpoint to the rest of the graph; if we find no points where it branches out, it must be a simple path with just 2 endpoints). This means that using the checkers we'd kept before, we can cover all the yet uncovered squares except at most 2, so $3k \geq n^2-2$, which is what we wanted to prove.

Let us place the checkers on the board as follows: place one checker on the board to start, and then continually place a checker adjacent to one that has already been placed. Since any arrangement of checkers that satisfies the problem must be connected (i.e. (b)), we can form any arrangement of checkers in this manner.

Call a square "on" if it has a checker on it or it is next to a square that has a checker. By (a), we require an arrangement for which every single square is on. When we place the first checker, at most five squares are turned on. For every subsequent checker we place, at most three squares are turned on (the place we just filled and the filled space next to it being already on).

If we fill the board with $ k$ checkers, we have turned on at most $ 5 \plus{} 3(k\minus{}1) \equal{} 3k\plus{}2$ checkers. Since there are $ n^2$ squares on the board, we need $ 3k\plus{}2 \ge n^2$ or $ k \ge \frac{n^2\minus{}2}{3}$.

How about the result if this just has only the first restriction? Is it 12 in the case 5*5 blackboard? :roll:

What about when n=2? Then the problem requires that we need at least 2/3, or 1 checker. But no matter where we put it the square diagonally from that square doesn't share a side with that square with the checker.

That's true and doesn't contradict the statement of the problem.  (It does show that the statement is not sharp for $n = 2$, since in that case at least 2 checkers are needed.)

@ghjk, two years later: no, 12 is much too large for the $5 \times 5$.  For example, my first attempt did it in 8.  In general, the best case should be approximately $\frac{n^2}{5}$ (but slightly larger due to edge effects).

Suppose there are $f(n)$ checkers on the board (and $n^2-f(n)$ empty squares), so from (b), the checkers form a connected graph $G$.

Now let $X$ denote the number of pairs of neighboring squares both containing checkers and $Y$ denote the number of pairs of neighboring squares with exactly one containing a checker. Then $X$ is simply the number of edges in $G$, so $X\ge f(n)-1$. Furthermore, since each checker borders at most four other squares, we have $2X+Y\le4f(n)$, so $Y\le4f(n)-2X$.

But (a) tells us that
\[n^2-f(n) \le Y\le4f(n)-2X,\]so
\[5f(n)\ge n^2+2X\ge n^2+2f(n)-2\implies f(n)\ge\frac{n^2-2}{3},\]as desired.

Solution with vsathiam.
[hide=solution]
Let an arrangement of checkers [i]complete[/i] a shape if it satisfies the conditions stated in the problem, wrt the shape.
Consider some shape with edges on the grid lines that can be completed by putting $k$ checkers in a line. These $k$ checkers cover $3k+2$ squares, as each checker individually covers three squares, that is the two adjacent squares not covered by another checker as well as the square that the checker itself is on, while the first and last squares additionally cover another square each. Note that no square is covered twice. It follows that $3k+2=mn=A$, where $A$ is the area of the shape. We can then calculate that $k=\frac{A-2}{3}$.
Now note that a line is the most efficient way to cover as many squares as possible. Since an arrangement of checkers that completes an $n$ by $n$, where $n>3$, must contain some turn, where a turn consists of a checker, such that there exists no line(of grid squares) that goes through the checker as well as it's two neighboring checkers, any arrangement of checkers completing a $n$ by $n$ grid must be less efficient a line. Since a line could hypothetically complete the $n$ by $n$ square with $\frac{n^2-2}{3}$ checkers, we have that it will take greater than $\frac{n^2-2}{3}$, where $n\not \in (1,2,3)$. $\square$

Remark: The motivation for considering a line of checkers, was the $-2$ and the denominator of three in the fraction. The three suggested looking at how many new squares each new checker in a "snake" of checkers would cover, while the $-2$ reminded us of the first and last checker in the line.

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

Take a spanning tree on the set $V$ of checkers where the $|V|-1$ edges of the tree are given by orthogonal adjacency. By condition (a) we have \[ \sum_{v \in V} (4-\deg v) \ge n^2 - |V| \] and since $\sum_{v \in V} \deg v = 2(|V|-1)$ we get \[ 4|V| - \left( 2|V|-2 \right) \ge n^2 - |V| \] which implies $|V| \ge \frac{n^2-2}{3}$.

statement (b) says that the squares with checkers on them form a polyomino

finally, statement (a) says that if we "extend" the polyomino outward (this probably doesn't make sense oops) the area must be $n^2$

its easy to see that extend the polyomino, maximum area is $3k+2$ given $k$ checkers (because we form a $1$ by $k$ rectangle)

therefore number of checkers satisfies $3k+2\geq n^2$ and we are done

to show max is $3k+2$ area, note that the first tile adds $5$ squares, and each successive one must add only $3$ at most

Let $m$ be the number of checkers. We say that a cell is good if it either contains a checker or is adjacent to a cell that contains a checker. Since the order we place them in does not matter, we placed them in an order such that each checker after the first is placed next to an existing checker (this is possible due to the connected component condition).\\

Note that the first checker can make at most 5 cells good. Each additional checker can make at most 3 cells good (since the cell its placed on is already good and it is "blocked" in one of the directions by an existing checker). Therefore, we must have $5+3(m-1)\geq n^2$, so $$m\geq \frac{n^2-2}{3}$$

Remark: this was somewhat motivated by the very peculiar condition $\frac{n^2-2}{3}$. in particular, this is never an integer due to 2mod3, so this motivates a method where each checker placed counts as multiple "things" (in this case good cells)

Let $k$ be the number of squares with checkers.

First, notice that the number of pairs of adjacent squares with one filled and one empty is at least $n^2-k$ (the number of empty squares). On the other hand, the number of pairs of adjacent squares, both filled, is at least $k-1$. However, in total, there are at most $4k$ pairs of adjacent squares, at least one of which is filled. Therefore, we have the inequality $$n^2-k+2(k-1) \leq 4k \iff k \geq \frac{n^2-2}3,$$ as required.

We count the pairs of adjacent squares such that one is filled and one is empty. If there are $k$ filled squares, the maximum perimeter of the filled squares is $2k+2$ as each new block after the original one adds at most $2$ to the perimeter of the filled squares. The minimum perimeter is $n^2-k$ as each non-filled square must share a side with a filled square. Hence, $2k+2\ge n^2-k$, as desired. $\square$

Consider a square to be "covered" if it is covered by a checker or shares a side with a square covered by a checker. First place down one checker. This checker can "cover" at most $5$ squares. For each checker placed down after that, by problem conditions, it must share a side with another checker, meaning that it can "cover" at most $3$ new squares. This means that we need at least
\[1+\frac{n^2-5}{3},\]
or $\frac{n^2-2}{3}$ checkers.

We count the total number of cells in two ways:

First, it is obviously $n^2$.

On the other hand, it is the number of cells with checkers along with the number of cells that are adjacent to checkers. Let the number of cells with checkers be $k$. Then from the second condition, all checkers must be able to form a path, and so each checker must be adjacent to two others except for the "end" checkers if they exist, giving a maximum of $2k+2$ in this case. Thus, we have
\[ 3k+2 \leq n^2 \implies k \geq \frac{n^2-2}{3} \]
completing the proof. $\blacksquare$

Let there be $k$ squares with checkers in them. It is clear that they border at most $2(k-2)+3 \cdot 2=2k+2$ squares because of the first condition. Thus,

\[3k+2 \ge n^2 \iff k \ge \frac{n^2-2}{3}. \ \square\]

This is a funny way to solve.
Connect squares with checkers in them to become a spanning tree $V$
Consider the sum $\sum_{v \in V} (4-\text{deg}(v))$
Note that in this sum each square without a checker on it gets counted so
$$\sum_{v \in V} (4-\text{deg}(v)) \geq n^2- |V|$$
$$4|V|-2(|V|-1) \geq n^2-|V|$$
$$|V| \geq \tfrac{n^2-2}{3}. \square$$

Let $w$ be the number of white cells and let $b$ be the number of black cells. Then each white cell has at least one black neighbor.

Consider the graph on the black cells where there is an edge between two black cells if they share an edge. Since this graph is connected, it must have at least $b-1$ edges. Therefore, the black cells can satisfy at most $2b+2$ white cells, so $w\le 2b+2$. Therefore, $n^2=w+b\le 3b+2$, which implies the wanted result.

Let $A$ and $B$ be the set of squares with and without a checker, respectively. A upper bound for $|B|$ is \[\sum_{a \in A} (4-\deg a) = 4|A| - \sum_{a \in A} (\deg a)\].
The graph of $A$ is connected, so it contains at least $|A|-1$ edges, and thus
\[4|A| - \sum_{a \in A} (\deg a) \ge 4|A| - 2(|A|-1) \ge |B| = n^2 - |A| \implies |A| \ge \frac{n^2-2}{3}. \quad \blacksquare\]


Graph Airlines $ (GA)$ operates flights between some of the cities of the Republic of Graphia. There are at least three $ GA$ flights from each city, and it is possible to travel from any city in Graphia to any city in Graphia using $ GA$ flights. $ GA$ decides to discontinue some of its flights. Show that this can be done in such a way that it is still possible to travel between any two cities using $ GA$ flights, yet at least $ 2/9$ of the cities have only one flight.


Looking for help on this problem

suppose an object is thrown upward from a height of 6.5 feet at a velocity of 72 feet per second. Use the position equation p = -16t^2+vt+s, where v is the initial velocity and s is the initial position, to write a funciton that represents the situation.
p =

Find the average rate of change in position from t1 = 0 and t2 = 4.  ______feet per second

Find the equation of the secant line through t1 and t2. p=_________

I really appreciate any help on this

[hide="hints"]

Finding the function is simply an exercise in reading the problem. v and s are givens. The equation is given to you.

"rate of change in position" means velocity.
Average velocity is $ \frac{total distance}{total time}$
[/hide]

The secant-line question doesn't make sense because the problem involves 1 dimensional motion. There aren't really any lines to be drawn.

chickendude, I think you may be over-thinking the question.

The secant refers to the line that intersects the two given points on the graph of p(t).

But this appears to be just a homework question; nitram I think you should post what specifically you’re having trouble with if you wish for any help.


geometry
trigonometry
area of a triangle
Heron\u0027s formula

I need help with this problem:

The sides $ a,b,c$ of a triangle satisfy $ 0\le{a}\le1\le{b}\le2\le{c}\le3$. Find the largest possible area of such a triangle.

The answer is $ 1$, but the solution uses trigonometry. Is it possible to solve this problem without trigonometry (as I haven't learned it yet)? If so, I would appreciate it if someone posted a solution.

I started likewise:

Assume the sidelengths are $ 1,2,(3 \minus{} k)$ for any positive real number $ k$. Then, I plugged the values into Heron's Formula, which, apparently, got me nowhere.

I edited the problem now.

Wouldn't the area be maximized when a=b=c=3? (Area = 9sqrt3/4 in that case)

[hide="Solution (kinda uses trig)"]Assume that $ c$ doesn't have a restriction. Then the area of the triangle with two sides $ a$ and $ b$ is maximum when $ a$ and $ b$ are maximized and they are perpendicular to each other (to prove this, you'll need trig). So, let $ a \equal{} 1$ and $ b \equal{} 2$ and calculate $ c$ to be $ \sqrt {5}$. Lucky for us, $ \sqrt {5}\le 3$ anyway since $ 5\le 9$. Hence, the max area of the triangle is $ \frac12 \times 1 \times 2 \equal{} 1$.
[hide="the trig part"]Area of triangle $ \equal{} \frac12 ab \sin \theta$ where $ \theta$ is the angle between $ a$ and $ b$. Obviously, the area is maximized whenever $ \sin\theta$ is maximum, which occurs when $ \theta \equal{} 90^\circ$.[/hide]
[/hide]


AMC
USA(J)MO
USAMO
geometry unsolved
geometry

In an acute-angled triangle $ ABC$, let $ M$ be the midpoint of $ BC$ and $ N$ be a point in the interior of the triangle such that $ \angle NBA \equal{} \angle BAM$ and
$ \angle NCA \equal{} \angle CAM$. Prove that $ \angle NAB \equal{} \angle MAC$.

In the fact, it is the second problem in USAMO 2008, http://www.mathlinks.ro/viewtopic.php?p=1116181#p1116181