function
probability
parameterization
integration
real analysis
real analysis unsolved

$ u(x,t) \equal{} u$, $ v(x,t) \equal{} v$,
\[ \frac {\partial u}{\partial t} \equal{} \frac {\partial }{\partial x}\left[\left(x \minus{} \gamma\right)u\right] \plus{} \alpha v \minus{} \beta u \\
 \\
 \\
\frac {\partial v}{\partial t} \equal{} \frac {\partial }{\partial x}\left(x v\right) \minus{} \alpha v \plus{} \beta u \\
 \\
 \\
u\left(0,t\right) \equal{} v\left(\gamma ,t\right) \equal{} 0
\]
Any ideas?

$ \frac {\partial}{\partial x}[(x \minus{} \gamma)u] \equal{} u \plus{} (x \minus{} \gamma)\frac {\partial u}{\partial x}$

$ \frac {\partial}{\partial x}(xv) \equal{} v \plus{} x\frac {\partial v}{\partial x}$

From second equation:

$ u \equal{} \frac {1}{\beta}\left\{\frac {\partial v}{\partial t} \minus{} x\frac {\partial v}{\partial x} \plus{} (\alpha \minus{} 1) v\right\}$

Put this into first:

$ \frac {1}{\beta}\left\{\frac {\partial^2 v}{\partial t^2} \minus{} x\frac {\partial^2 v}{\partial x\partial t} \plus{} (\alpha \minus{} 1) \frac {\partial v}{\partial t}\right\} \equal{} \frac {1}{\beta}\left\{\frac {\partial v}{\partial t} \minus{} x\frac {\partial v}{\partial x} \plus{} (\alpha \minus{} 1) v\right\} \plus{}$
$ \plus{} \frac {x \minus{} \gamma}{\beta}\left\{\frac {\partial^2 v}{\partial x\partial t} \minus{} \frac {\partial v}{\partial x} \minus{} x\frac {\partial^2 v}{\partial x^2} \plus{} (\alpha \minus{} 1) \frac {\partial v}{\partial x}\right\} \plus{} \alpha v \minus{} \left\{\frac {\partial v}{\partial t} \minus{} x\frac {\partial v}{\partial x} \plus{} (\alpha \minus{} 1) v\right\}$

or

$ \frac {\partial^2 v}{\partial t^2} \minus{} x\frac {\partial^2 v}{\partial x\partial t} \plus{} (\alpha \minus{} 1) \frac {\partial v}{\partial t} \equal{} \frac {\partial v}{\partial t} \minus{} x\frac {\partial v}{\partial x} \plus{} (\alpha \minus{} 1) v \plus{}$
$ \plus{} (x \minus{} \gamma)\left\{\frac {\partial^2 v}{\partial x\partial t} \minus{} \frac {\partial v}{\partial x} \minus{} x\frac {\partial^2 v}{\partial x^2} \plus{} (\alpha \minus{} 1) \frac {\partial v}{\partial x}\right\} \plus{} \alpha\beta v \minus{}$
$ \minus{} \beta\left\{\frac {\partial v}{\partial t} \minus{} x\frac {\partial v}{\partial x} \plus{} (\alpha \minus{} 1) v\right\}$

After simplifying we get:

$ \frac {\partial^2 v}{\partial t^2} \minus{} (2x \minus{} \gamma)\frac {\partial^2 v}{\partial x\partial t} \plus{} (\alpha \plus{} \beta \minus{} 2)\frac {\partial v}{\partial t} \plus{} x(x \minus{} \gamma)\frac {\partial^2 v}{\partial x^2} \minus{}$
$ \minus{} ( \; (\alpha \minus{} 2)(x \minus{} \gamma) \plus{} (\beta \minus{} 1)x \; )\frac {\partial v}{\partial x} \minus{} (\alpha \plus{} \beta \minus{} 1)v \equal{} 0$

Anyway if $ \gamma \equal{} 0$ then after adding two equations and setting $ F(x,t) \equal{} u \plus{} v$, we get:

$ \frac {\partial F}{\partial t} \equal{} F \plus{} x\frac {\partial F}{\partial x}$

Set $ F(x,t) \equal{} f(x)g(t)$:

$ fg' \equal{} fg \plus{} xf'g$

or

$ \frac {g'}{g} \equal{} 1 \plus{} x\frac {f'}{f} \equal{} A$, where $ A$ - some constant.

Get:

$ g(t) \equal{} g(0)e^{At}$ and $ f(x) \equal{} Bx^{A \minus{} 1}$, $ B$ - alsow constant.

So, $ F(x,t) \equal{} T x^{A \minus{} 1}e^{At}$, where $ T \equal{} g(0)B$.

And $ u \equal{} F \minus{} v$. Put this into second equation:

$ \frac {\partial v}{\partial t} \minus{} x\frac {\partial v}{\partial x} \plus{} (\alpha \plus{} \beta \minus{} 1)v \equal{} \beta T x^{A \minus{} 1}e^{At}$, which again can be solved if $ \beta\cdot T \equal{} 0$ by the same method. :)

Unfortunately $ \gamma \equal{} 0$ is not a physically interesting case.  And it's mathematically straightforward since, as you noted, the system just becomes a single PDE for $ u \plus{} v$.

Maybe try to solve equation with some conditions on the $ \alpha$ and $ \beta$ to destroy some members in that general equation for $ v$. Hard to imagine it helps... From what part of physics you get that equation, heat conductivity?

We can alsow write new functions: $ a(x,t)\equal{}\frac{u(x,t)}{x\minus{}\gamma}$ and $ b(x,t)\equal{}\frac{v(x,t)}{x}$, maybe something changes...

The PDEs are related to the pdf of the random variable $ V$ satisfying $ \frac {dV(t)}{dt} \equal{} X(t) \minus{} V(t)$ where $ X$ is a two-state variable ($ 0$ or $ 1$) switching back and forth at exponentially distributed times.  The problem arises in a stochastic model of mRNA synthesis.

I just wonder what exactly you want from your system. Obviously, $ u\equal{}v\equal{}0$ is a solution. Are you looking for non-trivial solutions, for the proof that no such solutions can exist, or for what? :?

Yes, sorry I wasn't clear, I'm looking for nontrivial solutions.  Assuming no mistake has been made in the derivation of the equations, they ought to correspond to probability distributions, so I do expect nontrivial solutions to exist.  It doesn't seem too far-fetched -- the corresponding steady-state problem does admit nontrivial solutions (and ones that seem to make physical sense).

Well, when $ \alpha\equal{}0$ or $ \beta\equal{}0$, the trivial solution is the only one (this is almost obvious because in this case one of the equations includes only one function and you can easily solve it by the method of characteristics). Does it still agree with what you are saying about existence? If it does, here are 2 more questions:

1) What is known about $ \alpha$ and $ \beta$?
2) What class do you expect your solutions to belong to (e.g., should they be bounded near the endpoints $ x\equal{}0,\gamma$ and/or as $ t\to\infty$, or anything else like that)?
3) What exactly do you mean by the "corresponding steady state problem"?

I'm asking because so far I failed to construct any non-zero solutions and I started to suspect that they may just not exist at all or be just physically meaningless :(. If I could get a clearer picture of what you have in mind, it might help.

1) They should be positive, but that's essentially it -- they are related to the switching rate of the random variable $ T$, but are independent of each other and other parameters in the problem (you're right about trivial solutions for the case $ \alpha \equal{} \beta \equal{} 0$, but it's physically uninteresting as it corresponds to deterministic and constant forcing).
2) Well to be continuous and satisfy the boundary conditions, yes $ u$ should certainly be bounded near $ 0$ and $ v$ near $ \gamma$.  Since I expect these things to be related to probability distributions (after some normalization), I moreover expect them to be in $ L^1\left(0,\gamma\right)$, and of one sign.
3) $ \frac {\partial u}{\partial t} \equal{} \frac {\partial v}{\partial t} \equal{} 0$.

Existence of non-trivial solutions for the time-independent case (or, if you like, the $ t\to \infty$ 'steady state' case):

$ \boxed{u(x) \equal{} c x^{\alpha}(\gamma \minus{} x)^{\beta \minus{} 1}, \ \ v(x) \equal{} cx^{\alpha \minus{} 1}(\gamma \minus{} x)^\beta}$ where $ c\in \mathbb{R}$ is arbitrary.  Notice for $ \alpha,\beta > 0$ these solutions are $ L^1$ and of one sign (moreover for $ \alpha,\beta > 1$ they are nice continuous bounded functions).

Edit: Having given it a second thought, I realize I need to pin down more conditions to specify the time-dependent behavior.  I'll go over the derivation and see if I can come up with the correct conditions. Edit 2: Hah looks like you pointed that out a minute earlier, I'll think about it.

Great, I see. Now, what dependence of $ t$ are you looking for? (Your functions that do not depend on $ t$ definitely solve the system, so why do you need anything else?)

Ok, so it should be something like: $ u(x,0) \equal{} 0$, $ v(x,0) \equal{} \phi(x)$ for $ x\in (0,\gamma)$ where $ \int_0^\gamma \phi (x)dx \equal{} 1$ and $ \phi(\gamma) \equal{} 0$.  $ \phi$ shouldn't matter, but if it makes it any easier it's physically sensible to have it be a $ \delta$ function centered somewhere in the interval $ (0,\gamma)$.

OK, so do you now give up the conditions $ u(0,t)\equal{}v(\gamma,t)\equal{}0$? (they will be quite hard to maintain in the new version of the problem when both $ u$ and $ v$ are prescribed for all $ x$ at the moment $ 0$)

No, those boundary conditions must stay.  But why do you think they are necessarily inconsistent?

[b]JoeBlow[/b], you tried to solve it in Maple? It can solve it. Answer consist hypergeometric series. And if $ \alpha\ne \minus{} \beta$, then from condition $ v(\gamma,t) \equal{} 0$ we get that $ v$ is constant in time, it is a function only $ x$.

[hide="image"][img]http://pic.ipicture.ru/uploads/090206/fzVMk8WQml.png[/img][/hide]

I solved the thing numerically in Mathematica for $ \gamma \equal{} 1$, $ \alpha \equal{} 4$, $ \beta \equal{} 3$, $ \phi(x) \equal{} 6\left(\frac {1}{4} \minus{} \left(\frac {1}{2} \minus{} x\right)^2\right)$ and it returns curves that settle into the correct asymptotic behavior, and satisfy the PDE everywhere to within precision except on what seems to be a shock wave.

I'm guessing the point is that Maple returns classical solutions, whereas solutions to the PDE with those BCs may only exist in a weak sense.

Here is one trick that may be useful or useless depending on what you really want to do next. If you set
\[ \begin{aligned} u(x,t) & \equal{} f(e^t)e^{i\lambda e^t(x \minus{} \gamma)} \\
v(x,t) & \equal{} g(e^t)e^{i\lambda e^tx} \end{aligned}
\]
with any $ \lambda\in \mathbb R$, you get the system
\[ \begin{aligned} T f'(T) & \equal{} (1 \minus{} \beta)f \plus{} \alpha e^{i\lambda\gamma T}g \\
T g'(T) & \equal{} (1 \minus{} \alpha)g \plus{} \beta e^{ \minus{} i\lambda\gamma T}f \end{aligned}
\]
where $ T \equal{} e^t$ is a new variable.
(check it!)

Thus, you have a system of first order ODE's for each frequency (notice that the frequency itself changes with time, so to get the coefficients at frequency $ \lambda$ at time $ t$, you need to start with the coefficients at frequency $ \lambda e^{ \minus{} t}$ at time $ 0$).

The system of ODE's can be "solved" by modified "power series technique" but I'm not sure this is of any value whatsoever.

So, in a sense, the "explicit" solution is here: take the Fourier transforms of $ u$ and $ v$ at $ t \equal{} 0$ and solve the above system of ODE's to get the Fourier transforms at time $ t$. The big problem is whether the next question you were going to ask is tractable in these terms.

Thanks very much for your ideas, I'll mess around with it.

You are welcome. :) Feel free to ask more questions any time. The things that can be easily concluded from the "Fourier method" I proposed are that if $ u$ and $ v$ were $ C^\infty$ on the real line and supported in $ [0,\gamma]$ at $ t\equal{}0$, then they will remain $ C^\infty$ for all times $ t>0$ and will be supported in the same interval (so, there are no "shock waves" or other singularities in this case; the solution is classical; also, the boundary conditions you requested will hold automatically). It seems like with some extra effort one can show that the solutions tend to the steady state ones exponentially fast but I'll have to check many details to ensure that I, indeed, have a proof of it. Since I have no idea what you are really after, I'll stop here and forget about the problem for the time being...


Since $abc=def,$ we can write
\[ \frac{ab}{de}=\frac{f}{c}=\frac{x}{y},  \]
with $\gcd\left( x,y\right) =1.$ There exist the positive integers $\alpha,\beta$ such that $ab=\alpha x,$ $de=\alpha y,$ $f=\beta x$ and $c=\beta y.$

Similarly, writing
\[ \frac{ac}{df}=\frac{e}{b}=\frac{z}{t},  \]
with $\gcd\left( z,t\right) =1,$ we can find the positive integers $\gamma,\delta,$ such that $ac=\gamma z,$ $df=\gamma t,$ $e=\delta z$ and $b=\delta t.$

We have
\[ a\left( b^{2}+c^{2}\right) +d\left( e^{2}+f^{2}\right) =ab\cdot b+ac\cdot c+de\cdot e+df\cdot f=  \]
\[ =\alpha x\cdot\delta t+\gamma z\cdot\beta y+\alpha y\cdot\delta z+\gamma t\cdot\beta x=\left( \alpha\delta+\beta\gamma\right) \left( xt+yz\right) ,  \]
a composite number.

[quote="tranthanhnam"]let a,b,c,d,e,f be six positive integers satisfying abc=def.
prove that : a(b^2+c^2)+d(e^2+f^2) is a composite number[/quote]
tranthanhnam, why do you often post problems in ussue of recent Math & Magazine ??  :mad:
This problem isn't hard but you don't permit to do this.
Stop it right now

He did not claim it of his own. Perhaps he is seeking a different or at least a solution.  I don't see a problem in that. If a solution is meant to be submitted and the due date is not reached yet, then there's a problem.  I don't know for sure, so I won't judge anyone.  Other than that, it's perfectly fine.  Besides, I see the problems :D

[quote="tranthanhnam"]Let $a,b,c,d,e,f$ be six positive integers satisfying $abc=def$.

Prove that : $a(b^2+c^2)+d(e^2+f^2)$ is a composite number.[/quote]$a(b^2+c^2)+d(e^2+f^2)=\frac{(de^2+ac^2)(de^2+ab^2)+(def+abc)(def-abc)}{de^2}$

$=\frac{(de^2+ac^2)(de^2+ab^2)}{de^2}.$

@Ji Chen: but how do you know that $\frac{(de^2+ac^2)(de^2+ab^2)}{de^2}$ is not a prime?

That is easy, each of the factors $de^2+ac^2$ and $de^2+ab^2$ at the numerator is strictly larger than the denominator $de^2$, so none can disappear when dividing the numerator by the denominator.

Oh thanks... Didn't notice that...  :blush:


The following proposition is true or false:
Let n is given positive integer, we can find 8 positice integer numbers a, b, c, d, e, f, g, h so that: $n=\frac{2^a-2^b}{2^c-2^d}.\frac{2^e-2^f}{2^g-2^h}$

I don't think it is from the German TST 2005. Check here: http://www.mathlinks.ro/Forum/resources-64-54-2005.html

It's German TST 2006 actually  :) http://www.mathlinks.ro/Forum/viewtopic.php?p=383965#p383965


Prove that there exist [b]ten[/b] triangles in a plane, no two of them having all the vertices same, such that all of them have a common circumcircle.
Deduce the same problem for [b]twenty[/b] triangles also.

I'm not sure, but could we construct a circle, and draw 10 distinct triangles in the circle? (30 random points on the circumference connecting groups of 3)

Really? Are you absolutely sure.....cause I have a solution. Here it follows:
Consider any circle in the plane. You can take [b]five[/b] distinct points on it. With these point you can construct exactly ten triangles, no two the same(5C3=10), and hence is the problem.
For twenty triangles, take [b]six [/b]points and continue(6C3=20)
I am not sure if there is any gross mistake in this proof. Please point out if there is.

Sunkern_sunflora: From what you say, I think you meant that there should be exactly ten triangles possible through the points, and all of them must have a common circumcircle. Or it might be what most ppl including SimonM thought. In either case the problem looks trivial.

I guess I understand. Sorry everyone. Thanx Akashnil.


[color=blue]Hay una desigualdad que encontré empíricamente por casualidad, pero no logro demostrarla.

Sean [b][i]a[/i]1, [i]a[/i]2, … [i]an[/i][/b] reales positivos tales que el producto de todos ellos es igual a [b]1[/b]. Demostrar que su sumatoria es mayor que [b][i]n[/i]-1[/b].[/color]

Bueno, una desigualdad facil.
Por $MG-MA$ tenemos que
(a1+a2+...+an)/n >=  (a1a2...an)^1/n = 1
entonces a1+a2+...+an >= n >n-1
 :D

[color=blue]¡Gracias! ¿Cómo no me pude dar cuenta de lo sencilla que era la solución?

De paso, te agradezco nuevamente porque también conjeturaba que era mayor o igual que [b][i]n[/i][/b] dicha suma.

Gracias, nuevamente.[/color]

Uno facil.
sean, $a,b,c,d$ numeros reales positivos.
donde: $a,b,c < d$.
demostrar que :
                         $d^{3}> d(ab+bc+ca)-3abc$.

[quote="R.G.A.M."][color=blue]Estuve intentando resolver tu problema, aev5peru, pero no llego a nada útil. Mis acotaciones no me permiten llegar a la solución. ¿Me puedes dar alguna pista?

Saludos.[/color][/quote]
ese problema se me ocurrio desps de ver el siguiente.
si $d>a,b,c>0$ demostrar que $d^{2}>d(a+b+c)-(ab+bc+ca)$,
espero que esta ayuda sea suficiente.
chau.

[quote="R.G.A.M."][color=blue]Discúlpame por no haber preguntado antes. Cuando dices que $d>a;b;c>0$, ¿estás afirmando que es $d$ es estrictamente mayor que los demás términos? Hasta ahora supuse como cierto lo que acabo de mencionar.

Muchas gracias.[/color][/quote] SI, eso sirve vastante en mi demostracion.


Let $x,y,z$ be positive integers. Consider the value of

$\alpha = |\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-x}{z+x}|$

a) Prove that $\alpha <1$

b)Suppose now that $x,y,z$ are the side lengths of a triangle (which may be degenerate).

(i) Prove that $\alpha <\frac{1}{8}$

(ii) Find the absolute maximum of $\alpha$.

[quote="courtrigrad"]1. you mean prove $0< \alpha < 1$[/quote]

no.....consider $x=y=z$

Also the >0 is implied since it is absolute value....


calculus
number theory proposed
number theory

Let $ m,n,k$ be positive integers such that
$ mn\equal{}k^{2}\plus{}k\plus{}3$
Prove that there exist odd integers $ x,y$ satisfy one of the equations
$ x^{2}\plus{}11y^{2}\equal{}4m$ and $ x^{2}\plus{}11y^{2}\equal{}4n$

[quote="ThangTongHopK41"]Let $ m,n,k$ be positive integers such that
$ mn = k^{2} + k + 3$
Prove that there exist odd integers $ x,y$ satisfy one of the equations
$ x^{2} + 11y^{2} = 4m$ and $ x^{2} + 11y^{2} = 4n$[/quote]
Indeed it is a very nice problem . You can see a solution of iura in this links :
http://www.mathlinks.ro/Forum/viewtopic.php?p=407830#407830
Although I want to post an other solution .
[i]
Note 1: If $ p$ is a prime satisfy condition :exist two relative prime number $ (a,b)$ such that $ p|a^2+11b^2$ then $ 4p$ can represent in the form $ x^2+11y^2$
[/i]
Proof : Because $ p|a^2+11a^2$ so exist an integer $ k$ such that $ p|k^2+11$
Let $ m=[\sqrt{p}]$
Consider set :
\[ A=\{x+ky|x,y\in (0,..,m)\}
\]
Because $ |A|>p$ so exist two different elements $ (x_1,y_1),(x_2,y_2)$ in A such that
$ x_1+ay_1\equiv x_2+ay_2 (\mod p)$
Let $ u=|x_1-x_2|,v=|y_1-y_2|$ and easy to check that $ u,v$ is not all zero .
Thus $ u^2\equiv k^2v^2(\mod p)$
And because $ k^2\equiv -11 (\mod p)$ so we have $ p|u^2+11v^2$ .And because $ u^2,v^2$ are less than $ p$ so it can take only one follow value $ {p,2p,3p,...,10\}}$ .By considering modulo $ 4$ and modulo  $ 7$ we get $ u^2+11v^2\in \{p,3p,4p,5p,9p\}$ . If $ u^2+11v^2$ is equal $ p,4p$ then statement true. We only need to consider other cases . It can be done by some algebra calculus . For example when $ 3p=u^2+11v^2$
Then we have $ u=3t+1,v=3j+1$ (notice that sign $ +$ can be replace by $ -$ )
Thus  $ p=3k^2+2k+11(3j^2+2j)+4$ . It is easy to check that :
\[ 4p=(k+11j+4)^2+11(k-j)^2
\]
In case $ u^2+11v^2=5p$ ,it can be done by same method as above represent . The case $ 9p=u^2+5v^2$ is the longest ,but you can check that it is true .
Therefore notice 1 has been proved .
Note 2 : If $ 4m$ and $ 4n$ can represent in the form $ u^2+11v^2$ then $ 4mn$ can represent in this form .
Now consider our problem . $ 4mn=(2k+1)^2+11$
By the lemma $ 4m,4n$ can represent in the form $ u^2+11v^2$
Suppose that $ 4m=u_1^2+11v_1^2$ and $ 4n=v_2^2+11u_2^2$
If at least of two pairs $ (u_1,v_1)$ $ (v_2,u_2)$ has odd elements then statement is true . And because $ u_1,v_1$ and $ u_2,v_2$ has same parity so we only need to consider the case :
$ u_1\equiv v_1 \equiv 0 (\mod 2)$ and $ v_2\equiv u_2\equiv 0 (\mod 2)$
Thus $ m,n$ can represent in the form $ u^2+11v^2$ .From condition it is easy to check that $ m,n$ are odd . Therefore exist two odd integer $ x,y$ such that $ 4m=x^2+11y^2$
Problem claim .

In principle, this is just an easy question of arithmetic of $ \mathbb Z[\frac{1\plus{}\sqrt{\minus{}11}}{2}]$ and it's suborder $ \mathbb Z[\sqrt{\minus{}11}]$.
Much further ,it is possible to classify all numbers of type $ x^2\plus{}xy\plus{}3y^2$ or $ x^2\plus{}11y^2$. For how to do this, see http://www.mathlinks.ro/viewtopic.php?t=232507 . This directly applies to $ x^2\plus{}xy\plus{}3y^2$, and $ x^2\plus{}11y^2$ is just a little more work.

Let $ m,n,k$ be positive integers such that
$ mn \equal{} k^{2} \plus{} k \plus{} 3$
Prove that there exist odd integers $ x,y$ satisfy one of the equations
$ x^{2} \plus{} 11y^{2} \equal{} 4m$ and $ x^{2} \plus{} 11y^{2} \equal{} 4n$

Indeed it is a very nice problem . You can see a solution of iura in this links :
http://www.mathlinks.ro/Forum/viewtopic.php?p=407830#407830
Although I want to post an other solution .
[i]
Note 1: If $ p$ is a prime satisfy condition :exist two relative prime number $ (a,b)$ such that $ p|a^2 \plus{} 11b^2$ then $ 4p$ can represent in the form $ x^2 \plus{} 11y^2$
[/i]
Proof : Because $ p|a^2 \plus{} 11b^2$ so exist an integer $ k$ such that $ p|k^2 \plus{} 11$
Let $ m \equal{} [\sqrt {p}]$
Consider set :
\[ A \equal{} \{x \plus{} ky|x,y\in (0,..,m)\}\]
Because $ |A| > p$ so exist two different elements $ (x_1,y_1),(x_2,y_2)$ in A such that
$ x_1 \plus{} ky_1\equiv x_2 \plus{} ky_2 (\mod p)$
Let $ u \equal{} |x_1 \minus{} x_2|,v \equal{} |y_1 \minus{} y_2|$ and easy to check that $ u,v$ is not all zero .
Thus $ u^2\equiv k^2v^2(\mod p)$
And because $ k^2\equiv \minus{} 11 (\mod p)$ so we have $ p|u^2 \plus{} 11v^2$ .And because $ u^2,v^2$ are less than $ p$ so it can take only one follow value $ \{p,2p,3p,...,10p \}$ .By considering modulo $ 4$ and modulo  $ 7$ we get $ u^2 \plus{} 11v^2\in \{p,3p,4p,5p,9p\}$ . If $ u^2 \plus{} 11v^2$ is equal $ p,4p$ then statement true. We only need to consider other cases . It can be done by some algebra calculus . For example when $ 3p \equal{} u^2 \plus{} 11v^2$
Then we have $ u \equal{} 3t \plus{} 1,v \equal{} 3j \plus{} 1$ (notice that sign $ \plus{}$ can be replaced by $ \minus{}$ )
Thus  $ p \equal{} 3k^2 \plus{} 2k \plus{} 11(3j^2 \plus{} 2j) \plus{} 4$ . It is easy to check that :
\[ 4p \equal{} (k \plus{} 11j \plus{} 4)^2 \plus{} 11(k \minus{} j)^2\]
In case $ u^2 \plus{} 11v^2 \equal{} 5p$ ,it can be done by same method as above represent . The case $ 9p \equal{} u^2 \plus{} 11v^2$ is the longest ,but you can check that it is true .
Therefore notice 1 has been proved .
Note 2 : If $ 4m$ and $ 4n$ can represent in the form $ u^2 \plus{} 11v^2$ then $ 4mn$ can represent in this form .
Now consider our problem . $ 4mn \equal{} (2k \plus{} 1)^2 \plus{} 11$
By the lemma $ 4m,4n$ can represent in the form $ u^2 \plus{} 11v^2$
Suppose that $ 4m \equal{} u_1^2 \plus{} 11v_1^2$ and $ 4n \equal{} v_2^2 \plus{} 11u_2^2$
If at least of two pairs $ (u_1,v_1)$ $ (v_2,u_2)$ has odd elements then statement is true . And because $ u_1,v_1$ and $ u_2,v_2$ has same parity so we only need to consider the case :
$ u_1\equiv v_1 \equiv 0 (\mod 2)$ and $ v_2\equiv u_2\equiv 0 (\mod 2)$
Thus $ m,n$ can represent in the form $ u^2 \plus{} 11v^2$ .From condition it is easy to check that $ m,n$ are odd . Therefore exist two odd integer $ x,y$ such that $ 4m \equal{} x^2 \plus{} 11y^2$
Problem claim .

   There are too many mistakes in that solution . I have edited some of them


A right triangle has an inscribed circle cutting the hypotenuse into lengths of 7 and 11. What is the area of the triangle?

Can someone show me how to do this hard problem?

Hint: From a point, the two tangents to the same circle are of equal length

[quote="Iversonfan2005"]From warut suk's hint, I guess then, since the tangents from one point is equal, you can set up the equation as $(7+x)^2+(11+x)^2=18^2$. Then solve for $x$ and then find out each leg right? Thus, you have $x^2+18x-77=0--->x=3.569805...--->\frac{(14.569805...)(10.56980...)}{2}$, thus the area is $77$.[/quote]

I write $(7+x)^2+(11+x)^2=18^2$ as $(11+x-7-x)^2+2(7+x)(11+x)=18^2$

Hence the area $\frac{1}{2}(7 + x)(11 + x)=\frac{{(18 - 4)(18 + 4)}}{4}=77$

I think this works out to be simpler. :)

Ok, i guess if the circle would have cut the hypotenuse in 9:9 the sides would have been equal....and 45 degrees.
the smaller angle=a
hypotenuse=c
a=45-2*45/9=35 deg
Thats why i guess that the area is:
c*cos35*c*sin35/2=76,11510235

It may be terribly wrong, but its almost close to 77 anyway.

I think vidyamanohars's way is the easiest. Same as mine.

[quote="warut_suk"]Yes, Iversonfan2005, you should think carefully before solving for x!  :)[/quote]

oh well, still got the right answer :P even though it got messy :?

Yeah, my method was similar...
Let the length of the shorter leg be $a$.  Then we're trying to find the area, which is $\frac{a(a+4)}{2}$.  We know that
$a^2+(a+4)^2=324$
$2a^2+8a+16=324$
$2a(a+4)=308$
$\frac{a(a+4)}{2}=\frac{308}{4}=77$


assume that $C$ is a convex polygon ,for any edge like $AB$ we consider the vertex which has the most distance to this edge like $D$ ,and then prove that $\sum_{A,B are vertex} <ADB=\pi$

Consider these angles together with vertical angles to them (angles are subsets of plane). Then each point outside the polygon belongs to exactly one angle (or vertical angle). This claim may be reformulated as follows: any point outside the polygon belongs to the continuation of unique affina diamtre, i.e. the chord of a polygon, in ends of which parallel support lines exist. Then by standard area in large disk argument we get the claim of problem.

It's not Sankt-Petersburg, it's ARO 1996.


The following is an excerpt from [i]A Void[/i]. What is unusual about it?

----

Noon rings out. A wasp, making an ominous sound, a sound akin to a klaxon or a tocsin, flits about. Augustus, who has had a bad night, sits up blinking and purblind. Oh what was that word (is his thought) that ran through my brain all night, that idiotic word that, hard as I'd try to pun it down, was always just an inch or two out of my grasp - fowl or foul or Vow or Voyal? - a word which, by association, brought into play an incongruous mass and magma of nouns, idioms, slogans and sayings, a confusing, amorphous outpouring which I sought in vain to control or turn off but which wound around my mind a whirlwind of a cord, a whiplash of a cord, a cord that would split again and again, would knit again and again, of words without communication or any possibility of combination, words without pronunciation, signification or transcription but out of which, notwithstanding, was brought forth a flux, a continuous, compact and lucid flow: an intuition, a vacillating frisson of illumination as if caught in a flash of lightning or in a mist abruptly rising to unshroud an obvious sign - but a sign, alas, that would last an instant only to vanish for good.

***Spoiler Ahead ***
.
.
.
.
.
.
.
.
.
.
.
..
It does not contain the letter "e."

It is devoid of it.

@ boxorocks: you can use the hide button to hide your answer if you want.

Is that the entire book without the letter e in it that I heard about? I didn't realize that it was so.... wordy

I will try to remember to experiment for the first time with
the "hide" button in an upcoming thread.[/hide]

The novel [i]Gadsby[/i] has a similar approach.

http://www.spinelessbooks.com/gadsby/

and e a poe had a story rather like that


A car arrives at an intersection from the east every six minutes, and the first car arrives from this direction at 9:05 a.m. A car arrives from the north every seven minutes, and the first car arrives from that direction at 9 a.m. What is the earliest time after 9 a.m. when cars arrive from both of these directions at the same time?

Write out when a car arrives from the East, and when it is a multiple of 7 in minutes after 9:00, that is your answer.

9:11, 9:17, 9:23, 9:29, 9:35. This tells you that the answer is then $ \boxed{9: 35}$


algebra polynomial search quadratics AMC

Is there a pretty way to do this? I'm not sure of any laws that dictate the properties of a prime numbers.

mathfanatic wrote:Source: WMTS (Wisconsin Math Talent Search) 1995 Round 5 #4

Find all positive integers n such that each of the numbers 6n:^2: + 5, 2n:^2: + 3, and n:^2: + 1 is prime.

[hide]The only possible numbers are 1 and 2. Respectively they give the prime triples, (11, 5, 2) and (29, 11, 5). Consider the quadratic residues mod 5. They are 1, 0, and -1.  A quadratic residue of 1 will make the second number divisible by 5; a quadratic residue of 0 will make the first number divisible by 5, and a quadratic residue of -1 will make the last number divisible by 5. Therefore, the only chance any number will have of making a prime triple is if one of the above expressions is exactly equal to 5. Solving, I get possible numbers as 0, 1, and 2, but only the last 2 work. [/hide]


topology
geometry
number theory
prime numbers
group theory

I have been thinking about this question for some time.
Could anybody enlighten me on some of the applications, if any, of number theory on any of the sciences?
Directions to useful webpages would also be useful while personal comments are preferable.

Well, cryptography, which is largely based on ideas from number theory, has major applications in the world of e-commerce (though not really science).  This is probably the most important application, as internet buisness would be very difficult without secure public-key cryptography. This type of encryption is called RSA, after the inventors.  You can read more about the math of RSA [url=http://mathworld.wolfram.com/RSAEncryption.html]here[/url] (though it will probably be sort of hard to understand unless you have a little knowledge about both number theory and cryptography).  Public-key encryption, of which RSA is the major (and only?) type, is really cool because you can send a secret message to someone with out currently sharing a secret (in other forms of encryption this priorly held secret is known as the key).  Basically every one who uses RSA has two keys: a public key and a private key.  If I want to send you a message, I encode the message with your public key and then you decode it with your private key.  It may seem a bit odd that a different key is used for encryption than is used for decrypttion.  This is where the math comes into play.  RSA basically relies on the fact that it is much easier to multiply two prime numbers than to factor the resulting product.  More math basically transforms the act of encrypting into multiplying the numbers and the act of decrypting into factoring the product, creating a secure system.

I'm sure you can find a wealth of information about this topic by using google.  If you become interested in cryptography, I would recommend reading "The Code Book" by Simon Singh.

Thanks a lot for the information on cryptography. I know about this application since I just read "Gift of Fire" by Saarah Baase as a part of IB course and this has a lot of details on things you talk about.

Further, I especially wanted to know about applications leaving out cryptography, particularly in the sciences.

I recommend this site: http://www.maths.ex.ac.uk/~mwatkins/zeta/physics.htm

A great book on cryptography is The Code Book by Simon Singh.

It has lots of information.

I highly recommend the recently published book [i]Music of the Primes[/i] by the Oxford Mathematician [i]Marcus de Sautoy[/i]. It talks about:
-The application of number theory in cryptography
-More surprisingly, the recent discovery of the link between number theory and quantum chaos in physics. Some people even speculate that the link with physics may shed light on the most important open problem of number theory: the Riemann hypothesis

What is wrong with cryptology? ok other applications: coding theory and I have heard quantum mechanics is based on group theory.

[quote="chess64"]Is cryptology the only real-world application of Number Theory?  :( I love number theory! :D[/quote]
I love it also and hope there is not very much real use, because all applied mathematics are ugly  ;)

[quote="coolamit"]I have heard quantum mechanics is based on group theory.[/quote]
Well, group theory is [u]not[/u] number theory  :P

I would say that number theory is a subset of group theory. after all, (subsets of) natural numbers are a special class of groups with respect to addition and multiplication.

See if you can get hold of "Number Theory in Science and Communication: with applications in cryptography, physics, digital information, computing and self-similarity" (3rd edition) by M.R. Schroeder (Springer Verlag, 1997) [url=http://www.amazon.co.uk/exec/obidos/ASIN/3540620060/qid%3D1125000877/202-4429708-9439818]3-540-62006-0[/url].  It has 30 chapters of applications ...

Thanks a lot everybody. I believe the most interesting info is about the use of number theory in physics and computing.
Oh, nice to know about the possibility of application of group theory in quantum physics. I hope I can learn more about that.

PS: I started this topic in July. I am lucky to find it again. Thanks once more.

[quote="coolamit"][quote="ZetaX"]Group theory is obviously algebra!
Your argumentation only shows that (some kind of) number theory is a special case of algebra, not the inverse!
Additionally one can say by your argument that number theory is also part of analysis or topology or geometry or... ;)[/quote]

I want examples of non-trivial groups that cannot be represented using numbers to be convinced of that[/quote]
1. group theory is unaccurate, at least the whole term Algebra should be used (rings are more than just groups).
2. to be represented by numbers means for me that we can write numbers for the group's elements which is clearly possible everytime but says nothing.
3. more generally: what exactly you want¿ I've problems to understand (could be possible that we mean the same but have problems in telling the other).


$\text{Find all real } x \text{ such that } \sqrt[5]{x^4-4x^3+6x^2-4x+1} + \sqrt[5]{x^2-2x+1} - 20 = 0$

[hide]Try to simplify the stuff under the radicals[/hide]

Yep, looks good to me... buttttttttttttttttt, there's one more solution!  :D

Yep, he's got it. Your other answer will be [hide]-31[/hide]

$\displaystyle x^4-4x^3+6x^2-4x+1 = (x-1)^4$
$\displaystyle x^2-2x+1 = (x-1)^2$

so... the equation becomes

$(x-1)^\frac{4}{5} + (x-1)^\frac{2}{5} - 20 =0$

let's denote $(x-1)^\frac{2}{5}$ as N.

then $N^2 + N - 20 =0$

$N = -5 \text{ or } 4$
$(x-1)^\frac{2}{5} = -5 \text{ or } 4$
-5 is impossible since square of any number cannot be negative, even if its pentarooted...

so $(x-1)^\frac{2}{5} = 4$
$(x-1)^2 = 1024$
$x-1 = \pm 32$
$x = 33, -31$

gah!  Yeah, I should have factored it the normal way instead of getting it to $y(y-1)=20$  I guess that's what you get when you stay up later than usual just to solve math problems.


function
induction
algebra proposed
algebra

Let x :cap: y=$\frac{x+y}{1+xy}$
Find the value of (....((2 :cap: 3) :cap: .....) :cap: n

Not too hard, indeed. This is a general form of problem 1, the 12th Balkan Math Olympiad:

Find the value of the expression
\[
(\ldots(((2\star3)\star)4)\star\ldots)\star1995,
\]
where
\[
x\star y=\frac{x+y}{1+xy},
\]
for all positive numbers $x,y.$

Consider the function $f:\left(  0,+\infty\right)  \rightarrow\left(
-1,1\right)  ,$ $f\left(  x\right)  =\frac{1+x}{1-x}$ and observe that
\[
f\left(  x\star y\right)  =\frac{1+\frac{x+y}{1+xy}}{1-\frac{x+y}{1+xy}}%
=\frac{\left(  1+x\right)  \left(  1+y\right)  }{\left(  1-x\right)  \left(
1-y\right)  }=f\left(  x\right)  f\left(  y\right)  .
\]
Therefore, if
\[
a=(\ldots(((2\star3)\star)4)\star\ldots)\star1995,
\]
then

\[f\left(  a\right)  =f\left(  2\right)  \cdot f\left(  3\right)  \cdot
\ldots\cdot f\left(  1995\right)  =\]
\[
=\frac{3}{-1}\cdot\frac{4}{-2}\cdot\frac{5}{-3}\cdot\ldots\cdot\frac
{1994}{-1992}\cdot\frac{1995}{-1993}\cdot\frac{1996}{-1994}=\]
\[=\frac{1995\cdot1996}{2}.\]

The equation
\[
f\left(  a\right)  =\frac{1+a}{1-a}=\frac{1995\cdot1996}{2}=1991010
\]
has the solution $a=\frac{1991\,009}{1991\,011}.$

Thanks enescu  :)
Your solution is much more beautiful than mine

I think enescu's solution is very nice . This is from Balkan MO 1995


function
real analysis
real analysis unsolved

Let $ A \subset [0,1]^n$ be a Borel set with positive Lebesque measure.

Show that there is a Borel set $ B \subset A$ with $ 2 \lambda^n(B) \equal{} \lambda^n(A)$, where of course $ \lambda^n$ is the $ n$-dimensional Lebesque measure.

$ f(r)\equal{}\lambda^n(A \cap B_r)$,where $ B_r$ is the ball centered at the origin with radius $ r$, is a continuous function with $ f(0)\equal{}0$ and $ f(\sqrt{n}) \equal{} \lambda^n(A)$.


The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, 100 workers can produce 300 widgets and 200 whoosits. In two hours, 60 workers can produce 240 widgets and 300 whoosits. In three hours, 50 workers can produce 150 widgets and m whoosits. Find m.

i posted a thread about this before, but i forgot to write the problem. i need help on this one, don't make fun of me...

oohh! now i see. i went about it using the fact that # of things were directly proportional to # of workerhours and so on. whoops.  :blush:

[quote="Ignite168"][hide]Setup the equations
$100=300a+200b, 120=240a+300b, 150=150a+mb$.
These simplify to
$1=3a+2b, 2=4a+5b, 1=a+mb$
where you want to find 150m.
Working with the first two equations, you get that $a=\frac{1}{7}$ and that $b=\frac{2}{7}$. It's easy to see that m must be 3 and therefore the answer must be $\boxed{450}$.
[/hide]

Editted to make my work more readable as Latex doesn't seem to register line breaks.[/quote]

Use eqnarray (look it up in the guide) and \\ for line breaks.

[hide="A little faster"]It's a little faster to just take $2=4a+5b$ and subtract $1=3a+2b$ to get $1=a+3b$, so $m=450$.[/hide]

How do we get the equations, ignite? I don't get how you formed those equations... Can somebody explain? It is not d=rt, I've never seen anything like it. My mind is exploding!  :wow:

$a$ is the number of hours it takes 1 worker to create 1 widget and $b$ is the number of hours it takes 1 work to create 1 whoosit.

The reason why Ignite multiplied the hour and the number of workers is because since the production rate is identical for all workers. For example, if there are 100 workers working for 3 hours, it would equal 300 workers working for an hour.

We try to find for the unit rate (a and b) and at the end, we just have to substitute to get a value for m.

Hope this clears it up.

Right, I see that. But why does $100=300a+200b?$ I don't see how that was derived. (I'm sorry I'm being so dumb; I've thought about this for an hour straight to no avail... :()

Can somebody help me? Why can't I do this problem?

Think of it like 300a where a is the number of workers per hour and 200b where b is the number of workers per hour.  Adding these two together (for 1 hour) gives 100 workers needed to get this widgets and whoosits for 1 hour.

Apply these to all 3 equations:
300a + 200b = 100
240a + 300b = 120 (2 hours)
150a + Mb = 150 (3 hours)

This gives:
3a + 2b = 1
4a + 5b = 2
150a + Mb = 150

12a + 8b = 4
12a + 15b = 6
7b = 2 so b = 2/7

Therefore a = 1/7.
150/7 + 2M/7 = 150

So M = 450.

Wait hold on I am a bit confused. We have $300 \quad \text{widgets} \cdot \frac{a \quad  \text{hours}}{\text{widget}}=300a\quad  \text{hours}$ right?

$300 \text{widgets} \cdot \frac{a \text{workers}}{\text{hour*widgets}}$

I don't think that is stated in the problem. What do we define $a$ as?

[quote=gamjawon]$a$ is the number of hours it takes 1 worker to create 1 widget and $b$ is the number of hours it takes 1 work to create 1 whoosit[/quote]

Well, I guess a is just the proportion of how many workers it would take to do 300 widgets in a given hour

If $a$ is the number of hours it takes one worker to build 1 widget then $100a$ is how many widgets 100 workers build in 1 hour right?
What am I doing wrong...

In the first part of your sentence, you defined $a$ as a number of hours (time). But then in the second part, you say that $100a$ is a number of widgets.

[hide=Solution]Let it take $x$ hours for a worker to create one widget and let it take $y$ hours for a worker to create one widget. Then $100=300x+200y$ and $120=240x+300y$, so $x=\frac{1}{7}, y=\frac{2}{7}$. Then we have $150=\frac{150}{7}+\frac{2}{7}m$, so $m=\boxed{450}$.[/hide]


A man was numbering the pages of his book with consecutive positive integers, 1,2,3,..., in decimal notation. When he wrote a page number with three digits, he wrote the hundreds digit first, then the tens digit, and finally the units. He was interrupted after writing the 1832nd digit. When he comes back, what digit should he write next?

I can do it, but is there any fast way?

What I would do is this: 9 pages have 1 digit.  90 digits have 2, 900 have 3.  So if he gets up to page 100, he's written $ 9\times 1\plus{}90\times 2\equal{}189$ digits.  Now if he gets up to 200 (including 200), there are 101 digits, so $ 101\times 3\equal{}303$.  Up to and including 300 yields another 300.  Up to an including 400 yields another 300.  Up to 500 is another 300 so we are up to $ 189\plus{}303\plus{}900\equal{}1392$ digits.  Up to 600 brings us to 1692.  Up to 640 brings us another 120 up to 1812.  Up to 645 brings us 1827 digits.  646 is 1830.  Then the 1832nd digit is on page 647.  So his 1833rd digit will be for 648, in other words, $ \boxed{6}$.


Here's a cute one (although kind of easy):

Wedivide $n$ by the numbers $1,2,\ldots n$. Let the sum of the residues we obtain be $r(n)$. Show that there are infinitely many $n$ s.t. $r(n)=r(n+1)$.

I distinctly remember that amfulger has solved it u can take look on solved problem

Sorry, I didn't know that. Anyway, it's a cute problem :).


geometry circumcircle trigonometry ratio AMC AIME ARML

Some I could not figure out...

 Find the least positive integer n such that 1/(sin 45sin 46) + 1/(sin 47sin 48) + ... + 1/(sin 133sin 134) = 2/sin n.

 Find the area of the large square divided by the area of the small square.
[img]http://www.kalva.demon.co.uk/aime/soln/diag016.jpg[/img]

The triangle ABC has AB = 24. The median CE is extended to meet the circumcircle at F. CE = 27, and the median AD = 18. Find area ABF.

[quote="xxreddevilzxx"]

 Find the area of the large square divided by the area of the small square.
[img]http://www.kalva.demon.co.uk/aime/soln/diag016.jpg[/img]

[/quote]

For this one, without the loss of generality, let the radius of the circle be sqrt(2).  Therefore the side length of the large square is 2.  Next, draw the right triangle containing the center of the circle, midpoint of the top side of the small square, and either one of the points on the top of the small square.  By letting the side length of the small square be s, this gives the right triangle of sides (1+s), (s/2) and sqrt(2).  Using Pythagorean’s theorem, solve for s to get s=2/5 or s=-2.  Throw out s=-2 and from here, its 4/(4/25) or 25.

[quote="xxreddevilzxx"]
The triangle ABC has AB = 24. The median CE is extended to meet the circumcircle at F. CE = 27, and the median AD = 18. Find area ABF.[/quote]

This question has come up a couple of times. I remember the solution =P.

[hide]
Since $CE$ is a median, $AE = BE = 12$.
First do Power of a Point on E. We get $(12)(12) = (27)(EF) \Rightarrow EF = \frac{16}{3}$. We note that the $[AFB] = 2[BFE]$ because the bases of each half are equal. So we want to find $[BFE]$. We know that $[BFE] = \frac{1}{2}(12)(\frac{16}{3})\sin{\angle BEF}$. So we want to find $\angle BEF = \angle AEG = \theta$, where $G$ is the centroid. Using the property that medians divide into a $2:1$ ratio, we see that $EG = 9$ and $AG = 12$. Using Law of Cosines on $\triangle AEG$, we have that $12^2=12^2+9^2-2(9)(12)\cos{\theta} \Rightarrow \cos{\theta} = \frac{3}{8}$. This means $\sin{\theta} = \frac{\sqrt{55}}{8}$. So $[BFE] = \frac{1}{2}(12)\frac{16}{3}\sin{\theta} = 4\sqrt{55}$. So $[AFB] = 8\sqrt{55}$. [/hide]

Power of a Point : If two chords intersect in a circle, then the product of the two sides (in relation to the intersection) of one chord is equal to the product of the two sides of the other chord.

[quote="xxreddevilzxx"]Find the least positive integer n such that 1/(sin 45sin 46) + 1/(sin 47sin 48) + ... + 1/(sin 133sin 134) = 2/sin n.[/quote]

[hide]We have
$\cot(a)-\cot(a+1)\\
=\frac{\cos(a)}{\sin(a)}-\frac{\cos(a+1)}{\sin(a+1)}\\
=\frac{\sin(a+1) \cos(a) - \cos(a+1) \sin(a)}{\sin(a) \sin(a+1)}\\
=\frac{\sin(1)}{\sin(a) \sin(a+1)},$
which means we can express our sum as
$\frac{1}{\sin 1}(\cot45-\cot46+\cot47-\cdots-\cot134)$.
Since $\cot45=1$ and all other terms cancel out since $\cot(x)+\cot(180-x)=0$ and $\cot90=0$, the sum is $\frac{1}{\sin(1)}$. So, I'm not convinced about the problem. I think it should be $\frac{1}{\sin n}$, in which case the answer would be 1. Suggestions?[/hide]

[quote="tetrahedr0n"]The theorems dealing with intersecting chords or tangents in a circle. I'm sure you know it, probably not with that name.[/quote]

yeah, I had the worst geometry teacher ever. He would never teach nething and hw was extra credit if u did it. on tests, he would write the answers on the overhead. Good thing I learned it by myself beforehand. But I did get like 110% :D

well... it still says 2/sin n. but by the way, zabelman, how did you think of using cotangent in solving that problem? Does that kind of "hints" or "ways of solving" come quickly when you are actually taking aime?

because i did something like that, but it took me 30 minutes and a book on trig to figure out.

[quote="xxreddevilzxx"]
 Find the area of the large square divided by the area of the small square.
[img]http://www.kalva.demon.co.uk/aime/soln/diag016.jpg[/img]
[/quote]

hehe i jus eyeballed that one....
just like in arml individual last year with the small circles inside the big circles
[url]http://www.arml.com/2004contest.htm[/url]

are we allowed rulers on the AIME jus like the AMC?

Thought process on the trig one:

[hide]Well, I was certain that we could rewrite 1/(sin(a)sin(a+1)) as (something) \pm (something else), which would then make it nice or telescoping or something. Also, I was pretty certain that it would be like x/sin(a)+y/sin(a+1), since that would give us the right denominator. So I started playing with x and y that would make this come close to 1/(sin(a)sin(a+1)), and eventually the idea of sin(p-q)=sin p cos q - sin q cos p popped into my head. It all worked out nicely from there.[/hide]

[quote="xxreddevilzxx"]
Find the area of the large square divided by the area of the small square.
[img]http://www.kalva.demon.co.uk/aime/soln/diag016.jpg[/img]
[/quote]

It's a shame there are no names on the vertices, so label them like this (E is the midpoint):

[code]  A B

C DEF G

H     I[/code]

Solution:

[hide]Let the big square have dimensions 2a*2a, and the small one 2b*2b. Since A, E, and I are collinear, EA*EI=EC*EG, i.e. $(b\sqrt5)(a\sqrt5)=a^2$, which is equivalent to $5b=a$. The ratio of areas is therefore 25.[/hide]


For which values of $ a$ equation
         $ \sqrt{x^4\plus{}4}\plus{}|x|\plus{}3\equal{}(a\minus{}1) \frac{x^{10}\plus{}1}{5}\minus{}x^2$
have  $ 1$ or $ 3$ or $ 5$ or $ 7$ or $ 9$ real roots?

If $ x_1\equal{}b$, then $ x_2\equal{}\minus{}b$, that means that equation has even number of real roots untill $ 0$ is one of them:

$ \sqrt{4\plus{}0^4}\plus{}|0|\plus{}3\equal{}(a\minus{}1)\frac{0^{10}\plus{}1}{5}\minus{}0^2$

$ 5\equal{}(a\minus{}1)\frac{1}{5}$

$ a\equal{}24$

From here we need to find how many real roots (1,3,5,7 or 9) this equation: $ \sqrt{x^4\plus{}4}\plus{}|x|\plus{}3\equal{}5x^{10}\plus{}5\minus{}x^2$ has.

Here is the graphic:
[img]http://www.part.lt/img/ce8b441112ddc2ffeb6bd1fb0eb5a45235.JPG[/img]

Answer: equation has 3 real roots, then a=24.

But in this situation we must take $ 0$ two times. Because function is tangent to x axes

No, function is not tangent to x axes:

[img]http://www.part.lt/img/e7c944f5a832da522a887724e65e6af7295.JPG[/img]

First $ a$ must be $ 26$.  Then if you take  $ a \equal{} 26$ , you  will see the fuction is tangent to x axes

The function above is then a=26. I got that 24-1=25  :D

So the answer a=26, number of real roots - 3, is not correct?  :maybe:


calculus
integration
analytic geometry
conics
parabola
geometry
Vieta

1.Let $ x \equal{} \alpha ,\ \beta \ (\alpha < \beta )$ are $ x$ coordinates of the intersection points of a parabola $ y \equal{} ax^2 \plus{} bx \plus{} c\ (a\neq 0)$ and the line $ y \equal{} ux \plus{} v$.

Prove that the area of the region bounded by these graphs is $ \boxed{\frac {|a|}{6}(\beta \minus{} \alpha )^3}$.

2. Let $ x \equal{} \alpha ,\ \beta \ (\alpha < \beta )$ are $ x$ coordinates of the intersection points of parabolas $ y \equal{} ax^2 \plus{} bx \plus{} c$ and $ y \equal{} px^2 \plus{} qx \plus{} r\ (ap\neq 0)$.

Prove that the area of the region bounded by these graphs is $ \boxed{\frac {|a \minus{} p|}{6}(\beta \minus{} \alpha )^3}$.

what's happen with $ a\equal{}1$, $ m\equal{}0$, $ n\equal{}\minus{}1$?

[quote="bookworm_vn"]what's happen with $ a \equal{} 1$, $ m \equal{} 0$, $ n \equal{} \minus{} 1$?[/quote]

I have edited in general version, but how can we find the area? Your question doesn't make sense in case of $ b\equal{}0,\ c\equal{}0.$

[quote="Carcul"][hide="For a start"]Since there are two points of intersection between the line and the parabola, then we have the following two equations:

$ \alpha m \plus{} n \equal{} a \alpha^2 \plus{} b \alpha \plus{} c$
$ \beta m \plus{} n \equal{} a \beta^2 \plus{} b \beta \plus{} c$.

Solving these two equations for m and n we get

$ m \equal{} a(\alpha \plus{} \beta) \plus{} b$ and $ n \equal{} c \minus{} a \alpha \beta$.

Now consider the several possible cases, integrate between alpha and beta, and see what happens.[/hide][/quote]

If you start like this, you can use Vieta's formula directly.

[hide="#1 Solution"]The area under the parabola is $ \int_\alpha^\beta ax^2+bx+c\,\mathrm{d}x$, and the area under $ y=ux+v$ is $ \int_\alpha^\beta ux+v\,\mathrm{d}x$.

As Carcul has already said, $ u=a\left(\alpha+\beta\right)+b$ and $ v=c-a\alpha\beta$.

The second integral now becomes $ \int_\alpha^\beta x\left(a\left(\alpha+\beta\right)+b\right)+c-a\alpha\beta\,\mathrm{d}x$.

\begin{align*}\int_\alpha^\beta x\left(a\left(\alpha+\beta\right)+b\right)+c-a\alpha\beta\,\mathrm{d}x&=\left[\frac{x^2\left(a\left(\alpha+\beta\right)+b\right)}{2}+cx-ax\alpha\beta\right]_\alpha^\beta\\\\
&=\frac{\beta^2\left(a\left(\alpha+\beta\right)+b\right)}{2}+c\beta-a\alpha\beta^2-\frac{\alpha^2\left(a\left(\alpha+\beta\right)+b\right)}{2}-c\alpha+a\alpha^2\beta\\\\
&=\frac{a\alpha\beta^2+a\alpha\beta^3+b\beta^2}{2}+c\beta-a\alpha\beta^2-\frac{a\alpha^3+a\alpha^2\beta+b\alpha^2}{2}-c\alpha+a\alpha^2\beta\qquad (1)\end{align*}
\begin{align*}\int_\alpha^\beta ax^2+bx+c\,\mathrm{d}x&=\left[\frac{ax^3}{3}+\frac{bx^2}{2}+cx\right]_\alpha^\beta\\\\
&=\frac{a\beta^3}{3}+\frac{b\beta^2}{2}+c\beta-\frac{a\alpha^3}{3}-\frac{b\alpha^2}{2}-c\alpha\qquad(2)\end{align*}
(1) can be re-written as $ \frac{3a\alpha\beta^2+3a\beta^3+3b\beta^2+6c\beta-6a\alpha\beta^2-3a\alpha^3-3a\alpha^2\beta+3b\alpha^2-6c\alpha+6a\alpha^2\beta}{6}$.

(2) can be re-written as $ \frac{2a\beta^3+3b\beta^2+6c\beta-2a\alpha^3-3b\alpha^2-6c\alpha}{6}$.

Subtracting the second overly complicated expression from the first yields $ \frac{3a\alpha\beta^2+a\beta^3-6a\alpha\beta^2-a\alpha^3-3a\alpha^2\beta+6a\alpha^2\beta}{6}$.

This can then be further simplified to $ \frac{a\left(-3\alpha\beta^2+\beta^3-\alpha^3+3\alpha^2\beta\right)}{6}$.

Finally, factoring the numerator gives $ A=\frac{a\left(\beta-\alpha\right)^3}{6}$.[/hide]

Here's the way Archimedes phrased this problem:

A line intersects a parabola at points $ A$ and $ B.$ This line cuts off a bounded region within the parabola whose area we want to find. Now find point $ C$ on the parabola such that a line through $ C$ parallel to the axis of the parabola passes through the midpoint of $ \overline{AB}.$ The area of the parabolic segment is exactly $ \frac43$ times the area of triangle $ ABC.$

The essence of the "official" proof given by Archimedes (not the infinitesimals and law-of-the-lever argument in the [i]Method[/i]):

The parabolic segment cut off by $ \overline{AB}$ consists of triangle $ ABC$ plus two smaller parabolic segments, cut off by $ \overline{AC}$ and $ \overline{CB}.$

Find points $ D$ and $ E$ so that $ D$ is related to $ A$ and $ C$ exactly as $ C$ is related to $ A$ and $ B$, and $ E$ is related to $ C$ and $ B$ in the same fashion.

We need to show that $ \text{area}(\triangle ADC)\equal{}\text{area}(\triangle CEB)\equal{}\frac18\text{area}(\triangle ACB).$ As a first step in showing this, we show that the distance from $ D$ to the midpoint of $ \overline{AC}$ is $ \frac14$ of the distance from $ C$ to the midpoint of $ \overline{AB}.$

Once we've done the geometric work in the previous step, the procedure iterates. Take out those two triangles and we have four new parabolic segments, and so on. We get that the area of the original parabolic segment is

$ \text{area}(\triangle ACB)\left(1\plus{}2\cdot\frac18\plus{}4\cdot\left(\frac18\right)^2\plus{}\cdots\right).$

This is a geometric series; summing it gives us our result.

[quote="i_like_pie"][hide="#1 Solution"]The area under the parabola is $ \int_\alpha^\beta ax^2 + bx + c\,\mathrm{d}x$, and the area under $ y = ux + v$ is $ \int_\alpha^\beta ux + v\,\mathrm{d}x$.

As Carcul has already said, $ u = a\left(\alpha + \beta\right) + b$ and $ v = c - a\alpha\beta$.

The second integral now becomes $ \int_\alpha^\beta x\left(a\left(\alpha + \beta\right) + b\right) + c - a\alpha\beta\,\mathrm{d}x$.
\begin{align*}\int_\alpha^\beta x\left(a\left(\alpha + \beta\right) + b\right) + c - a\alpha\beta\,\mathrm{d}x & = \left[\frac {x^2\left(a\left(\alpha + \beta\right) + b\right)}{2} + cx - ax\alpha\beta\right]_\alpha^\beta \\
 \\
& = \frac {\beta^2\left(a\left(\alpha + \beta\right) + b\right)}{2} + c\beta - a\alpha\beta^2 - \frac {\alpha^2\left(a\left(\alpha + \beta\right) + b\right)}{2} - c\alpha + a\alpha^2\beta \\
 \\
& = \frac {a\alpha\beta^2 + a\alpha\beta^3 + b\beta^2}{2} + c\beta - a\alpha\beta^2 - \frac {a\alpha^3 + a\alpha^2\beta + b\alpha^2}{2} - c\alpha + a\alpha^2\beta\qquad (1)\end{align*}

\begin{align*}\int_\alpha^\beta ax^2 + bx + c\,\mathrm{d}x & = \left[\frac {ax^3}{3} + \frac {bx^2}{2} + cx\right]_\alpha^\beta \\
 \\
& = \frac {a\beta^3}{3} + \frac {b\beta^2}{2} + c\beta - \frac {a\alpha^3}{3} - \frac {b\alpha^2}{2} - c\alpha\qquad(2)\end{align*}
(1) can be re-written as $ \frac {3a\alpha\beta^2 + 3a\beta^3 + 3b\beta^2 + 6c\beta - 6a\alpha\beta^2 - 3a\alpha^3 - 3a\alpha^2\beta + 3b\alpha^2 - 6c\alpha + 6a\alpha^2\beta}{6}$.

(2) can be re-written as $ \frac {2a\beta^3 + 3b\beta^2 + 6c\beta - 2a\alpha^3 - 3b\alpha^2 - 6c\alpha}{6}$.

Subtracting the second overly complicated expression from the first yields $ \frac {3a\alpha\beta^2 + a\beta^3 - 6a\alpha\beta^2 - a\alpha^3 - 3a\alpha^2\beta + 6a\alpha^2\beta}{6}$.

This can then be further simplified to $ \frac {a\left( - 3\alpha\beta^2 + \beta^3 - \alpha^3 + 3\alpha^2\beta\right)}{6}$.

Finally, factoring the numerator gives $ A = \frac {a\left(\beta - \alpha\right)^3}{6}$.[/hide][/quote]

Good job, i_like_pie :)  You showed in case of $ a>0.$ How about $ a<0$?

Let's wait others solution.

[quote="kunny"][quote="i_like_pie"][hide="#1 Solution"]The area under the parabola is $ \int_\alpha^\beta ax^2 + bx + c\,\mathrm{d}x$, and the area under $ y = ux + v$ is $ \int_\alpha^\beta ux + v\,\mathrm{d}x$.

As Carcul has already said, $ u = a\left(\alpha + \beta\right) + b$ and $ v = c - a\alpha\beta$.

The second integral now becomes $ \int_\alpha^\beta x\left(a\left(\alpha + \beta\right) + b\right) + c - a\alpha\beta\,\mathrm{d}x$.
\begin{align*}\int_\alpha^\beta x\left(a\left(\alpha + \beta\right) + b\right) + c - a\alpha\beta\,\mathrm{d}x & = \left[\frac {x^2\left(a\left(\alpha + \beta\right) + b\right)}{2} + cx - ax\alpha\beta\right]_\alpha^\beta \\
 \\
& = \frac {\beta^2\left(a\left(\alpha + \beta\right) + b\right)}{2} + c\beta - a\alpha\beta^2 - \frac {\alpha^2\left(a\left(\alpha + \beta\right) + b\right)}{2} - c\alpha + a\alpha^2\beta \\
 \\
& = \frac {a\alpha\beta^2 + a\alpha\beta^3 + b\beta^2}{2} + c\beta - a\alpha\beta^2 - \frac {a\alpha^3 + a\alpha^2\beta + b\alpha^2}{2} - c\alpha + a\alpha^2\beta\qquad (1)\end{align*}

\begin{align*}\int_\alpha^\beta ax^2 + bx + c\,\mathrm{d}x & = \left[\frac {ax^3}{3} + \frac {bx^2}{2} + cx\right]_\alpha^\beta \\
 \\
& = \frac {a\beta^3}{3} + \frac {b\beta^2}{2} + c\beta - \frac {a\alpha^3}{3} - \frac {b\alpha^2}{2} - c\alpha\qquad(2)\end{align*}
(1) can be re-written as $ \frac {3a\alpha\beta^2 + 3a\beta^3 + 3b\beta^2 + 6c\beta - 6a\alpha\beta^2 - 3a\alpha^3 - 3a\alpha^2\beta + 3b\alpha^2 - 6c\alpha + 6a\alpha^2\beta}{6}$.

(2) can be re-written as $ \frac {2a\beta^3 + 3b\beta^2 + 6c\beta - 2a\alpha^3 - 3b\alpha^2 - 6c\alpha}{6}$.

Subtracting the second overly complicated expression from the first yields $ \frac {3a\alpha\beta^2 + a\beta^3 - 6a\alpha\beta^2 - a\alpha^3 - 3a\alpha^2\beta + 6a\alpha^2\beta}{6}$.

This can then be further simplified to $ \frac {a\left( - 3\alpha\beta^2 + \beta^3 - \alpha^3 + 3\alpha^2\beta\right)}{6}$.

Finally, factoring the numerator gives $ A = \frac {a\left(\beta - \alpha\right)^3}{6}$.[/hide][/quote]

Good job, i_like_pie :)  You showed in case of $ a > 0.$ How about $ a < 0$?

Let's wait others solution.[/quote]
edit-[hide="solution"]
take the integral of the parabola minus the integral of the line for $ a<0$
i don't know how to do integral  notation in latex.
i got $ \frac {a(\alpha - \beta) ^3}{6}$ for the case of $ a < 0$
[/hide]

No one seems to be posted solution any more, so I post the solution of problem 1.

From the condition, we can write $ ax^2 \plus{} bx \plus{} c \minus{} (ux \plus{} v) \equal{} a(x \minus{} \alpha)(x \minus{} \beta)\ (\alpha < \beta)$, thus desired area is $ S \equal{} \minus{} a\int_{\alpha }^{\beta} (x \minus{} \alpha )(x \minus{} \beta)dx \equal{} \frac {a}{6}(\beta \minus{} \alpha )^3$ for $ a > 0.$ Similarly we have $ S \equal{} \minus{} \frac {a}{6}(\beta \minus{} \alpha )^3$ for $ a < 0.$ As the result, We can find directly the area as $ S \equal{} \left|\int_{\alpha}^{\beta} a(x \minus{} \alpha )(x \minus{} \beta)\ dx\right| \equal{} \frac {|a|}{6}(\beta \minus{} \alpha )^3$.

Note: $ \boxed{\int_{\alpha}^{\beta} (x \minus{} \alpha )(x \minus{} \beta)\ dx \equal{} \minus{} \frac {1}{6}(\beta \minus{} \alpha )^3},$ which is written in Japanese high school mathematics text today.

So can we for Problem 2.


Fie $\ a,b,c,d$ naturale consecutive.Sa se arate ca
 $\ [\frac{\sqrt a+\sqrt b+\sqrt c+\sqrt d}{2}]$=$\ [\sqrt{a+b+c+d}]$, unde $\ [x]$ este partea intreaga a numarului real $\ x$.
     $\ Dana\ Piciu,\ Craiova$

enuntul e lafel cu
$\left[\frac{\sqrt{n} + \sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}}{2}\right]=\left[\sqrt{4n+6}\right]$ pentru n natural.
dar oricare n natural el e cuprins intre 2 patrate
$k^2\leq n \leq (k+1)^2$ (se observa $[\sqrt{n}]=k$)
se poate observa ca $LHS \geq 2k$
se poate observa deasemeni ca $RHS \geq 2k$
P.R.A. daca $RHS=2k+1 \ \Rightarrow (2k+1)^2 \leq 4k^2 +6 \ \Rightarrow$
 contradictie  pentru k suficient de mare.
P.R.A daca $LHS=2k+1 \ \Rightarrow (2k+1)^2 \leq LHS \Rightarrow$
  -//-

deci de la un k suficient de mare(maximul din cele doua parti de mai sus)
rezulta ca $LHS=RHS=2k$.
sper sa fie bine

Sorry, dar ce inseamna $\ LHS$ si $\ RHS$?


Given $ x \equal{} \bar{x} \plus{} \beta \sqrt{D^2 \plus{} \bar{x}^2}$, how does one obtain $ \bar{x} \equal{} \gamma^2x \minus{} \beta\gamma\sqrt{D^2\plus{}(\gamma x)^2}$? The simplification $ \gamma \equal{} \frac {1}{\sqrt{1 \minus{} \beta^2}}$ may be used.

The problem i'm facing is that i don't get the correct required expression..

[quote="aidan"]Given $ x \equal{} \bar{x} \plus{} \beta \sqrt {D^2 \plus{} \bar{x}^2}$, how does one obtain $ \bar{x} \equal{} \gamma^2x \minus{} \beta\gamma\sqrt {D^2 \plus{} (\gamma x)^2}$? The simplification $ \gamma \equal{} \frac {1}{\sqrt {1 \minus{} \beta^2}}$ may be used.

The problem i'm facing is that i don't get the correct required expression..[/quote]
Writing $ x \equal{} \bar{x} \plus{} \beta \sqrt {D^2 \plus{} \bar{x}^2}$ as
$ x \minus{} \bar{x} \equal{} \beta \sqrt {D^2 \plus{} \bar{x}^2}$  and squaring on both sides, we get
[hide]$ (x \minus{} \bar{x})^2 \equal{} \beta ^2 ({D^2 \plus{} \bar{x}^2})$
collecting all the like terms, we get
$ (1\minus{}\beta ^2) \bar{x}^2 \minus{}2x\bar{x}\plus{}x^2\minus{}\beta ^2 D^2\equal{} 0$
Dividing throughout by $ (1\minus{}\beta ^2)$ and using $ \gamma \equal{} \frac {1}{\sqrt {1 \minus{} \beta^2}}$, we have
$ \bar{x}^2 \minus{}2x\gamma ^2\bar{x}\plus{}\gamma ^2(x^2\minus{}\beta ^2 D^2)\equal{} 0$
Completing the perfect sqaure, we have
$ (\bar{x} \minus{}x\gamma ^2)^2\equal{}\gamma ^4x^2\minus{}\gamma ^2x^2\plus{}\gamma ^2\beta ^2 D^2\equal{}\gamma ^2x^2(\gamma ^2\minus{}1)\plus{}\gamma ^2\beta ^2 D^2$
Observing $ \gamma ^2\minus{}1\equal{}\gamma ^2\beta ^2$, we can write
$ (\bar{x} \minus{}x\gamma ^2)^2\equal{}\gamma ^4\beta ^2x^2\plus{}\gamma ^2\beta ^2 D^2$
Further simplification gives
$ (\bar{x} \minus{}x\gamma ^2)^2\equal{}\gamma ^2\beta ^2((\gamma x)^2\plus{} D^2)$
Taking the square roots on both sides, we can get [/hide]the required expression as one of the answers.

Oh.. I didn't divide throughout by $ \frac {1}{\sqrt{1 \minus{} \beta^2}}$ first..I just applied the quadratic formula directly. Thanks :)


Giving: $ F(x)\equal{}\frac{ln(1\plus{}x)}{x}$ $ \forall x\neq0$ and $ F(0)\equal{}1$.

evaluate this limit by using that: $ e^{x}\equal{}1\plus{}x\plus{}\frac{x^{2}}{2}\plus{}x^{2}t(x)$ with $ \lim_{x\to0}t(x)\equal{}0$ :

$ \lim_{x\to0}\frac{ln(x\plus{}1)\minus{}1}{x}$

thanx!.

hello, by the rule of L'Hospital we get
$ \lim_{x \to 0}\frac{\ln(x\plus{}1)\minus{}1}{x}\equal{}\lim_{x \to 0}\frac{1}{1\plus{}x}\equal{}1$.
Sonnhard.

$ \lim_{x\to0}\frac{ln(x\plus{}1)\minus{}1}{x}  \equal{} \left|\begin{array}{l} y\equal{}\ln(1\plus{}x) \to 0 \\ \end{array}\right| \equal{} \lim_{y\to 0} \frac{y\minus{}1}{e^y\minus{}1} \equal{} \lim_{y\to 0} \frac{y\minus{}1}{1\plus{}y\plus{}y^2/2 \plus{} y^2o(1)\minus{}1} \equal{} \lim_{y\to 0} \frac{y\minus{}1}{y\plus{}yo(1)} \equal{} \lim_{y\to 0} \frac{1\minus{}\frac{1}{y}}{1\plus{}o(1)} \equal{}1$

This limit does not exist, does it? It goes to $ \plus{}\infty$ from the left and $ \minus{}\infty$ from the right. I thought L'Hopstal's rule was only applicable for limits of the indeterminate forms $ 0/0$ or $ \infty/\infty$. This one is not even an indeterminate form. That said, I am quite sure I am missing something important. Could someone point it out?

Haaaaaaa)))))))
[b]KBriggs[/b] you are right )))
this limit is not exists. What a stupid mistake i made )
\[ \lim_{y\to 0}\frac{y\minus{}1}{e^{y}\minus{}1} \equal{} \frac{\minus{}1}{0} \equal{}\infty\]

i garble this limit with $ \lim_{x\to0}\frac{ln(1\plus{}x)}{x}$... and i was inattentive ))  :blush:
Thanks [b]KBriggs[/b]

I don't understand why defining F(x) was necessary.


trigonometry
geometry
algebra
polynomial
logarithms
analytic geometry
graphing lines

1) What is the leftmost digit of $2006^{2006}$

2) In $\triangle{ABC}$, $AD$ and $AE$ trisect $\angle{BAC}$. $BD=2, DE=3, EC=6$, what is the length of the shortest side of $\triangle{ABC}$?

3) Find the largest value of $\frac{y}{x}$ if $(x-3)^2+(y-3)^2=6$

4) $z^4+az^3+bz^2+cz+d$ has only complex roots which lie on a circle in a complex plane centered at $0+0i$ and has radius $1$. What is the sum of the reciprocals of the roots in terms of $a$, $b$, $c$, or $d$?

[quote="joml88"][hide="Number 2"]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=578[/img]
By the angle bisector theorem we have  \[ 3x=2e \ \text{and} \ y=2d.  \]

From the Law of Cosines and since $\cos \angle AEC=-\cos \angle AEB$ and $\cos \angle ADB=-\cos \angle ADC$

\begin{eqnarray} x^2 &=& d^2+4+4d\cos\angle ADC\\ e^2 &=& d^2+9-6d\cos \angle ADC\\ y^2 &=& e^2+36-12e\cos \angle AEC\\ d^2 &=& e^2+9+6e\cos \angle AEC\\ \end{eqnarray}

So then

\[ 3(1)+2(2) = 3x^2+3e^2=5d^2+30  \]
\[ (3)+2(4) = y^2 + 2d^2 = 3e^2 + 54.  \]

Then plug $x=\frac 23e$ in the first and $y=2d$ in the second to get

\begin{eqnarray*} 2d^2-e^2 &=& 18\\ -5d^2 +\frac{10}3e^2 &=& 30 \Rightarrow e = 3\sqrt{10} && d=3\sqrt{6} \end{eqnarray*}

So then $x=\frac 23 e=2\sqrt{10}=\sqrt{40}$ and $y=2d=6\sqrt{6}=\sqrt{216}$.  Clearly $AB=x=2\sqrt{10}$ is the smallest.
[/hide]

I'll leave the rest for other people to solve...[/quote]

 :yup: except I didn't use trig to solve it, I simply used angle bisector theorem, and found the lengths of all the bisectors, etc..

[quote="joml88"]How'd you find the lengths of the bisectors....Stewart's?  If so, I wouldn't realy call that simpler :P  Look at where I added up the numbered equations (i.e. $3(1)+2(2)$, etc).  Look at what those equal...they look suspiciously like what Stewart's would give you (in fact they are what you would get :D ).  Indeed, you can actually prove Stewart's with LoC (just do what I did but with variables instead).[/quote]

[hide="length of bisectors"]Let $AB, AD, AE, AC$ be $w,x,y,z$ respectively. Using angle bisector theorem, we know that $2y=3w$, $x^2+6=wy$, $z=2x$, $y^2+18=xz$, so we have a bunch of equations now. We can get the system of equations: $\frac{2}{3}y^2-x^2+6=0$ and $y^2+18-2x^2=0$. So we can get $x=\sqrt{54}, y=\sqrt{90}$, thus $w=2\sqrt{10}$ and $z=6\sqrt{6}$, so $\boxed{2\sqrt{10}}$ is the answer[/hide]

I would've used Stewart's, but I almost never use it in geometry problems (hard to memorize anyway)

Here's a problem gauss202 posted but noone solved yet..

$1+11+111+....+11....11111$ (2002 ones) has how many ones?
[hide="solution"]

Notice that it is a finite geometric series for each one of the terms $(1+10+100...)$ etc. So we can put this as $\frac{1+10}{1+10}+\frac{1+10+100}{1+10}+....+\frac{1+10+...1000...}{1+10}=\frac{2002+11111...11110}{11}$, which contains 2002 ones. So we have $182+\frac{11111...110}{11}$ and then doing long division, we see that the quotient has 2002 digits and is in the form $101010101...10$ which contains $1001$ ones. and adding 182, it still has $\boxed{1001} ones$[/hide]

Oh, well...I don't consider $x^2+6=wy$ to be the angle bisector theorem.  I consider it to be a combination of angle biscector and stewart's/LoC.  But, whatever, it doesn't really matter which way you look at it :P

If anybody needs hints, here they are:

[hide="hint problem 1"]logs[/hide]

[hide="hint problem 3"]graph[/hide]

[hide="hint problem 4"]transform the polynomial[/hide]

[quote="joml88"]About number 4...does it give you extraneous information.  Because I got an answer but I don't think I used the facts that the roots are on the unit circle... (I might've missed something while doing the problem)[/quote]

Um..to be honest, it was an old AHSME problem that I remembered but I don't think it gave anything else :? , what exactly did you get because now that I think about it, there can be more than one way to express the answer.
[hide="what I got"]
$\frac{-c}{d}$ or since it's a unit circle $-c$ or just $c$[/hide]

I forgot the answer choices and I don't remember which test or question this was :?

[quote="joml88"]Yeah, that's what I got as well.[/quote]

should i/you post the solution for it or should we let others solve it (if anyone else even cares)

The thing I wasn't sure about is whether the transformation of the polynomial for the reciprocal roots works for polynomials with only complex roots as well :?:

[quote="arkmastermind"][hide="Number 1"] Well...technically if this were an AMC type problem i suppose it could be done without a calculator, but caclulators are allowed...so.

Let $n$ be the number of digits in $2006^{2006}$. Then $n=[\log{2006^{2006}}]+1$. Thus $n=6225$. If we divide $2006^{2006}$ by $10^{6224}$, we will be left with the first digit and then the first few numbers following the decimal place after that..

Denote $q$ to be the quotient we are looking for, then $[q]$ will be the leftmost digit of $2006^{2006}$.

$q=2006^{2006} / 10^{6224}\\ \log{q}=2006\log{2006} -6224\\ \log{q} = 0.47584294...\\ q=2.99118...$

The leftmost digit is thus $\boxed{2}$.[/hide][/quote]

 :yup: , nice to see you posting here Asif, and no it wasn't an AMC problem, I just made it up..but I don't think you can do it without a calculator.

[quote="arkmastermind"]for number 3, i didnt really work it out, but im guessing $3+\sqrt{6}$?[/quote]

[hide="mysolution"]Obviously we have a circle centered at $(3,3)$ with radius $\sqrt{6}$, the largest solution point is tangent to the circle so by drawing that tangent, we form a right triangle with base $\sqrt{6}$ and hypotenuse $3\sqrt{2}$ (from the $45,45,90$ triangle that contains the center of the circle at it's top vertex). Notice that $\frac{y}{x}$ seems strangely falmiliar, and it involves $\tan$, so by calling the small angle of the right triangle with sides $3\sqrt{2}, \sqrt{6}, 2\sqrt{2}$ as $\theta$, we want to know the entire angle, which is $\theta+45$, so $\frac{y}{x}=\tan{(\theta+45)}=\frac{\frac{\sqrt{2}}{2}+1}{1-\frac{\sqrt{2}}{2}}=\boxed{2\sqrt{2}+3}$[/hide]

hmm, i didn't get what you got so one of us is wrong..so how did you solve it?  :?

EDIT: Didn't rationalize the denominator right

i dont seem to be able to visualize your solution that well...

i didnt really work it out, i just took a guess that the largest value for y/x would occur either at either of the circles four extreme points (topmost, bottom most, left most, right most). upon inspection it seemed that the leftmost point on the circle would have the greatest slope from the origin to it, and thus the greatest y/x. so the y value was 3 and the x value was $3-\sqrt{6}$ so, dividing those i got $3+\sqrt{6}$.

But seeing as you posted the problems...you probably have the right answer  :P

Hmm, I disagree with both of you! lol

But I checked with Geometer's Sketchpad to see if my answer agrees, and it does :D

First, let $\frac yx=k$ so that $y=kx$.  This is just a series of lines through the origin with different slopes.  We are looking for the one with the largest slope (i.e. $k$).  The line has to intersect the circle though.  So imagine picking a point on the circle and drawing a line through it and the origin.  Then just imagine moving this point around so that the line will be getting either more or less steep.  When will it be tilting upwards the most?  This will occur when the line is tangent to the circle (actually there are two places where it will be tangent...the other is the minimum).  The reason that Asif's reasoning is wrong is that, although we are moving the point on the circle as far to the left as possible (in a sense), we have to be aware of the fact that once we pass the point of tangency we will still be moving left but there will be another point of interesection with the circle to the right of the point of tangency.  So instead of moving the point as far left as possible, we have to consider the farthest right intersection point and move it as far left as possible (this occurs when it [i]is[/i] the other intersection point).

Ok, so now we know what we're after.  Let $(x,y)$ be the point of tangency.  The slope between the line through $(0,0)$ and $(x,y)$ and the line through $(x,y)$ and $(3,3)$ are perpendicular.  Thus:

\[ \frac{y-3}{x-3}=-\frac xy \\ \Rightarrow x^2+y^2 = 3x+3y.  \]

Then expand $(x-3)^2+(y-3)^2 = 6$ and plug in from above, cancel, simplify, etc. and get $x+y=4\Leftrightarrow y=4-x$.  Plug this back into the original equation for the circle and we find that $x=2\pm\sqrt{2}$ which makes $y=2\mp\sqrt{2}.$  Therefore, the minimum $\frac yx$ is $\frac{2-\sqrt{2}}{2+\sqrt{2}}=3-2\sqrt{2}$ and the max is its reciprocal: $3+2\sqrt{2}$.

Whew!

Oh, and Shobhit, how can you have a 45-45-90 right triangle with legs $\sqrt{6}$ and hypotenuse $3\sqrt{2}$ ;)

ah duh. of course it had to be the tangent line. *hits self on head*

lol. maybe i should have actually worked it out instead of guessing around  :roll:

[quote="joml88"]Hmm, I disagree with both of you! lol
Oh, and Shobhit, how can you have a 45-45-90 right triangle with legs $\sqrt{6}$ and hypotenuse $3\sqrt{2}$ ;)[/quote]

Joe, I didn't mean that was the triangle, i got $3\sqrt{2}$ FROM the hypotenuse of the 45,45,90 triangle, and I guess I did it right since I got the same answer as you, I just didn't rationalize $\frac{\frac{\sqrt{2}}{2}+1}{1-\frac{\sqrt{2}}{2}}$ correctly since it's equal to $2\sqrt{2}+3$ not $2\sqrt{2}+6$ :blush: , I made a diagram to explain myself, the $45,45,90$ triangle comes from the $x$ and $y$ values of the center of the circle...

[quote="joml88"]Oh gothca.  I was just quickly scanning your solution to see where you might have gone wrong and once I got there I stopped reading  :oops: .

Btw, no offense or anything but that diagram is pretty horrendous :D   You know that paint has line and ellipse tools, right?  You can even hold down shift and it will make the lines at straight angles, i.e. the lines will go in one of the four direction or one of the four intermediate directions.  Holding the shift key down for the ellipse tool draws circles.  Paint is far from great but you can still make semi-decent diagrams with it.[/quote]

I know, i was just doing the graph quickly and didn't really think about it's neatness :rotfl: . I cared more about getting the picture to work, since it was the first picture i posted on Aops.


Finally I find a problem worth calling "MATHCOUNTS [b]Challenge[/b]" :)

Either prove or disprove that Associative Property is valid under exponent: a^(b^c) equal or not equal to (a^b)^c

also, do the same for Commutative Property under Exponent.

hmm...its a bit of common sense...

[quote="Tare"]a^(b^c)[/quote]
That equals a^b^c...

[quote="Tare"](a^b)^c [/quote]
equals abc...

Clearly, they're not equal...or am i misunderstanding something?

I'd put it in spoiler...but its probly not hard enuf for that...

Hmm...that's what I thought before my math teacher confuted me :lol: by raising tons of examples...
sorry, just had to say that or else I'll go crazy :lol:

you know, maybe, just maybe I used "confuted" correctly this time...*Apollo 13 theme song plays* :)
Onto MithsApprentice! :lol: </jk>

Are you sure? I think he's right. a^b^c = a^bc if and only if bc = b^c, and that's clearly not the case all the time.

Well...I'll leave it up to you guys prove it (or disprove it). Remember, I'm not saying which is right because that's part of the question.

Well, I mean, it's trivial. Here's a case where that doesn't work.

a = 4 b= 2, c = 3

4^2^3 = 4 ^8
4^2*3 = 4^6

4^6 /= 4^8.

[quote="MysticTerminator"]remember...to disprove something, you need only offer one counter-example. For example, if we wanted to disprove the conjecture that n:^2:+n+41 outputs a prime for each n. we need only show that it doesn't for n=41.[/quote]

n=40 is actually the smallest positive integer such that it's not a prime.

Any proofs/disproofs for the Commutative Property?

...is "disproof" a word? :)
whew, it would've been embarrasing if I made up a word unintentionally (like the time my teacher misheard me say "coordinate system" and thought I said "para-coordinate system", which later on was defined as "coordinate systems in which instead of points, there are poligonal "boxes" in which numbers fit, as is the Pascal's Triangle"...oh well :) )

Yes, it is. Often on problems you will be asked to prove or disprove.

wouldnt $ a^{(}b^{c})$ be $ a^{bc}$ and $ (a^{b})^{c}$ be $ a{}^{b}{}^{c}$ which is $ a^{bc}$ so they are equal.


What is the 100th digit to the right of the decimal point in the
decimal representation of $ \frac{13}{90}$?

After the initial $ 1$, all digits to the right are $ \fbox{4}$ in the decimal representation of $ \frac{13}{90}$ due to properties of fractions in the form $ \frac{n}{9}$ being expressed with a $ 0$ after the $ 9$ in the denominator.


Sequence
bounded
algebra
IMO
imo 1991
construction
Harm Derksen

An infinite sequence $ \,x_{0},x_{1},x_{2},\ldots \,$ of real numbers is said to be [b]bounded[/b] if there is a constant $ \,C\,$ such that $ \, \vert x_{i} \vert \leq C\,$ for every $ \,i\geq 0$. Given any real number $ \,a > 1,\,$ construct a bounded infinite sequence $ x_{0},x_{1},x_{2},\ldots \,$ such that
\[ \vert x_{i} \minus{} x_{j} \vert \vert i \minus{} j \vert^{a}\geq 1
\]
for every pair of distinct nonnegative integers $ i, j$.

I'm not sure if this works:

After we have found $ x_0, x_1, ... x_{n\minus{}1}$ satisfying the conditions, we now choose an $ x_n$ as follows. Draw open intervals of lengths $ 2/1^a, 2/2^a, 2/3^a, ... , 2/n^a$ around $ x_{n\minus{}1}, x_{n\minus{}2}, ... x_1, x_0$ (so that each of these points are the midpoints of the intervals. It is clear that if $ x_n$ does not lie in any of these intervals, it satisfies the conditions of the problem. Thus we choose the $ x_n$ with smallest absolute value that does not lie in these $ n$ open intervals.

Now, because the series $ 1/1^a \plus{} 1/2^a \plus{} ...$ converges for all real $ a > 1$, these intervals cover a total area that approaches a limit, say c. Then, we can always find $ x_n$ with absolute value less than c, so the sequence is bounded.

Could you not simply do x_i = -1 ^ i? It doesn't seem like a very interesting problem unless you're looking for a monotonic bounded infinite sequence, in which case x_i = [sum from 1 to i] 2/k^{1 + epsilon} would suffice, where epsilon is small.

Yesterday I was solving this problem and I thought about exactly the same solution as alkjash's solution. But I don't know if it's really correct, beacuse we should "construct" such a sequence and I don't know if we could say it's construction. What's your opinion?

Actually, the minimal $a$ for which such a sequence exists is $a=1$, which follows from the problem [url=http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A624&l=en]Kömal A.624[/url]. ([hide=Explanation.]
Of course, in the IMO problem, the interval $[-C;C]$ can be dilated into interval $[0;1]$, hence such a sequence exists for some $a$ if and only if there is a sequence $x_1,x_2,\dots\in[0;1]$ satisfying $|x_i-x_j|\ge \frac{c}{|i-j|^a}$ for all $i,j\ge 1$ ($i\neq j$) with a suitable constant $c$.

Part b) in the Kömal (or CIIM) problem is in fact equivalent to giving a sequence satisfying $|x_i-x_j|\ge \frac c{|i-j|}$ for some suitable small $c$ - if we give a sequence $x_1,x_2,\dots\in[0;1]$ satisfying the conditions in part b), then we can simply take $y_i=x_{ri}$. (The converse is also true, but takes a bit more effort to prove.)

Part a) in the Kömal problem says that if $x_1,x_2,\dots\in[0;1]$, then for suitable $K>0$, and any positive integer $r$, we can find $i,j$ with $|i-j|\ge r$ and $|x_i-x_j|<\frac{K}{|i-j|}$. Suppose that we have a sequence $x_1,x_2,\dots\in[0;1]$ that satisfies $|x_i-x_j|\ge \frac{c}{|i-j|^a}$ $\forall i,j\ge 1$ ($i\neq j$) for some $c>0$ and $a<1$, and let's take the appropriate $K$. Then if $|i-j|\ge r$ and $|x_i-x_j|<\frac{K}{|i-j|}$, then from the condition we get
$$\frac{K}{|i-j|}\ge \frac{c}{|i-j|^a}\implies \frac{K}{c}\ge |i-j|^{1-a}\ge r^{1-a},$$
which gives a contradiction for $r>(K/c)^{1/(1-a)}$.
[/hide])

I have posted a solution to the Kömal/CIIM problem [url=http://artofproblemsolving.com/community/c7h621441]here[/url].

There's an easy construction for the IMO problem as well, in the case $a=1$ actually. Just let $x_n=\{n\sqrt 2\}$ and use the well-known estimate
$$|a\sqrt 2-b|=\frac{|2a^2-b^2|}{|a\sqrt 2+b|}\ge \max\left(1,\frac1{a\sqrt 2+(a\sqrt 2+1)}\right)\ge \frac1{4a}$$
for positive integer $a,b$. (This is a Liouville-type estimate such as [url=http://www.artofproblemsolving.com/community/c2927h1057645_liouvilles_approximation_theorem]here[/url].)


Recall that the Fibonacci numbers F(n) are defined by F(0) = 0, F(1) = 1,
and F(n) = F(n−1)+F(n−2) for n ≥ 2. Determine the last digit of
F(2006)

[hide="Less brute force and more generalizable"]Find the units digits of each term until one is a multiple of 10. This occurs at $ F_{15}$. This sequence will repeat itself, but each time it is multiplied by $ F_{14}$. In this case the units digit of that is 7.

So $ F_{n}\equiv 7F_{n\minus{}15}\pmod{10}$, and $ F_{2006}\equiv 7^{133}F_{11}\equiv 7 F_{11}\pmod{10}$ (note $ 7^{4}\equiv 1\pmod{10}$). The units digit of $ F_{11}$ is 9, and $ 7\cdot 9\equiv\boxed{3}\pmod{10}$.[/hide]


Find X :
- X is a natural number
- X has a most number of divisol
- 9<X<100
Thank you for reading. :lol:

What do you mean "X has a most number of divisol"?

Intuitively, is the answer 60?
and of course, i'm assuming "divisol" in your language, is "natural divisors" :)

$ 60,72,84,90$ were the answers I found. I believe it is not possible to get more than $ 12$ divisors, but the only solution I have is casework and I haven't checked it thoroughly.

$ \{(2^5)(3),(2^3)(3^2),(2)(3^2)(5),(2^2)(3)(5),(2^2)(3)(7)\}\equal{}\{96,72,90,60,84\}$ is, I believe, the complete list (they all have 12 divisors). Those numbers with more than 3 prime divisors are ruled out as they only have at most 6 divisors. Prime powers are also ruled out as they have at most 7 divisors. So it lefts to check with the 2 and 3 prime divisors cases.


Find all positive integers w, x, y, z satisfying the following:

(a) w^5=x^2
(b) y^3=z^2
(c) y - w=49

how do you type in latex? =/

edit: how do you delete posts? i think i posted on the wrong forum. =/

This topic should probably be in High School Basics.

[hide="A Summary on How to Type in Latex"]
A first step and very important step would be to just use the $ \$$ sign (on the $ 4$ key buttom) in between expressions you want to post. Example: if you type "w^5=x^2", just finish off by posting one $ \$$ sign on both ends of the expression like $ \$$w^5=x^2$ \$$, to get $ w^5 \equal{} x^2$.
Also, you can add other stuff like type in the expression \frac{n+1}{n-1}, and using $ \$$ signs to get the fraction $ \frac {n \plus{} 1}{n \minus{} 1}$.
You can learn more codes like [b]\sum[/b] to get $ \sum$, [b]\binom{a}{b}[/b] to get $ \binom{a}{b}$, etc.
Notice the use of {}. With that, instead of typing w^(2x+1), which gives this $ w^(2x \plus{} 1)$, use w^{2x+1} to get $ w^{2x \plus{} 1}$.
Characters like alpha, beta, sigma, and gamma can easily be typed by: \alpha for $ \alpha$, \beta for $ \beta$, \infty for $ \infty$,etc.
Two other commands that will be important in addition with those I've given is the use of _ and ^. For example, x_2 gives $ x_2$. Also, type \sum_{n=0}^{\infty}{n+1} to get $ \sum_{n \equal{} 0}^{\infty}{n \plus{} 1}$.
A way to learn more codes and ways to type an expression, one can always refer to clicking an expression from someone and see the codes he/she used (notice they don't show the $ \$$ sign). Overall, the $ \$$ sign is a must use.
Give it a try. This method had worked for me when I learned to type in latex. :D  [/hide]

[quote="Patterns_34"][hide="A Summary on How to Type in Latex"]
A first step and very important step would be to just use the $ \$$ sign (on the $ 4$ key buttom) in between expressions you want to post. Example: if you type "w^5=x^2", just finish off by posting one $ \$$ sign on both ends of the expression like $ \$$w^5=x^2$ \$$, to get $ w^5 = x^2$.
Also, you can add other stuff like type in the expression \frac{n+1}{n-1}, and using $ \$$ signs to get the fraction $ \frac {n + 1}{n - 1}$.
You can learn more codes like [b]\sum[/b] to get $ \sum$, [b]\binom{a}{b}[/b] to get $ \binom{a}{b}$, etc.
Notice the use of {}. With that, instead of typing w^(2x+1), which gives this $ w^(2x + 1)$, use w^{2x+1} to get $ w^{2x + 1}$.
Characters like alpha, beta, sigma, and gamma can easily be typed by: \alpha for $ \alpha$, \beta for $ \beta$, \infty for $ \infty$,etc.
Two other commands that will be important in addition with those I've given is the use of _ and ^. For example, x_2 gives $ x_2$. Also, type \sum_{n=0}^{\infty}{n+1} to get $ \sum_{n = 0}^{\infty}{n + 1}$.
A way to learn more codes and ways to type an expression, one can always refer to clicking an expression from someone and see the codes he/she used (notice they don't show the $ \$$ sign). Overall, the $ \$$ sign is a must use.
Give it a try. This method had worked for me when I learned to type in latex. :D  [/hide][/quote]

If you need to look up $ \text{\LaTeX}$ code you can do it here: http://en.wikipedia.org/wiki/Help:Displaying_a_formula

Or maybe you could actually look for a few seconds and find the forum's guide, which is probably one of the best on the web?

[[LaTeX]] <-- Ooh, hard to find.

EDIT: Actually, shouldn't dio equations be in Intermediate Topics?

[hide="solution"]Since w through z are positive integers, w and y are perfect squares. Therefore, the difference of two squares is a different square, and we have a pythagorean triple: $ (7,\sqrt{w},\sqrt{y})$. The only pythagorean triple with 7 in it is $ (7,24,25)$. That gives us two solutions for w and y are (576,625) and (625,576). Those give us two solutions for $ (w,x,y,z)$:  $ (576, 24^5, 625, 25^3), (625, 25^5, 576, 24^3)$.
[/hide]


calculus
integration
trigonometry
derivative
calculus computations

Find the minimum value of the following definite integral.

$ \int_0^{\pi} (a\sin x \plus{} b\sin 3x \minus{} 1)^2\ dx.$

[hide]
Hi kunny-hun  :lol:

can I use this relation? Let $ \int_{\alpha }^{\beta } f(x,a,b) \, dx \equal{} G(a,b)$

Since $ f$ is continuosly and differentiable in the interval, we can get the system:

$ 0 \equal{} \frac {\partial G(a,b)}{\partial a} \equal{} \int_{\alpha }^{\beta } \frac {\partial f(x,a,b)}{\partial a} \, dx$ , $ 0 \equal{} \frac {\partial G(a,b)}{\partial b} \equal{} \int_{\alpha }^{\beta } \frac {\partial f(x,a,b)}{\partial b} \, dx$

If so, is rather simply to show:

$ \frac {\partial (a \sin (x) \plus{} b \sin (3 x) \minus{} 1)^2}{\partial a} \equal{} 2 a \sin ^2(x) \plus{} 2 b \sin (3 x) \sin (x) \minus{} 2 \sin (x)$
$ \frac {\partial (a \sin (x) \plus{} b \sin (3 x) \minus{} 1)^2}{\partial b} \equal{} 2 b \sin ^2(3 x) \plus{} 2 a \sin (x) \sin (3 x) \minus{} 2 \sin (3 x)$

$ \int_0^{\pi } \sin (x) \sin (3 x) \, dx\equal{}0$ and the others is simply

So the integral become very basic, and is easy to show:

$ \int_0^{\pi } \frac {\partial (a \sin (x) \plus{} b \sin (3 x) \minus{} 1)^2}{\partial a} \, dx \equal{} a \pi \minus{} 4$ , $ \int_0^{\pi } \frac {\partial (a \sin (x) \plus{} b \sin (3 x) \minus{} 1)^2}{\partial b} \, dx \equal{} b \pi \minus{} \frac {4}{3}$

$ \left\{a \equal{} \frac {4}{\pi },b \equal{} \frac {4}{3 \pi }\right\}$

The rest is just show that this is the minimum.
But this solutions is to big, that i prefer someone post a simple one  :blush: [/hide]

We don't learn partial differentiation in high school in my country. :P


Find all polynomials $p(x)$ ( with real coefficients ) such that for every $x \in R$ :

             $1+p(3x) = \frac{p(x-1)+p(x)+p(x+1)}{3}$


Why is there still a "Join Jam" button next to the Count Down math jam that happened on May 5?

No big deal...they probably just forgot to remove the link or close the classroom or something...


number theory
greatest common divisor
modular arithmetic

1) Find all positive integers , representable uniquely as $ \frac {x^2 \plus{} y}{xy \plus{} 1}$ where $ x$ and $ y$ are postive integers.

2) Determine all ordered pair $ (m,n)$ of positive integers such that $ \frac {n^3 \plus{} 1}{mn \minus{} 1}$  is a integer.

3) Determine all pairs of $ (a,b)$ such that $ \frac {a^2}{2ab^2 \minus{} b^3 \plus{} 1}$ is a positive integer.

4) Find all pairs of positive integers  $ m,n\ge 3$ for which there exists infinitely many positive integers $ a$ such that $ \frac {a^m \plus{} a \minus{} 1}{a^n \plus{} a^2 \minus{} 1}$ is itself an integer.

Are these IMO problems? :o   :oops:

Better if I edit all this

[hide=" 1"]
  If the GCD of the two is 1, the two must be the same number, and it is easy to see that when $ x\equal{}1$, $ 1$ is reached.  If the GCD of the two is greater than 1, then it must equal $ xy\plus{}1$, since otherwise we would not get an integer.  $ x^2\minus{}xy\plus{}y\minus{}1$ is the GCD by Euclidean algorithm.  Therefore, $ x^2\minus{}2xy\equal{}\minus{}y\plus{}2 \rightarrow (x\minus{}y)^2\equal{} y^2\minus{}y\plus{}2$.
$ y^2\minus{}y\plus{}2\equal{}n^2\rightarrow y^2\minus{}y\minus{}2\equal{}n^2\minus{}4 \rightarrow (y\minus{}2)(y\plus{}1)\equal{}(n\plus{}2)(n\minus{}2)$.
Examine modular 3.  $ y \equiv 2 \pmod 3$.  $ y\equal{}2$ yields a solution of 2.  If $ y\minus{}2\equal{}a, n\minus{}2\equal{}b$, then $ a(a\plus{}3)\equal{}b(b\plus{}4)$.  Notice that the modular 3 analysis forces the left side to be $ 3([a]\plus{}[a\plus{}1])$.  The factors of $ a$ and $ a\plus{}1$ cannot be switched since they are relatively prime.

Thus, only 1 and 2 can be reached. [/hide]

I hope I didn't make any false assumptions.

Problems that are IMO #3s (your 4th one) belong nowhere near intermediate topics. This topic should be moved.


it is wel known that :
$\prod (1+a_{i})\geq \left( 1+\sqrt[n]{\prod a_{i}}\right)^{n}$

just put $a_{1}=\frac{p}{2005}$...

[quote="Albanian Eagle"]it is wel known that :
$\prod (1+a_{i})\geq \left( 1+\sqrt[n]{\prod a_{i}}\right)^{n}$
[/quote]

I tried proving it.

By AM-HM,

$\frac{1+\frac{1}{a_{i}}}{2}\ge \frac{2}{1+a_{i}}$
$\Leftrightarrow 1+a_{i}\ge \frac{4}{1+\frac{1}{a_{i}}}$
$\Leftrightarrow 1+a_{i}\ge \frac{4a_{i}}{a_{i}+1}$
$\Leftrightarrow (1+a_{i})^{2}\ge 4a_{i}$
$\Leftrightarrow 1+a_{i}\ge 2\sqrt{a_{i}}$

$\prod (1+a_{i})\geq \left( 1+\sqrt[n]{\prod a_{i}}\right)^{n}$
$\Leftrightarrow 2^{n}\sqrt{\prod a_{i}}\geq ( 1+\sqrt[n]{\prod a_{i}})^{n}$
$\Leftrightarrow 2\sqrt[2n]{\prod a_{i}}\geq 1+\sqrt[n]{\prod a_{i}}$

Let $k=\sqrt[2n]{\prod a_{i}}$

$2k\ge 1+k^{2}$
$\Leftrightarrow 0\ge (k-1)^{2}$

Where did I go wrong?  :huh:

What makes you think that applying AM-HM right off the bat (which severely weakens the LHS) preserves the inequality?

Anyways, it is a trivial consequence of Holder's inequality.

[quote="blahblahblah"]Anyways, it is a trivial consequence of Holder's inequality.[/quote]

Oh, gotta learn that inequality.

well if you don't know Holder than just open the brackets of the LHS and AMGM the symmetric sums :wink:

@ Splash D
$\Leftrightarrow 1+a_{i}\ge 2\sqrt{a_{i}}$
that was your conclusion of AM-HM, which is just AM-GM on 1 and a_i...
umm also see AoPS vol. 2 chapter 14 for inequalities, it will really help you ;P
a problem very similar to this is in there as a challege problem... edit:oh no 666 posts

@Alb. Eagle:
[hide="don't peek splash d, sorry"]that's how the book i referred to proved this problem!![/hide]

and one last edit:
in the problem we are given:
$pqrs=tuvw=1.$
don't we only need to know that
$pqrstuvw=1.$? like we are given more than enough info?


You should get these correct, hopefully.

1) How many times does the minute hand and the hour hand form $180\deg$ angle in one full day?

2) 20 people takes off their hats to go in a party. When the party is finished, 19 people got their correct hats. What's the probability that the last person gets the wrong hat?

3) There are 2004 points around a circle. Four random points are chosen and label them A,B,C,D. Find the probability that the quadrilateral ABCD is convex.

[/hide][hide] For number one, I think it is 24, once for every hour. For number 2 it should be 0 because there is only person left so he will obviously choose the correct one.[/hide]

#2 would be [color=white]0 since the person had only their hat to choose from[/color].

#1 is[color=white]24[/color].

Right???

1.    24 - because only once in an hour it's possibel to be 180degrees.
             24 hours are in a day..

2.    0%- because everyone in the party got the correct hat and the last
              person should get the last hat which is supposed to be his...

3.    0%- even u choose a dot randomly they have to be connected(even the
              on that's right next the point)when they be connected, they have to form
              line segments...(if u get what I'm talking about)

on second thoughts I think it is 24, as when the hour hand goes past 06:00 it happens twice, once at about 05:55 and once at about 06:05, and so although it happens 24 times a day it does not happen regularly.

It would be interesting to investigate that. how do you work out the positions?

does the frequency of this occurrence vary like a sine wave against time? :\

sounds difficult :cool:

Here are some of the hints to get your heads on the right track:

1) Answer is [b]not[/b] 24. Study closely the positions of the hands (as Mikey suggested, even though he got the wrong answer  :D )

2) Too easy, no comments

3) The problem itself already contains a hint.

If anyone has the [b]urge[/b] to see the answers, I'll gladly reveal it.

You should get these correct, hopefully.

1) How many times does the minute hand and the hour hand form  angle in one full day?

2) 20 people takes off their hats to go in a party. When the party is finished, 19 people got their correct hats. What's the probability that the last person gets the wrong hat?

3) There are 2004 points around a circle. Four random points are chosen and label them A,B,C,D. Find the probability that the quadrilateral ABCD is convex.

[hide]

1) 23? I'unno.

2) 0%

3) Dunno...

[/hide]

I think I made a mistake..

I think it should be 26 because like Mikey said there are two 6:00's on a day..
so I think it's 26

other problemos... I think I got them right...

[hide]Answers:

1) 22. During every hour before 5 o'clock and after 7 o'clock, the hands form 180 angle every hour, that accounts for 20 times. But between 5 and 7, the hands are 180 angle exactly when it hits 6 o'clock and that happens twice in one day, for a total of 22 times.

2) 0. Easy, self-explanatory.

3) 1/3. Consider the following combinations: ABCD, ACBD, ACDB. Only 1 of them is convex, the rest are self-intersecting. The trick to this problem is the labelling of the points mentioned in the problem.[/hide]

1.) l[hide]ogically speaking it's 24, but truthfully, it depends on the clock. so the answer varies, or the answer can be 0.[/hide]

2.) [hide]the person gets the correct hat no matter what.[/hide]

3.) [hide]the points are convex no matter what[/hide]

Treething wrote:[hide]
Why they have to be convex:

To have a concave polygon, one angle must be greater than 180 degrees. However, when a quadrilateral is inscribed in a circle, opposite sides add up to 180 degrees. (Try proving this for a fun proof.) Obviously, it's impossible to have an angle greater than 180 degrees.

[/hide]

3X.lich said 'self-intersecting', not concave. The other combinations would result in bow ties, which aren't polygons at all.

1)
well we are looking for when the minute hand and hour hand form a straight line.  Since the clock goes through 2 cycles every day, and 12 180degree angles are formed per cycle, 24 times a day a 180 degree angle is formed

2) 2 is not possible since is all but 1 person has the right hat, then that person would have the right hat which is a contradiction so the chance of having the wrong hat os 0%.  It is possible if the hat was lost or destroyed :?

[quote="demji"]1)
well we are looking for when the minute hand and hour hand form a straight line.  Since the clock goes through 2 cycles every day, and 12 180degree angles are formed per cycle, 24 times a day a 180 degr1)ee angle is formed
[/quote]
Its been said quite a few times now that 24 is [b]not[/b] the correct answer :)

i didn't read any previous ones.  but i think i see why now.  The hands are never 180 in hour 5.  The min hand chases the hour hand and gets the 180 exactly on the 6 so that is one less line per cycle which then brings the total to 22.


Let $n \in N;n \geq 2$
Suppose $a;b \in R$ and satisfying $a^{n}=a+1;b^{2n}=b+3a$
Prove that $a>b$


Consider a $m*n$ board. On each box there's a non-negative integrer number assigned. An operation consists on choosing any two boxes with $1$ side in common, and add to this $2$ numbers the same integrer number (it can be negative), so that both results are non-negatives.
What conditions must be satisfied initially on the assignment of the boxes, in order to have, after some operations, the number $0$ on every box?.

If we color the cells of the board like those of a chessboard (alternatively black and white) then the sum of the numbers in the white cells must be equal to that of the numbers in the black cells.

The necessity is clear: in each move, you make equal contributions to both sums (white and black), and if in the end you reach $0$, i.e. both sums equal to $0$, then they must have been equal in the beginning as well.

If, conversely, the two sums are equal, we can clear the board one column at a time. When we're left with only one column, we continue clearing up cells one by one until we're left with a $2\times 1$ board. By the hypothesis (white and black sums are equal) the two numbers we're left with are equal and we can finish.

Why the black sum must be equal to the white sum?

Thanks