function
inequalities
algebra unsolved
algebra

Let $ f  :  [ 0 ; 1] \to\mathbb{R}$ be a strictly increasing function. We know that $ f(0) \equal{} 0$ and $f(1) \equal{} 1$. Moreover,
\[ \frac{1}{2} \le \frac{ f(x\plus{}y) \minus{} f(x)}{ f(x) \minus{} f(x\minus{}y)} \le 2, \] for all $x$ and $y$ such that $ 0 < y \le x < x\plus{}y \le 1$. Prove that \[ f \left( \frac{1}{3} \right) \le \frac{76}{135}.\]

[i]My lovely teacher , sir . Tran Nam Dung gave me a very good solution as follows :lol:

  Solution :
      From the hypothesis ,     notice that :  $ \frac {1}{3} f(x \plus{} y) \plus{} \frac {2}{3}f(x \minus{} y) \le f(x) \le \frac {1}{3} f(x \minus{} y) \plus{} \frac {2}{3}f(x \plus{} y)$

 for all $ 0 \le y \le x \le x \plus{} y \le 1 \ \ (*)$

   From that property , we can have some relative inequalities :

  $ a/ \ \ \frac {3}{2} f \left( \frac { 2 }{3 } \right) \minus{} \frac {1}{2} f \left( \frac {7 }{12 } \right) \le f \left( \frac { 3 }{ 4} \right) \le \frac {2}{3} \plus{} \frac {1}{3} f \left( \frac { 1}{ 2 } \right)$ ( in $ (*)$ put $ x \ \equal{} \ \frac {2}{3} ; y \ \equal{} \ \frac {1}{6}$ and $ x \equal{} \frac {3}{4} \ ; \ y \ \equal{} \ \frac {1}{4}$ )

   $ b/ \ \ f \left( \frac { 7 }{ 12 } \right) \le 3f \left( \frac {1 }{ 2 } \right) \ \minus{} \ 2 f \left( \frac { 5 }{12 } \right)$ ( in $ (*)$ put $ x \ \equal{} \ \frac {1}{2} ; y \ \equal{} \ \frac {1}{12}$ )
   $ c/ \ \ f \left( \frac { 5 }{12 } \right) \ge \frac {3}{2} f \left( \frac { 1 }{ 3 } \right) \ \minus{} \ \frac {1}{2} f \left( \frac { 1 }{ 4 } \right)$( in $ (*)$ put $ x \ \equal{} \ \frac {1}{3} ; y \ \equal{} \ \frac {1}{12}$ )

     $ d/ \ \ f \left( \frac { 1 }{ 4 } \right) \le \frac {2}{3} f \left( \frac {1 }{ 2 } \right)$

     $ e/ \ \ f \left( \frac { 1 }{ 3 } \right) \le \frac {2}{3} f \left( \frac {2 }{ 3 } \right) \rightarrow \frac {3}{2}f \left( \frac { 1 }{ 3 } \right) \le f \left( \frac {2 }{ 3 } \right)$

    $ f/ \ \ f \left( \frac {1 }{ 2 } \right) \le \frac {2}{3}$
   Now , It's time to eat a cake  :blush:
From $ a/$ , we have : $ \frac {3}{2} f \left( \frac { 2 }{3 } \right) \minus{} \frac {1}{2} f \left( \frac {7 }{12 } \right) \le \frac {2}{3} \plus{} \frac {1}{3} f \left( \frac { 1}{ 2 } \right)$

   Thus , $ \frac {3}{2} f \left( \frac { 2 }{3 } \right) \le \frac {1}{2} f \left( \frac {7 }{12 } \right) \plus{} \frac {2}{3} \plus{} \frac {1}{3} f \left( \frac { 1}{ 2 } \right) \le \frac {2}{3} \plus{} \frac {1}{3} f \left( \frac { 1}{ 2 } \right) \plus{} \frac {3}{2} f \left( \frac {1 }{ 2 } \right) \ \minus{} \ f \left( \frac { 5 }{12 } \right)$ , due to $ b/$ :)

   Now , using $ c/$ , we have :

   Hence $ \frac {3}{2} f \left( \frac { 2 }{3 } \right) \le \frac {2}{3} \plus{} \frac {1}{3} f \left( \frac { 1}{ 2 } \right) \plus{} \frac {3}{2} f \left( \frac {1 }{ 2 } \right) \minus{} \frac {3}{2} f \left( \frac { 1 }{ 3 } \right) \ \plus{} \ \frac {1}{2} f \left( \frac { 1 }{ 4 } \right)$

  So that :
$ \frac {3}{2} f \left( \frac { 2 }{3 } \right) \plus{} \frac {3}{2} f \left( \frac { 1 }{ 3 } \right) \le \frac {2}{3} \plus{} \frac {11}{6} f \left( \frac { 1}{ 2 } \right) \ \plus{} \ \frac {1}{2} f \left( \frac { 1 }{ 4 } \right)$

$ \le \frac {2}{3} \plus{} \frac {11}{6} f \left( \frac { 1}{ 2 } \right) \ \plus{} \ \frac {1}{3} f \left( \frac { 1 }{ 4 } \right) \equal{} \frac {2}{3} \plus{} \frac {13}{6} f \left( \frac { 1}{ 2 } \right) \le \frac {2}{3} \plus{} \frac {13}{6} \cdot \frac {2}{3} \ \equal{} \ \frac {19}{9}$

  $ \rightarrow \frac {19}{9} \ge \frac {3}{2} f \left( \frac { 2 }{3 } \right) \plus{} \frac {3}{2} f \left( \frac { 1 }{ 3 } \right) \ge \frac {9}{4} f \left( \frac { 2 }{3 } \right) \plus{} \frac {3}{2} f \left( \frac { 1 }{ 3 } \right)$ , due to $ e/$

   Thus , $ f\left( \frac { 1 }{ 3 } \right) \le \frac { 76}{135}$ , Done

   So nice problem :love: [/i]

I got only $ f(\frac 13)\le\frac 47$ ...  :blush:

Here is my proof for my weaker result :

Setting $ x \equal{} y \equal{} \frac 13$ in the inequation, we get $ \frac 12\le\frac {f(\frac 23) \minus{} f(\frac 13)}{f(\frac 13)}\le 2$

And since $ f(x)$ is strictly increasing and $ f(0) \equal{} 0$, we get $ f(\frac 13) > 0$ and the above inequality becomes $ \frac 32f(\frac 13)\le f(\frac 23)\le 3f(\frac 13)$

Setting now $ x \equal{} \frac 23$ and $ y \equal{} \frac 13$, we get $ \frac 12\le\frac {1 \minus{} f(\frac 23)}{f(\frac 23) \minus{} f(\frac 13)}$ and so : $ \frac 32f(\frac 23)\le 1 \plus{} \frac 12f(\frac 13)$

And since we got in the beginning $ \frac 32f(\frac 13)\le f(\frac 23)$, we can conclude :

$ \frac 94f(\frac 13)\le \frac 32f(\frac 23)\le 1 \plus{} \frac 12f(\frac 13)$ and so $ \frac 94f(\frac 13)\le 1 \plus{} \frac 12f(\frac 13)$ and so : $ \boxed{f(\frac 13)\le\frac 47}$

and $ \frac 47$ is near of $ \frac {76}{135}$ but this is not enough for showing your request. ;)


if a rectangular block that is 4 inches by 4 inches by 10 inches is placed inside a right circular cylinder of radius 3 inches and height 10 inches the volumne of the unoccupied portion of the cylinder is how many cubic inches?

have no idea where to start....step by step answer please.

the volume of the unoccupied portion will be the differences of the 2 volumes.volume of the rect. block:4x4x10=160cubic inches. vol. of cyl. block=3.1416x(3^2)x10=283Cub. inches.the answer is 123cub inches.

[hide="Solution"]The unoccupied volume is the volume of the cylinder minus the volume of the rectangular block (do you see why?), so:
\begin{align*} V_{unoccupied} & = V_{cylinder} - V_{block} \\
& = \left(\pi r^2 h\right) - (l \cdot w \cdot h) \\
& = \pi \cdot \left(3 \text{ inches}\right)^2 \cdot 10 \text{ inches}- 4 \text{ inches}\cdot 4 \text{ inches} \cdot 10 \text{ inches} \\
& = \boxed{90 \pi - 160 \text{ cubic inches}} \\
& = \boxed{10\left(9\pi - 16\right) \text{ cubic inches}} \\
& \approx \boxed{122.743 \text{ cubic inches}} \end{align*}
[/hide]


algebra polynomial integration function trigonometry calculus quadratics

Let $ f(x)=e^{x^2}$ and write $ f^{(n)}(x)=P_n(x) f(x)$ where $P_n$ are polynomials. Then find an equivalent for $P_n(x)$ for every fixed positive $x$.

could you post some hints please ? I tried a lot to solve it yesterday but ..  :(

This is an extremely hard problem! The main idea is to find a second order differential equation that has as solution the polynomials in the problem. Next, use the fact that $ af(x)\leq f\"(x)\leq bf(x)$ implies some estimations on $f(x)$ and $x$. This is just a small  hint, but it is the mzin idea. After that there is some ugly work. I will  post a solution at the end of the week if it doesn't help much. Another hint: the equivalent is absolutely monstruous! I have never seen such an ugly equivalent!!!

I tried some simple "back of envelope" computation and got $\frac {n!e^{2xR+R^2}}{2\sqrt{2\pi}R^{n+1}}$ where $R=R(x,n)$ is the positive root of the equation $2xR+2R^2=n$, i.e., $R=\sqrt{\frac n2+\frac {x^2}4}-\frac x2$. Using Stirling's formula for $n!$ and Taylor series for $R$,  this formula can be simplified to
\[
\frac 1{\sqrt 2}\left(\frac{2n}{e}\right)^{\frac n2}e^{-\frac{x^2}2}e^{x\sqrt{2n}}
\]
Does it agree with your answer, Harazi? (If your answer looks differently and you are not completely sure whether or not it is asymptotically the same as what I got, just post it and I'll try to figure it out myself. As I said, I just did a quick estimate and cannot guarantee it, so I won't be surprised if I'm wrong :P )

Then I'll post my computation. It is different from Harazi's approach and uses a very classical technique. The problem is to verify some technical details, which may be not very hard but somewhat boring. I'll just present it the way I did it omitting formal justification of some steps.

The main tool is the formula $f^{(n)}(x)=\frac1{2\pi}n! R^{-n}\int_{[-\pi,\pi]} f(x+Re^{i\theta})e^{-in\theta}\,d\theta$ valid for any function $f$ analytic in the entire complex plane. Applying it to $f(z)=e^{z^2}$ and cancelling $e^{x^2}$, we get
\[
P_n(x)=\frac1{2\pi}n! R^{-n}\int_{[-\pi,\pi]} \exp\left\{ 2xRe^{i\theta}+R^2e^{2i\theta}-in\theta\right\}\,d\theta       \\
=\frac1{2\pi}n! R^{-n}\int_{[-\pi,\pi]} \exp\left\{ 2xR\cos{\theta}+R^2\cos{2\theta}\right\}  \exp\left\{i( 2xR\sin{\theta}+R^2\sin{2\theta}-n\theta)\right\}\,d\theta
\]
The first factor in the integral has a well-expressed maximum at $\theta=0$. The problem is that, for generic $R$, the second factor is oscillating fast. So, the idea is to choose $R$ for which the oscillation is slow. Since $\sin\theta=\theta+O(\theta^3)$, we see that, choosing $R$ satisfying the equation $2xR+2R^2=n$, we make the oscillation slow (cubic in $\theta$ for small $\theta$ instead of a linear one that takes place for other $R$). Now, $\cos\theta\approx 1-\frac{\theta^2}2$, so the decay is quadratic in $\theta$ with approximately the same (big) coefficient in exponent as that at $\theta^3$ in the oscillating term. This makes it believable (though one should accurately check some details) that the integral is almost the same as
$
e^{2xR+R^2}\int_{-\infty}^{\infty}e^{-(xR+2R^2)\theta^2}\,d\theta = e^{2xR+R^2}\sqrt{\frac{\pi}{xR+2R^2}}\approx \sqrt{\frac\pi 2}\frac{e^{2xR+R^2}}{R}
$.
This results in the first form of the answer in my post #4. Switching to the second form is a routine exercise.


geometry
3D geometry
sphere
analytic geometry
symmetry
number theory
real analysis

Call by $\mathbb{Z}$-ball the set of points of the form $\ds S=\{(x,y,z)\mid x^{2}+y^{2}+z^{2}\le R^{2},\ x,y,z,\in\mathbb{Z}\},\ R\in\mathbb{R}.$ Prove that there exists no $\mathbb{Z}$-ball which contains exactly $2005$ distinct points.

If one can prove that $7.1 \leq R < 8$, then it's simple from there:

The only points that matter $\bmod \, 8$ are the ones of the form $(a, \pm a, 0) \, \slash \, (a,0,0)$. By some computations, we get that $\left| S_{R}\right| \equiv 7 \left( \bmod \, 8 \right), \, \forall R \in \left[ 7.1,8 \right)$.

Thus, if one can prove that (hopefully correct) bound (especially the $R<8$ part), then the problem is solved. If not, then this is one useless post.

[quote="jmerry"]I'd work mod 6: the only points that matter are of the form $(\pm a,\pm a,\pm a)$. For $48\le R^{2}<75$, $|S|\equiv 3\mod 6$ and for $75\le R^{2}<108$, $|S|\equiv 5\mod 6$; if we can show that $|S|<2005$ when $R^{2}=48$ and $|S|>2005$ when $R^{2}=108$, we will be done.

If $R^{2}=108$, the points $(a,b,c)$ are inside the sphere whenever $|a|,|b|,|c|\le6$. There are $13^{3}=2197$ points of this form, and $|S|\ge 2197$ when $|R^{2}|=108$.

If $R^{2}=48$ and $(a,b,c)\in S$, we must have $|a|+|b|+|c|\le 11$ or $|a|=|b|=|c|=4$. There are $1561$ points of the first type and eight of the second, so $|S|\le 1569$ when $R^{2}=48$

This is number theory, not calculus.[/quote]
What about $(\pm 1,\pm 2,\pm 3)$?

I don't see the problem- there are 48 of those, not just 8. Remember that you can permute the elements as well as changing sign.

Assume such a ball $E_{R}$ exists. Obviously $R \geq 1$, and we must have $R < 10$, for otherwise $R$ contains a cube with integer coordinates and volume $14^{3}=2744>2005$.

By symmetry, $E_{R}$ contains the point $(0,0,0)$, $m \geq 1$ points on each of the $6$ branches determined by the usual axes, and $n$ points (strictly) inside each of the $8$ regions determined by those $6$ branches. The total number of points will be $2005=1+6m+8n$, whence $1002=3m+4n$, $m=2 \ mod \ 4$ and $3$ divides $n$ (*). Each of the $8$ regions can be divided by a line (with origin $(0,0,0)$ and taking one of the $8$ directions determined by the regions) into $3$ subregions (**) in the obvious fashion (for instance $S$ will be divided by the line $\{\alpha,\alpha,\alpha;\ \alpha \geq 0\}$). By symmetry, each of these lines will then contain $3k$ points, not counting $(0,0,0)$ . $E_{R}$ will contain the point $(3k,3k,3k)$, whence $\sqrt{3}+27k^{2}> R \geq 27k^{2}$, and then each of the $6$ branches will contain $E(\sqrt{3\sqrt{3}+27k^{2}}) \geq m \geq 5k$ points , i.e, since $R<10$, $k=1$ and each branch has $m=5$ points. This is impossible because then the ball would be inscribed inside a cube with integer coordinates and volume $12^{3}=1728<2005$ (or because we must have $m=1 \ mod \ 4$).


E - normed space over R(real numbers) , H - subspace E, dim E/H = 1

Are?
a) H is closed
b) H is dense

My prove

[i]lemma :

f linear is continous <=> ker f is closed[/i]

Because dim E/H = 1, so exist g isomorphic from E/H to  R (real numbers) and this isomorphic is continous because dim E/H is finite. Ker g = H, so H is closed.
If H is dense then clH = E. Because H is closed, thus H = clH = E. E/H = E/E, thus dim E/H = dim E/E = 0, contradiction

Is correct my prove?

Yes, your proof is correct. More generally, a subspace of finite codimension (or of finite dimension) is closed.


Many people will be taking the AMC10 or 12 A tomorrow or in the next few days. Good luck everyone. Hopefully you all will all get good scores.  :)

I'll be in the cheering section for all the AoPSers.

good luck guys! i hope you get the scores you want!


Suppose P1,P2,P3,P4 are 4 points on the plane.
Prove that ( \sum PiPj)/(min PiPj)  \geq 5+ \sqrt 3.


algebra
polynomial
trigonometry
algebra proposed

No one can solve this olympiad problem ?

If $ a_2$ was $ p^2 \minus{} 2q$ it would be easier. :P

I will post my progress tomorrow wherever I end up, but I do not expect to have a full solution.

[quote="thaithuan_GC"]Let a polynomial $ P(x) \equal{} rx^3 \plus{} qx^2 \plus{} px \plus{} 1$ $ (r > 0)$ such that the equation $ P(x) \equal{} 0$ has only one real root. A sequence $ (a_n)$ is defined by $ a_0 \equal{} 1, a_1 \equal{} \minus{} p, a_2 \equal{} p^2 \minus{} q, a_{n \plus{} 3} \equal{} \minus{} pa_{n \plus{} 2} \minus{} qa_{n \plus{} 1} \minus{} ra_n$.
Prove that $ (a_n)$ contains an infinite number of nagetive real numbers.[/quote]

Here is a very strange demo. I'm sure it exists something simpler, but this is the only demo I found :

Let $ b$ any root (real or complex) of the polynomial $ Q(x)\equal{}x^3 \plus{} px^2 \plus{} qx \plus{} r \equal{} 0$ (notice that since $ P(x)\equal{}rx^3\plus{}qx^2\plus{}px\plus{}1$ has one unique real root and two complex roots, so has $ Q(x)$ whose roots are the inverses of $ P(x)$).

Obviously, $ b\neq 0$. Let then $ u_{n\plus{}1}\equal{}a_{n\plus{}3}\plus{}(p\plus{}b)a_{n\plus{}2}\minus{}\frac{r}{b}a_{n\plus{}1}$ (sequence of complex number)

$ u_{n\plus{}1}\equal{}\minus{}pa_{n\plus{}2}\minus{}qa_{n\plus{}1}\minus{}ra_n\plus{}(p\plus{}b)a_{n\plus{}2}\minus{}\frac{r}{b}a_{n\plus{}1}$

$ u_{n\plus{}1}\equal{}ba_{n\plus{}2}\minus{}(q\plus{}\frac{r}{b})a_{n\plus{}1}\minus{}ra_n$

$ u_{n\plus{}1}\equal{}b(a_{n\plus{}2}\minus{}\frac{r\plus{}bq}{b^2}a_{n\plus{}1}\minus{}\frac{r}{b}a_n)$

But $ b^3\plus{}pb^2\plus{}qb\plus{}r\equal{}0$ $ \implies$ $ p\plus{}b\equal{}\minus{}\frac{r\plus{}bq}{b^2}$ and so $ u_{n\plus{}1}\equal{}b(a_{n\plus{}2}\plus{}(p\plus{}b)a_{n\plus{}1}\minus{}\frac{r}{b}a_n)\equal{}b u_{n}$
$ u_0\equal{}a_{2}\plus{}(p\plus{}b)a_{1}\minus{}\frac{r}{b}a_{0}\equal{}p^2\minus{}q\minus{}p(p\plus{}b)\minus{}\frac{r}{b}$ $ \equal{}\minus{}\frac{pb^2\plus{}qb\plus{}r}{b}$ $ \equal{}\frac{b^3}{b}$ $ \equal{}b^2$

And so $ u_n\equal{}b^{n\plus{}2}$ and so $ \boxed{a_{n\plus{}2}\plus{}(p\plus{}b)a_{n\plus{}1}\minus{}\frac{r}{b}a_{n}\equal{}b^{n\plus{}2}}$

This equation is true for the two complex roots of $ Q(x)$ : $ z\equal{}\rho e^{i\theta}$ and $ \overline{z}\equal{}\rho e^{\minus{}i\theta}$. So :
$ a_{n\plus{}2}\plus{}(p\plus{}z)a_{n\plus{}1}\minus{}\frac{r}{z}a_{n}\equal{}z^{n\plus{}2}$

$ a_{n\plus{}2}\plus{}(p\plus{}\overline{z})a_{n\plus{}1}\minus{}\frac{r}{\overline{z}}a_{n}\equal{}\overline{z}^{n\plus{}2}$

Subtracting these two equalities :
$ (z\minus{}\overline{z})a_{n\plus{}1}\minus{}r(\frac{1}{z}\minus{}\frac{1}{\overline{z}})a_{n}\equal{}z^{n\plus{}2}\minus{}\overline{z}^{n\plus{}2}$

$ a_{n\plus{}1}\plus{}\frac{r}{\rho^2}a_{n}\equal{}\frac{z^{n\plus{}2}\minus{}\overline{z}^{n\plus{}2}}{z\minus{}\overline{z}}$

$ a_{n\plus{}1}\plus{}\frac{r}{\rho^2}a_{n}\equal{}\rho^{n\plus{}1}\frac{\sin((n\plus{}2)\theta)}{\sin(\theta)}$

The right part of this equality reaches infinitely many negative values when $ n$ increases. So do the left part.
So, since $ \frac{r}{\rho^2}>0$, at least one of $ a_n$ or $ a_{n\plus{}1}$ is negative for any such $ n$.

Hence the result.

How did you get that $ Q(x)$ has roots $ z\equal{}\rho e^{i\theta}$ and $ \overline{z}\equal{}\rho e^{\minus{}i\theta}$?

Also, how did you decide on your choice for $ u_n$?

[quote="Differ"]How did you get that $ Q(x)$ has roots $ z \equal{} \rho e^{i\theta}$ and $ \overline{z} \equal{} \rho e^{ \minus{} i\theta}$?[/quote]

If $ z\notin\mathbb{R}$ is a zero of $ P(x)\in\mathbb{R}[x]$, then $ P(z)\equal{}0$ $ \implies$ $ \overline{P(z)}\equal{}0$ $ \implies$ $ P(\overline{z})\equal{}0$ $ \implies$ $ \overline{z}$ is also a root

[quote="Differ"] Also, how did you decide on your choice for $ u_n$?[/quote]

I tried to find some $ \alpha$, $ \beta$ and $ \gamma$ such that $ a_{n\plus{}3}\plus{}\alpha a_{n\plus{}2}\plus{}\beta a_{n\plus{}1}$ $ \equal{}\gamma(a_{n\plus{}2}\plus{}\alpha a_{n\plus{}1}\plus{}\beta a_{n})$

[quote="pco"][quote="Differ"]How did you get that $ Q(x)$ has roots $ z \equal{} \rho e^{i\theta}$ and $ \overline{z} \equal{} \rho e^{ \minus{} i\theta}$?[/quote]

If $ z\notin\mathbb{R}$ is a zero of $ P(x)\in\mathbb{R}[x]$, then $ P(z) \equal{} 0$ $ \implies$ $ \overline{P(z)} \equal{} 0$ $ \implies$ $ P(\overline{z}) \equal{} 0$ $ \implies$ $ \overline{z}$ is also a root

[quote="Differ"] Also, how did you decide on your choice for $ u_n$?[/quote]

I tried to find some $ \alpha$, $ \beta$ and $ \gamma$ such that $ a_{n \plus{} 3} \plus{} \alpha a_{n \plus{} 2} \plus{} \beta a_{n \plus{} 1}$ $ \equal{} \gamma(a_{n \plus{} 2} \plus{} \alpha a_{n \plus{} 1} \plus{} \beta a_{n})$[/quote]

Thank you. :) I did not see the $ r > 0$ restriction in the original problem.

[quote="Differ"]$ x = 0$ does not lead to a root.

Then $ Q(x) = x^4P(\frac {1}{x}) = x^3 + px^2 + qx + r = (x - d)(x - e)(x - f) = 0$, where d is real and e and f are imaginary. This means $ - p = d + e + f$, and ${ q = de + ef + fe}$.
Our characteristic equation is also $ a^3 + pa^2 + qa + r = 0$. So $ a_n = Ad^n + Be^n + Cf^n$

$ A + B + C = 1$
$ Ad + Be + Cf = d + e + f$
$ Ad^2 + Be^2 + Cf^2 = p^2 - q = d^2 + e^2 + f^2 - de - ef - fd$

And now I do not even know how to show there are infinite real roots, not just negative roots...[/quote]

We can actually solve this equation!
Look at it - it's 3 variables A,B,C, and three equations lol

In fact, if you find the solution and substitute it into the original equation,
you get $ a_n = \frac{(e-f)d^{n+3} + (f-d)e^{n+3} + (d-e)f^{n-3}}{(d-e)(e-f)(d-f)}$
And by observing that d is negative and letting $ e=r(cos\theta + sin\theta ), f=r(cos\theta - sin\theta )$ the result follows.


Bác nào ủng hộ thì vào đây !  :)

Các bác có thể post theo mẫu sau :Bôi đen rồi copy vào bài post (Cho nó thống nhất)

[code][b][color=darkblue]TH.số thứ tự[/color] (Tác giả)[/b]
[hide]Đề bài[/hide]
[color=darkblue][b]Lời giải[/b][/color]
[b]Lời giải 1 (Tác giả)[/b]
[hide]Lời giải[/hide][/code]

Cac bac chung minh ho dong nhat thuc nay cai

[img]http://sweb.cz/Felix.Phan/combin.JPG[/img]


[quote="Tiger3"]Using nickels, dimes, quarters and/or half-dollars, how many ways can you make $ 75$ cents?

***I would just list all of them out, but that would take a long time.  Is there a faster way to do it?***

[hide="Answer"]22[/hide][/quote]
I can't think of faster ways of doing it without listing them out, but I could improve on the speed at which you count the ones listed out.
[hide]
Organize them by the amount of halfdollars and quarters used and just add the ways...
1 halfdollar: 4 ways
3 quarters: 1 way
2 quarters: 3 ways
1 quarter: 6 ways
0 quarter: 8 ways
$ 4\plus{}1\plus{}3\plus{}6\plus{}8\equal{}22$
[/hide]

[quote="xpmath"]There was a huge topic about a problem similar to this. For this, it's easy enough to list, but for problem where it asks for $ 750$ or $ 7500$?. Generating functions are the way to go.[/quote]

That does somewhat depent on if the problem sticks to our currency or if it uses some other thing, but I agree with the generating function idea.  However, further discussion should stay away from generating functions, I think it's gotten a lot of topics locked.

Was this from Mathcounts (If so, what round?) or from some other source?
Mathcounts wouldn't want you to list them out like that.  :D

Split them into groups based on how many quarters you're using.  I find that helps me.


...and, how can you prove that?maybe Chebishev? :?

[b]Problem:[/b] Let be given reals $x,y,z>0$ satisfy the condition $x+y+z=1$. Prove that:
\[ \frac{x}{xy+1}+\frac{y}{yz+1}+\frac{z}{zx+1} \ge \frac{36xyz}{1+13xyz}.  \]

[b]Solution:[/b]
First, we have:
\[ \frac{x}{xy+1} \ge x-\frac{x}{1+27z} \iff xuz \le \frac1{27}.  \]
Summing three ineqs, we obtain:
\[ \text{LHS} \ge x+y+z-\left(\frac{x}{1+27z}+\frac{y}{1+27x}+\frac{z}{1+27y}\right).  \]
We must prove:
\[ \frac{x}{1+27z}+\frac{y}{1+27x}+\frac{z}{1+27y} \le \frac{1-23xyz}{1+13xyz}.  \]
In other facts,
\[ \frac{x}{1+27z}+\frac{y}{1+27x}+\frac{z}{1+27y} \le \frac{1}{10} \le \frac{1-23xyz}{1+13xyz}.  \]
This is obvious, since $xyz \le \frac{1}{27}$ and the left part is equivalent to:
\[ 18+189(xy+yz+zx) \ge 3.27^2.xyz \\ \iff 18(x+y+z)^3+189(x+y+z)(xy+yz+zx) \ge 3.27^2.xyz,  \]
True by AM-GM.
Done.

[quote="Lovasz"]
In other facts,
\[ \frac{x}{1+27z}+\frac{y}{1+27x}+\frac{z}{1+27y} \le \frac{1}{10} \le \frac{1-23xyz}{1+13xyz}.  \]
This is obvious, since $xyz \le \frac{1}{27}$ and the left part is equivalent to:
\[ 18+189(xy+yz+zx) \ge 3.27^2.xyz \\ \iff 18(x+y+z)^3+189(x+y+z)(xy+yz+zx) \ge 3.27^2.xyz,  \]
True by AM-GM.
Done.[/quote]

I don't understand this last part because I think you used AM-GM inequality with reversed sign in last inequality and then your LHS cannot be less than $1/10$ if you put $x=1, y=0, z=0$

[quote="silouan"]It is equivalent to

$\frac{x^2y}{xy+1}+\frac{y^2z}{yz+1}+\frac{z^2x}{zx+1}\leq \frac{1-23xyz}{13xyz+1}$   (1)   [/quote]

After this form we have that
$\frac{x^2y}{xy+1}+\frac{y^2z}{yz+1}+\frac{z^2x}{zx+1}\leq \frac{1}{6}$ .
So (1)  holds for $xyz\leq\frac{5}{151}$
Now if $\frac{1}{27}\geq xyz\geq\frac{5}{151}$  then
the first inequality is obvious

Silouan, can you be more detailed, please. I don't understand your proof. I cannot solve it  :huh:

Thank you very much. :)

I worked exactly the same as here
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cyclic&t=21689

[quote="silouan"]
Now if $\frac{1}{27}\geq xyz\geq\frac{5}{151}$  then
the first inequality is obvious[/quote]

Can you please explain to me this step?

Thank you very much.

Sorry I thought it was an easy part and the other was the hard  but it is more  complicated .I have not a proof for
$\frac{1}{27}\geq xyz\geq \frac{1}{151}$

[quote="stancioiu sorin"]Let $\ x,y,z$ whith sum $\ 1$ be real positive numbers.Show that the following inequality holds
$\frac{x}{xy+1}+\frac{y}{yz+1}+\frac{z}{zx+1}\geq\frac{36xyz}{13xyz+1}$
   $\ Marius\ Ghergu,\ Romania$[/quote]
$\sum_{cyc}\frac{x}{xy+1}=\sum_{cyc}\frac{1}{y+\frac{1}{x}}\geq\frac{9}{1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{9xyz}{xyz+xy+xz+yz}.$
Hence, remain to prove that $\frac{9xyz}{xyz+xy+xz+yz}\geq\frac{36xyz}{13xyz+1}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0.$ :)


From this years MMPC:

What is the area of the region in the first quadrant bounded by the x-axis, the line y=2x, and the circles x^2+y^2=20 and x^2+y^2=30?

That isn't really a pleasant-looking number, is it?

no, but you can leave it unsimplified. these were the choices:

5 arctan 2
5/4 * pi
5/2 * pi
5
None of the Above

Mmm, not sure if trigonometry's appropriate for MATHCOUNTS...oh well :)

Hmm, the question is sorta weird. If you draw the picture, you see two different regions that are bounded by the given systems.  But assuming its the bigger one, you get

[hide]

Let the angle created by y=2x and the x axis be theta. tan theta = 2, so the angle is arctan 2. The region of the circle is arctan 2 (30pi - 20 pi) = arctan2 * 10pi, so thats my answer - none of hte above.

[/hide]


Ο εγγεγραμμένος κύκλος $ {\Omega_{ABC}}$ τριγώνου $ ABC$ εφάπτεται στις $ BC,CA,AB$ στα σημεία $ P,Q,R$ αντίστοιχα. Έστω
$ M = PQ\cap BA$ και $ N = PR\cap CA$. O εγγεγραμμένος κύκλος $ \Omega_{MNP}$ του τριγώνου $ MNP$ εφάπτεται στις $ MN, MP$ στα σημεία $ X,Y$ αντίστοιχα. Αν τα σημεία $ X,Y,B$ είναι συνευθειακά, τότε να αποδείξετε ότι:
(a) Οι κύκλοι $ {\Omega_{ABC}}$ και $ \Omega_{MNP}$ είναι ίσοι, και
(b) τέμνουν ο ένας τον άλλο σε τόξα $ 60^o$.

[i]Αφιερωμένη στο Στέφανο Αρετάκη (staretak)...[/i]

[color=red][EDIT] Παιδιά χίλια συγγννώμη για το typo.Τώρα το διόρθωσα.[/color]

Νίκο τα σημεία Χ,Υ,Β δεν γινεται να είναι συνευθειακά . (Μήπως εννοείς αλλα σημεία??) . Φαίνεται στο σχήμα . Πες μου αν κάνω λάθος

Πιστεύω είναι απλό typo κι ο Νίκος εννοεί ότι  $ Y$ είναι το σημείο που ο εγγεγραμμένος εφαπτεται στην $ PM$ κι όχι στην $ PN$ ή αλλιώς(συμμετρικό είναι εξάλλου) εννοεί ότι τα $ X,Y,C$ είναι συνευθειακά!

Πάντως Νίκο διευκρίνισε plz τι ακριβώς εννοείς!

[quote="Nick Rapanos"]Ο εγγεγραμμένος κύκλος $ {\Omega_{ABC}}$ τριγώνου $ ABC$ εφάπτεται στις $ BC,CA,AB$ στα σημεία $ P,Q,R$ αντίστοιχα. Έστω
$ M = PQ\cap BA$ και $ N = PR\cap CA$. O εγγεγραμμένος κύκλος $ \Omega_{MNP}$ του τριγώνου $ MNP$ εφάπτεται στις $ MN, NP$ στα σημεία $ X,Y$ αντίστοιχα. Αν τα σημεία $ X,Y,B$ είναι συνευθειακά, τότε να αποδείξετε ότι:
(a) Οι κύκλοι $ {\Omega_{ABC}}$ και $ \Omega_{MNP}$ είναι ίσοι, και
(b) τέμνουν ο ένας τον άλλο σε τόξα $ 60^o$.

[i]Αφιερωμένη στο Στέφανο Αρετάκη (staretak)...[/i][/quote]
Κατασκεύασα τα σχήμα ξεκινώντας αντίστροφα, δηλαδή από τους ίσους κύκλους με κοινή χορδή ίση με την ακτίνα τους.

Βλέπω κάποια ενδιαφέροντα αποτελέσματα με τον υπολογιστή, αλλά δεν έχω λύση και δεν βάζω hind για να μη σας μπερδέψω.

κώστας Βήττας.

Άρα μάλλον το Y είναι πάνω στην MP .

[quote="kritikos"]Άρα μάλλον το Y είναι πάνω στην MP .[/quote]

Ti ginetai re paidia? kaneis gi auth?

Για να μην παραπονιέται ο Nick βρήκα μια λύση ...
Έστω ότι η $ XY$ τέμνει την $ CN$ στο $ D$ . Έχουμε ότι $ (M,A,R,B) \equal{} (X,D,Y,B) \equal{} \minus{} 1$ (1) και επιπλέον αφού το $ B$ ανήκει στην πολική του $ N$ και το $ N$ ανήκει στην πολική του $ B$ και λόγω της (1) έχουμε ότι η $ CN$ είναι η πολική του $ B$ .

Eπισης ευθεία $ CN$ τέμνει τις εφαπτόμενες του κύκλου $ (K)$ από το σημείο $ M$ στα σημεία $ Q,\ N$ και επειδή το $ A$ είναι το αρμονικό συζυγές του $ C,$ ως προς τα $ Q,\ N$ συμπεραίνουμε ότι η ευθεία $ AM\equiv BM,$ είναι η πολική του $ C,$ ως προς τον κύκλο $ (K)$ και επειδή περνάει από το $ M$ προκύπτει ότι η ευθεία $ XZ,$ ως η πολική του $ M$ ως προς τον κύκλο $ (K),$ περνάει από $ C$ και έτσι έχουμε ότι τα σημεία $ X,\ Z,\ C,$ είναι συνευθειακά.

Αφού αποδείχτηκε ότι τα σημεία $ X,\ Z,\ C,$ είναι συνευθειακά, αποδεικνύεται με το θεώρημα Πάππου ότι
ότι και τα σημεία $ Y\equiv BX\cap PM,$ $ A\equiv BN\cap CM$ και $ Z\equiv CX\cap PN,$ είναι επίσης συνευθειακά

Αφού τώρα τα Y,A,Z είναι συνευθειακά από το θεώρημα Brianchon προκύπτει ότι το τετράπλευρο $ NMQR$ είναι περιγγράψιμο .
(Για να το δείτε αυτό απλά θεωρείστε ότι η εφαπτομένη από το $ R$ στον $ (K)$ τέμνει την $ MP$ στο $ Q'$ . Τότε οι ευθείες
NQ',MR,YZ συντρέχουν άρα $ Q\equiv Q'$ ) kai θεωρούμε  $ S$ το σημείο επαφής .

Το $ Q$ τώρα ανήκει στην πολική του $ B$ ως προς τον $ (K)$ επομένως $ Z,S,B$ συνευεθειακά και έστω $ F$ το σημείο τομής της
$ ZB$ με την $ YP$ . Τότε $ (X,G,Z,C) \equal{} (Y,R,F,P) \equal{} \minus{} 1$ (2)  και επιπλέον το $ Y$ ανήκει στην πολική του $ B$ ως προς τον $ (I)$
επομένως η πολική του $ Y$ ως προς τον $ (I)$ είναι η $ ZB$ άρα $ YI\bot ZF$ και αφού είναι $ QK\bot ZS$ θα είναι $ KE//YI$ (3)

Έστω ότι η $ KQ$ τέμνει τον $ (I)$ στο $ E$ . Τότε $ \angle{ERQ} \equal{} \angle{PQE} \equal{} \angle{ZQK} \equal{} \angle{KQR} \equal{} \angle{RPE}$
άρα το $ E$ είναι το μέσο του τόξου $ RP$ άρα $ IE\bot RP$ και αφού $ KY\bot PN$ είναι $ IE//KY$ (4) επομένως  λόγω (3),(4)
θα είναι $ KYIE$ παραλληλόγραμμο επομένως $ IE \equal{} KY$  , το ζητούμενο

Για το β) τώρα βλέπουμε ότι το $ K$ είναι το παράκεντρο του $ PQR$ και έχουμε από το θεώρημα Euler ότι
$ KI^2 \equal{} r(r \plus{} 2r_a)$ και αφού $ r \equal{} r_a$ θα είναι $ KI^2 \equal{} 3r^2$
Έστω ότι οι κύκλοι τέμνονται στα σημεία $ U,V$ τότε από το νόμο συνημιτόνων στο $ KVI$ έχουμε ότι
$ \cos\angle{ KVI} \equal{} \minus{} \frac {1}{2}$ επομένως $ \angle{KVI} \equal{} 120$ και προφανώς $ UVI \equal{} 60$  and so a big QED  :)

Ευχαριστώ τους Κώστα Βήττα , Τάσο Βαφείδη , Μιχάλη Σάββα με τους οποίους συζητήσαμε το problem .

Τι άσχετος που είσαι ρε Σιλουανέ... Σιγά τις γνώσεις γεωμετρίας που θέλει αυτή η λύση... :rotfl:

Συγχαρητήρια και για την ομάδα.

Καιρός να πάρεις αυτό το ρημάδι το παλιομετάλλιο ή μάλλον αυτά τα ρημάδια διότι θα είναι δύο. :)

Συγχαρητήρια και σε όλη την εξάδα. Νομίζω είναι dream team. Δε γίνεται καλύτερη. :D


Does there exist a sequence of natural numbers satisfying the relation
$\frac{1}{a_n}=\frac{1}{a_{n+1}}+\frac{1}{a_{n+2}}$?

I suppose your question is: does there exist [b]infinite[/b] sequence...

Then this question appeared on first round of Polish MO

Hint: use substitution $x_n=\frac{1}{a_n}$ then $(x_n)$ has to be convergent to zero if such a sequence would exist. Find explicit formula of $x_n$ and show that leads to a contradiction. (You'll get $\frac{x_2}{x_1}=\frac{\sqrt{5}-1}{2}$).

Thanks for help,but what i want is a solution using divisibility!
I have solved with the way you have written,and i have solved by
finding The first few terms!(using Fibonacci sequence)Any one?


hi ,
could any body solve this problem for me!!

Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24

If I misunderstand, please show me!

Let the average length, in inches, of the sticks in Box W is A
     the average length, in inches, of the sticks in Box V is B
      length of each red stick is X
According to the infomations, we have

X + 19 =A
X -  6  =B

=> A-B = 19+6 =25

I wonder why my solution isn't in these choices!  [/hide][/code]


A numbers is said to be a beautiful number if it is a composite number and but it is not a multiple of either $ 2;3$ or $ 5$(for example,the three smallest beautiful numbers are $ 49,77$ and $ 91$)
How many beautiful numbers which are less than $ 1000$

Er, and how do we know there are 168 primes less than 1000? :S
And why is the $ \plus{}3$ there?

[quote="Fermat's Little Turtle"]Er, and how do we know there are 168 primes less than 1000? :S
And why is the $ \plus{} 3$ there?[/quote]
These things are quite well known things. Like- we have $ 4$ primes $ \leq 10,$ $ 25$ primes $ \leq 100$ etc. and if you want reference just have a look at [url=http://primes.utm.edu/lists/small/]the list of the primes[/url]. (more precisely [url=http://primes.utm.edu/lists/small/1000.txt]here[/url])

And about the $ \plus{} 3$, we have already eliminated $ 2,3,5$ (i.e. three primes). So, I added them again.

I hope that, it is clear now. :)

[quote="Moonmathpi496"][hide="hint"]These type of problems can be solved easily with Inclusion and Exclusion principle.[/hide]
[hide="solution"]we can use that, $ 1000$ will be eliminated. So, for easier calculation we count the numbers $ \leq1000$ instead of $ < 1000$.
Using inclusion exclusion principle we get, the number of positive integers $ \leg 1000$ and multiple of $ 2,3,5$ is $ 734$.
Now, we know that, we have $ 168$ primes $ \leq 1000$.
So, the number of [i]beautiful numbers[/i] is $ 1000 - 734 - 168 + 3 = 101$

This problem is too easy for olympiad section, please post these types of problems in pre-olympiad section.[/hide][/quote]
hmm... you could not solve the problem in that way in a competition... So there are maybe a more elegant solution..

[quote="Mathias_DK"]hmm... you could not solve the problem in that way in a competition... So there are maybe a more elegant solution..[/quote] I think that many olympiad problems are solved using these information  :wink:

(My solution may not be that elegant at all, but I really don't think that without these type of work the problem can be solved. There is another much more ugly way,like taking the primes $ <32$ and their multiples which don't "collide" with those primes)

Anyway, any elegant solution is always appreciated. :lol:

[quote="Moonmathpi496"][quote="Mathias_DK"]hmm... you could not solve the problem in that way in a competition... So there are maybe a more elegant solution..[/quote] I think that many olympiad problems are solved using these information  :wink:

(My solution may not be that elegant at all, but I really don't think that without these type of work the problem can be solved. There is another much more ugly way,like taking the primes $ < 32$ and their multiples which don't "collide" with those primes)

Anyway, any elegant solution is always appreciated. :lol:[/quote]
I don't think your solution is wrong or anything! I just couldn't picture myself counting all the primes from 1 to 1000. (And i don't know these by heart) It would take awfully long time, so that's the reason i want to get a solution I could use in a contest.  :P

[quote="Moonmathpi496"][hide="hint"]These type of problems can be solved easily with Inclusion and Exclusion principle.[/hide]
[hide="solution"]we can use that, $ 1000$ will be eliminated. So, for easier calculation we count the numbers $ \leq1000$ instead of $ < 1000$.
Using inclusion exclusion principle we get, the number of positive integers $ \leg 1000$ and multiple of $ 2,3,5$ is $ 734$.
Now, we know that, we have $ 168$ primes $ \leq 1000$.
So, the number of [i]beautiful numbers[/i] is $ 1000 - 734 - 168 + 3 = 101$

This problem is too easy for olympiad section, please post these types of problems in pre-olympiad section.[/hide][/quote]
Because he want we help him fastly,a cheater
Here is a problem in newest MYM,so please lock it and warn negative


Hello

Does anyone here know how Schnapps - that irritating telepathy thing works...its driving me MAD!!!

Schnapps is the name of the game...

Thanks...either reply here or email me at astantine@gmail.com thanks

Yes, I know Schnapps! I was shown it once by my parents when I was younger, and it only surfaced in College recently! Technically, the rule is that you can't explain the rules of the game. Worse still, when I was shown it in College, the person followed the rule that you have to lie completely, even the guess is correct, and say no to everything!! The clicks after the "Schnapps is the name of the game" are hard to work out, so ignore them for the moment. Once you figure out the main part, that bit can be deduced. I don't particularly want to ruin the fun of the game, so the best I can say is to look carefully at the words/clicks, and at what the answer is. The examples on [url]http://www.mathlinks.ro/viewtopic.php?t=84627[/url] are almost right.

Incidentally, how did you find out about Schnapps? Because my friend learned it while in Croatia last summer at a Programming Olympiad. We've found it brilliant at first, because no one had a clue how the hell we were doing it, but now half our friends have worked out the secret!! Still, done at speed, it's quite impressive!!


find all solutions of the equation

$ x''\equal{}x$

We have $ x'' \minus{} x \equal{} 0$. The caracteristic equation for this second order d.e. is $ k^2 \minus{} 1 \equal{} 0$ whose solutions are +1 and -1. So, the general solution of the equation is $ x \equal{} C_1e^{t} \plus{} C_2e^{\minus{}t}$.

Let $ t$ be the independent variable. Then $ x'' \equal{} x$ is viewed as $ \frac{d^{2}x}{dt^{2}} \equal{} x(t)$.

Let $ p \equal{} \frac{dx}{dt}$, then $ \frac{dp}{dt} \equal{} p\frac{dp}{dx}$. Plugging this into the equation, we get

$ p\frac{dp}{dx} \equal{} x(t)$ or $ pdp \equal{} xdx$, which gives $ p^{2} \equal{} x^{2} \plus{} c^{2}$, where c is a constant

Now we have $ p \equal{} \frac{dx}{dt} \equal{} \pm\sqrt{x^{2} \plus{} c^{2}}$, which can be separated to give:

$ \frac{dx}{\sqrt{x^{2} \plus{} c^{2}}} \equal{} \pm dt$ By integrating both sides, we get:

$ \ln(x \plus{} \sqrt{x^{2} \plus{} c^{2}}) \equal{} \pm t \plus{}k$, where k is another constant

Or:

$ \frac{dx}{\sqrt{x^2 \plus{} c^2}} \equal{} \frac{dx}{\sqrt{c^2(1 \plus{} x^2/c^2)}} \equal{} \frac{dx/c}{\sqrt{1 \plus{} (x/c)^2}} \equal{} \pm dt$

$ \equal{}> argsinh(x/c) \equal{} \pm t \plus{} k \equal{}> \frac{x}{c} \equal{} \sinh(\pm t \plus{} k) \equal{} \frac{e^{\pm t \plus{} k} \minus{} e^{\minus{}(\pm t \plus{} k)}}{2}$

$ \equal{}> x \equal{} \frac{ce^k}{2}e^{\pm t} \minus{} \frac{ce^{\minus{}k}}{2}e^{\mp t} \equal{} C_1e^t \minus{} C_2e^{\minus{}t}$ or $ \equal{} C_1e^{\minus{}t}\minus{} C_2e^t$

or, in either case, $ x \equal{} K_1e^t \plus{} K_2e^{\minus{}t}$.


calculus
trigonometry
derivative
function
calculus computations

Hello, I'm trying to learn a bit of calculus over winter break (so I don't have much knowledge or experience on the subject)

My book gives a proof of $\frac{d}{dx}\text{Sin}^{-1}=\frac{1}{\sqrt{1-x^2}}$, and part of it I do not understand:
[quote]Let $y=\text{Sin}^{-1}x$.  Then $\sin y=x$, $-\frac{\pi}{2}<y<\frac{\pi}{2}$.  Differentiating both sides of $\sin y=x$ with respect to $x$ and using the chain rule on the left, we obtain $\cos y\frac{dy}{dx}=1$.[/quote]The last step is what I do not understand.  What is meant by "with respect to x" and how does he obtain $\cos y\frac{dy}{dx}=1$?

It is called "implicit differentiation" when you have an expression like $y^2 = x$ and differentiate both sides with respect to $x$ to get $2y'y = 1$. Two functions cannot be equal without having the same derivatives (this is obvious), and therefore implicit differentiation works ASSUMING that $y$ really is differentiable. Since the proof doesn't show that $\arcsin$ really is differentiable, it's not a fully correct proof.

If you just want an explanation of how it works: You differentiate both sides of the equation with respect to x. In my example, the derivative of x is of course 1, and for $y^2$ we use the chain rule to see that $\frac{d}{dx}y^2 = 2y'y$. Now do the same thing on $\sin y = x$.

oh okay, I understand now.  :D
What I didn't get at first was why the book used the chain rule, I did not notice that since $y=\text{Sin}^{-1} x$, $y$ represented a function and not solely a variable.

Thanks for your help!
as a side note, the book showed that $\arcsin$ is continuous earlier.


trigonometry
geometry
angle bisector
trig identities
Law of Cosines

Let $ \triangle ABC$ have $ \angle A\equal{}120^\circ$. Point $ P$ lies on the angle bisector of $ \angle A$ such that $ PA\equal{}AB\plus{}AC$. Prove that $ \triangle PBC$ is equilateral.

[hide]call $ PB$   $ x$, $ PC$    $ y$, $ BC$    $ z$, $ AB$    $ a$, $ AC$   $ b$, and $ PA$   $ (a\plus{}b)$
now apply the law of cosines three times
$ x^2 \equal{} a^2 \plus{} (a\plus{}b)^2 \minus{} 2 a (a\plus{}b) cos 60$           we all know that cos 60 is 1/2 so
$ x^2 \equal{} a^2 \plus{} a^2 \plus{} 2ab \plus{} b^2 \minus{}a^2 \minus{} ab$
$ x^2 \equal{} a^2 \plus{} ab \plus{} b^2$
$ y^2 \equal{} b^2 \plus{} (a\plus{}b)^2 \minus{} 2 b (a\plus{}b) cos 60$
$ y^2 \equal{} b^2 \plus{} b^2 \plus{} 2ab \plus{} a^2 \minus{}b^2 \minus{} ab$
$ y^2 \equal{} a^2 \plus{} ab \plus{} b^2$
$ z^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab cos 120$
because $ cos (90\plus{}p) \equal{} \minus{}sin p$
$ z^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab( \minus{}(sin 30))$             since we also know that sin 30 is 1/2
$ z^2 \equal{} a^2 \plus{} ab \plus{} b^2$
now we see that $ z^2, y^2$ and $ x^2$ are equal
so $ z \equal{} y \equal{} x$  which means $ \triangle PBC$ is equilateral[/hide]

I was trying to do that... Except I fail at trig so I couldn't do it properly  :P

There are other more geometric solutions. I'd encourage you to try those.

In particular two ideas that I had:

i) Indirect proof: Suppose $ P'$ is the point where the angle bisector hits the circumcircle. Show that $ P\equal{}P'$

ii) Construct a point $ B'$ on $ AP$ such that $ AB\equal{}AB'$. Then translate $ AC$.

[quote="PowerOfPi"]Isn't AwsomeMath for middle school? I don't think Trig is taught in middle school, so there should be a non-trig solution, like Altheman's.[/quote]

Awesome math is for grades 6-11, so it's for middle & high schoolers. and a lot of middle schoolers have a basic understanding of trig.

There are a lot of high schoolers...

and in middle school, you can learn trig...


Let ABCDE  be a regular pentagon with center O. Let P be a point on the extension of CD such that angleEAO is 3 times angleEAP. Let Q be the foot of the perpendicular to  PD  from E,  prove that PQ=QD.

i solved it in 20 minutes. still i took too long¡¡¡ :D

We look at the triangle $ADP$: it's easy to cpmpute its angles from the given info. We have $\angle DAP=18^{\circ},\angle ADP=108^{\circ}$, and $\angle APD=54^{\circ}$. At the same time, we have $\angle AED=2\angle APD=108^{\circ}$, and $EA=ED$, $E$ also being on the same side of $AD$ as $P$. All this shows that $E$ must be the circumcenter of $ADP$, so $ED=EP$, and from here the conclusion follows.

cool solution¡¡ its new to me. here is mine:

it is clear that there is only one point on CD  such that 3angleEAP=angleEAO. Let´s find that point, let P be the point on CD such that PED  is isoceles. with this it´s easy to calculate all the angles and obtain that angleEAP=18 and  angleEAO=54. SO WE ARE DONE¡¡¡ :D


I'm in AP math, and our teacher has given us a very challenging problem. I haven't worked it out very far, so some pointers would be nice. He's told us what the answer is, but it's up to us to show how to solve it.

if [i]a[/i], [i]b [/i]and [i]c[/i] belong to real numbers and $a^2+b^2+c^2=1$, what is the minimum value of ab+bc+ca?

I need to know the steps to solve this.
-Sarah

Someone else has already posted this  :rotfl:

[hide="Easy solution"]
We are given $a^2 + b^2 + c^2 = 1$.  Now, $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ (verify this for yourself), which equals $1 + 2(ab + bc + ca)$.

 Now, a property of squares is that they are all positive (:)), so we know

$1 + 2(ab + bc + ca) \ge 0$

Notice that we can rearrange this to form

$ab + bc + ca \ge - \frac{1}{2}$

This tells us that the minimum occurs at $ab + bc + ca = \boxed{- \frac{1}{2} }$, since we know it can never be less.
[/hide]

Even if this problem seems like MATHCOUNTS level, I ask you keep stuff where the source is.


Find the integer closest to $ \sqrt[3]{700}$.

We note that $ \sqrt[3]{729}\equal{}9$, so $ \sqrt[3]{700}$ is closest to $ \boxed{9}$.


Could someone provide me with a good definition of addition and the other basic operations? I was thinking about the things I assume when I do math problems and I realized I don't really understand the basic operations too well...

For example, what does 32+14 mean besides the fact that I know an algorithm to simplify it into one number?

Thanks.

32+14 means actually the 14th successor of the natural number 32. You should google for Peano axioms to find your answers ;)

Yes, follow up on Valentin's suggestion to explore the Peano axioms (sometimes called the Peano postulates) for more information. You can Google Peano and foundations and mathematics to get an interesting list of Web sites.

Or read Russell and Whitehead's Principia. :-)


calculus
derivative
function
limit
calculus computations

Let $ f: \mathbb{R}\rightarrow\mathbb{R}$ be a function with a continuous third derivative everywhere.  If $ f$ satisfies
\[ \lim_{x\rightarrow 0}\left( 1 \plus{} x \plus{} \frac {f(x)}{x}\right)^{\frac {1}{x}} \equal{} e^3,
\]
find $ f(0),f'(0),f''(0)$, and $ \lim_{x\rightarrow 0}\left(1 \plus{} \frac {f(x)}{x}\right)^{\frac{1}{x}}.$

$ x\plus{}\dfrac{f(x)}{x}\equal{}3x$

$ \Rightarrow f(x)\equal{}2x^2$

$ f(0)\equal{}0, f'(0)\equal{}0, f''(0)\equal{}4, \lim_{x\to 0}\left(1\plus{}\dfrac{f(x)}{x}\right)^{\frac{1}{x}}\equal{}e^2$  :|

How do we know that the solution is unique? For $ \exists c\in (0,x)$ such that $ f(x)\equal{}f(0)\plus{}xf'(0)\plus{}\frac{x^2f''(0)}{2}\plus{}\frac{x^3f'''(c)}{6}$

and there could be several functions satisfying
$ \lim_{x\rightarrow 0}\left( 1\plus{}x\plus{}\frac{f(0)}{x}\plus{}f'(0)\plus{}\frac{xf''(0)}{2}\plus{}\frac{x^2f'''(c)}{6}\right) ^{\frac{1}{x}}\equal{}e^3$


What is the greatest number of whole 4-inch by 6-inch notecards that can be cut from a 25-inch by 27-inch sheet of poster boart if every cut must be made parallel to a side of the poster board? You may not tape or glue.

The answer key says the answer is...

[hide]26[/hide]

[hide]
Well since an even + an even can never equal an odd, change the board into

24 inches by 26 inches After trying it out for a while, you find that they can fit in without any empty space. So you get

(24*26)/(4*6)

which is 26[/hide]

Thanks

I should stop multiplying every time there are two numbers  :rotfl:


Can somebody solve this ?

Some of the vertices of convex n-gon are conneted with line segments, no two of which meet at inner point. Prove that for random n points (no three collinear) exists bijection between them and the vertices of the n-gon, so that no two line segments, corresponding to line segments between the vertices of the n-gon meet at inner point


How many nine-digit permutations of 123456789 are divisible by 11?

[b]Hint[/b]
There is a simple test to determine if an integer is divisible by 11: take every digit of the number located in odd position and add them up, then take the remaining digits and add them up. If the difference between the two sums is a multiple of 11, including 0, then the number is divisible by 11. For instance, if the number is 65,637 then (6 + 6 + 7) - (5 + 3) = 19 - 8 = 11, so 65,637 is divisible by 11.

Also note that if you labled the digits $ a,b,c,d,e,f,g,h,i$ you will have

$ a+b+c+d+e+f+g+h+i=45$

and also $ a-b+c-d+e-f+g-h+i=0 \mod 11$ (from what Kouichi Nakagawa told us)

Then if you subtract the two equations we will have $ 2(b+d+f+h)=45 \mod 11$, which is congruent to $ 2(b+d+f+h)=12 \mod 11$, then $ b+d+f+h=6 \mod 11$

so $ b+d+f+h=17$- equation$ n_{1}$
or
$ b+d+f+h=28$- equation$ n_{2}$
(we do not need to worry about equaling any other $ 6 \mod 11$ because they wont fit our case since the bounds are at $ 1+2+3+4=10$ and $ 6+7+8+9=30$)

those two equations and the number of solutions to them such that $ b,d,f,h$ are all distinct and they are all less than or equal to $ 9$ and greater or equal to $ 1$ will give you the number of solutions (although that might take some time).  Then you multiply that answer by $ 5!$ for the rest spots..

(note once you find $ (b,d,f,h)$ you want to find ordered pairs so something like $ (1,2,6,8)$ is different than $ (2,1,6,8)$)

So its $ (5!) \cdot$(number of solutions of $ (b,d,f,h)$ to $ n_{1}$ $ +$ number of solutions of $ (b,d,f,h)$ to $ n_{2}$)

[quote="spell it out 11"]So whats the answer using that equation?[/quote]

just find the solutions then times 5! (you can do this by casework)


prove the inequality:

x 6+y6+z6+3x2y2z2 \leq 2(x^3y^3+y^3z^3+z^3x^3).

This is the "easy" part. I want to generalize this inequality.So... :(

Is the problem right??:?
Let z=0 then we must prove
x^6+y^6 \leq 2x^3y^3.
It must be
x^6+y^6+z^6+3x 2 y 2 z 2  \geq   2(x^3y^3+y^3z^3+z^3x^3) which can be proved by Schur .And here is the general inequality (Sleyfer inequality):
(n-2) \sum {1-->n}a^2n+na1 2*...*an 2   \geq  2\sum{i>j=1 to n}(ai*aj)^n
 :)

My mistake. your form is correct. :blush:


calculus
derivative
function
LaTeX
limit
calculus computations

The position of a particle moving in a straight line during a 5s trip is:

s(t)=t^2-t + 10cm

a) What is the avergae velocity of the entire trip?

b) Is there  a time at which the instantaneous velocity is equal to this average velocity. If so, find it?

So, I believe for (a) you just do s(5)-s(0)/(5-0) correct?

But for (b), I really don't know.

Thanks

The position of a particle moving in a straight line during a $ 5 \ \text{s}$ trip is:

$ s(t) = (t^2 - t + 10) \ \text{cm}$

a) What is the avergae velocity of the entire trip?

Yes, the average velocity is $ v_{avg} = \frac {\Delta s}{\Delta t}$.

b) Is there a time at which the instantaneous velocity is equal to this average velocity. If so, find it?

The instantaneous velocity is the time derivative of the position function, i.e., $ v(t) = s'(t)$.

Find the average velocity, then set the instantaneous velocity $ v(t)$ equal to that and find $ t$, i.e., solve for $ t$:

$ v(t) = v_{avg}$

BTW, I used $ \text{\LaTeX}$ to make it look better; you just need to type the appropriate code between dollar signs to use it.

There is a main result in analysis called Mean Value Theorem. If $ f$ is continuous on $ [a,b]$ and diferentiable at $ (a,b)$ then there exists some $ c \in (a,b)$ such that $ f'(c)\equal{}\frac{f(b)\minus{}f(a)}{b\minus{}a}$.

[url]http://en.wikipedia.org/wiki/Mean_value_theorem[/url]

[quote="Zakary"]The position of a particle moving in a straight line during a $ 5 \ \text{s}$ trip is:....

en dollar signs to use it.[/quote]

Thanks a lot!

I think I did it right.

I will try to make my future posts with Latex

S(t)=t^2-t+10   and S'(t)= 2t-1

and the Average velocity came out to be 4

so I just set S'(t)=4 and solve for t?

Thanks

Yes, that is what you should do.

A reference for future problems that involve a variation of the main value thereom is that the value of $ t$ that you find should be within the interval over which the mean value thereom is being applied (if the function is differentiable over that interval, of course).

$ \frac{\partial}{\partial t}\text{[}s\equal{}v_0t\plus{}\frac{1}{2}at^2\text{]}\to v_f\equal{}v_0\plus{}at$
Because there is so [i]little[/i] time in between $ v_0$ and $ v_f$ here, the idea is that $ v_0\equal{}v_f$.

This description is perceivable as instantaneous velocity, though the standard definition is $ v\equal{}\lim_{t\to 0} \frac{s}{t}$ (or d if your prefer, it really makes no difference). What I have only just realized is implied our standard kinematic equation is that the derivative of our initial function $ s(x)$ will always yield the same velocity no matter our initial height within $ s(x)$.

Aside from that you should probably know there are two standard definitions for $ s(x)$, the first and more common of which you have already dealt with. Here's the second:
$ s(x)\equal{}\frac{v_0\plus{}v_f}{2}t$
Now, now... that reminds me of average velocity! In fact, differentiating this we have:
$ s\prime(x)\equal{}v(x)\equal{}\frac{v_0\plus{}v_f}{2}$
A quick substitution of $ v_f\equal{}v_0$ then we'll just call it $ v$:
$ v(x)\equal{}v\equal{}\frac{s}{t}$
Now we can apply limits:
$ \boxed{v\equal{}\lim_{t\to 0} \frac{s}{t}}$

Even though I know this isn't exactly what one might call relevant, its what I thought when I saw this post and I hoped you would enjoy.  :)


modular arithmetic
algebra
binomial theorem
number theory

Find the remainder when $ 32^{1024}$ is divided by 7.

[hide="My solution"]Is there anything wrong here?
$ 32^{1024}\equal{}1024^{512}$

$ 1024\equiv 2\mod 7$

$ 512\equiv 1\mod 7$

So $ 1024^{512}\equiv 2\mod 7$

or $ 32^{1024}\equiv 2\mod 7$[/hide]

[quote="21 7 15"]my remainder is 4
32=4mod7;4^1024=4^1mod7[/quote]

Yes the answer is 4 but is there anything wrong in my solution?

Yes, there is. You must watch 512 mod 3 because $ 2^n$ mod 7 goes like this:
2,4,1,2,4,1,...(groups of 3, not 7). 1024 mod 7 is ok.

I didn't know you could do that in modular arithmetic.

[quote="21 7 15"]my remainder is 4
32=4mod7;4^1024=4^1mod7[/quote]

Why (and how) can you reduce the exponent from $ 4^{1024} \equiv 4^1 \pmod{7}$?

remainders of powers of mod 4 are 4,2,1,4,2,1........and so on
so1024 gives 1 remainder on divided by 3 . so its remainder is equal to remainder of 4^1(mod 7)

[b]Solution:[/b] $ 32^{1024} \equal{} 2^{5120} \equal{} (2^{3})^{1706} \cdot 2^2 \equiv 1^{1706} \cdot 2^2 \equiv 4$

[b]@Euclid:[/b] The reason why you can "replace" numbers in the base of exponents is a result of the binomial theorem:

In modulo $ k$, for integer $ n$, $ (a \plus{} nk)^m \equal{} \sum_{j \equal{} 0}^m \binom{m}{j} (nk)^j a^{m \minus{} j}$, and all terms except the $ j \equal{} 0$ term have a factor of $ k$, which makes them equivalent to $ 0$ in modulo $ k$.

A similar pattern does [u]not[/u] hold for numbers in the exponent. See Fermat's Little Theorem, Chinese Remainder Theorem, Euler's Totient Theorem, Carmichael's Theorem...

Come on Wickedestjr, stop spamming all topics!

What's spamming? Seriously, I've heard of it alot, but I have no clue what it means.

Wikipedia is good here.

http://en.wikipedia.org/wiki/Spam

[quote="Wickedestjr"]Well, before I was just saying that I liked this problem. What's useless about that. (or were you talking about canned meat when you said spam   :rotfl:)[/quote]

You can get banned for spamming up topics.  [b]Stop.[/b]


1=5
2=25
3=125
4=625
5= ______

Fill in the blank.
I know, it's lame.
If you can't do this one, shame on you.
Yeah, and there's obviously some mathematical flaws, but yeah....

does it have to do anything with the digits?

[quote="Nerd_of_the_Ages"]1=5

...

Yeah, and there's obviously some mathematical flaws, but yeah....[/quote]

So would $ 5\equal{}1$?

Yes. You just solved it. Told you it was lame.


1)Given $a_0=1,a_1=45,a_{n+2}=45a_{n+1}-7a_n\forall n\in N$
a. Find all positive divisor of $a_{n+7}^2-a_na_{n+2}$
b. Prove that: $1997 a_n^2+4.7^{n+1}$ is a perfect square
2)Let ${U_n}$ be a sequence satisfying $U_{n+1}=2U_n-n^2 \forall n\in N$
Find all $U_0$ such that $U_n>0 \forall n\in N$


Ring Theory
algebra
function
domain
superior algebra
superior algebra unsolved

Prove that a quotient ring $O/a$ of a Dedekind domain by an ideal $a\not = 0$ is a PID.

Write $a$ as product of prime ideals. By the Chinese Remainder Theorem, you just need to consider the case $n=p^{k}$, $p$ prime ideal. But then it's easy since all ideals of the quotient lift to ideals in the ring, and these liftings will be powers of $p$ (they divide $p^{k}$).


function
LaTeX
number theory proposed
number theory

Let $p$ be a prime and $N'=\{0,1,2,...\}$.suppose the function$f: N'\rightarrow N'$ satisfies;

$(i)\ \ f(n)=f([\frac{n}{p}])+f(p\{\frac{n}{p}\})$ for all $n \in N';$

$(ii)\ \ \text{there exists}\  n\in N'$ with $p\not | f(n)$.

If $c_0,c_1, ...  c_k$ are arbitrary integers, prove that there exist $a, d \in N'$ such that:

$f(a +di)\equiv  c_i\  (\mod p\ )$ for $i = 0,1, ... , k$.

in the special case where ${\{f(n-1)}\}_{n=1}^{\infty}$  was the Thue-Morse sequence mod 2

Does $\{\frac{n}{p}\}$ mean the fractional part of $\frac{n}{p}$ ?

Then I don't understand the problem :(
$\frac{n}{p}$ is integer because $f$ is from N to N. Thus we must have $p | n$ but this means $\{\frac{n}{p}\}=0$, and this looks a bit odd to me ... So we have $f(n)=f(\frac{n}{p})+f(0)$ and this means $f$ is constant, so $f$ must be 1. But finally this is absurd :( Were did I go wrong ? What I'm not understanding ?

I edited it .
I think I have to learn $\LaTeX$ again.


In your opinion, what's the best programming language for IOI: c or c++? Can you tell me some advantages each one has over the other?

I think that C++ is definitely better if you can use STL. Otherwise there's no real difference between the two, C++ is pretty much an extension to C and if we don't count STL then what it has that C does not is only a couple of convinient constructions, nothing more. But I don't think there are any drawback so why don't use C++?

[quote="quantum leap"]STL? :huh:[/quote]
Standard template library.

[quote="quantum leap"]STL? :huh:[/quote]

If you're pretty much confused as to what it is, check out this site........it's got some pretty interesting stuff on STL -- :)

http://www.infosys.tuwien.ac.at/Research/Component/tutorial/prwmain.htm

[quote="mahbub"]ya as BUS said there is no difference between C and C++. if u code at C that will also run in C++. But C++ have some extra fitures that does not work in C. So don't worry which one u r learning.... see what u r learning![/quote]

Yes, C should be a sub-set of C++, but the main point is the usage of the language.

By common programming practice in industry, C would be used in system programming like building Linux and Windows.

However, since many people superstitious belief that the same code compiled by C++ would be slower than C, none would use C++ to write system, even without using OO-style.

In my view, programming is the way to express the implementation of problem solving, not the way of problem solving.

I'd use C++ but try not to rely too much on pre-packaged data structures unless you're in a time crunch, because you have a lot more flexibility if you have the ability to write your own heaps, segtrees, queues, etc.

For competition programming C is usually enough (although I compile in C++), I've never needed the stl in a competition before.

[quote="salah"]ok then, now that its concluded that C++ is better, which way is the best way to learn it?
and does anyone here use DevCpp?[/quote]

I use Dev-Cpp.
To learn, I just got a book from Barnes & Noble.

Do the USACO, if you don't already.

I found these;
http://mindview.net/Books/TICPP/ThinkingInCPP2e.html
I dont think the basic stuff are that hard to learn, and besize I signed up for a course so that should be enough..

Dev-Cpp is great, I had some trouble uninstalling it after that I had messed with a bit, but now it works :)


Hi! Please, help me to solve the equation in real numbers:

$ x(x\minus{}2)(x\minus{}5)(x\minus{}6)(x\minus{}9)\equal{}\minus{}27$

It's too hard for me.
First of all, it's have 5 roots, and all of them are not rational.
May be we must group??:
$ x(x^2\minus{}11x\plus{}18)(x^2\minus{}11x\plus{}30)\equal{}\minus{}27$

This is a quintic equation. It is not solvable by radicals. However it is easy to sketch the graph of the left-hand-side and that of the right-hand-side to get first approximations to the $ 5$ roots.

Thank you very much, but I knew this. http://www67.wolframalpha.com/input/?i=x(x-2)(x-5)(x-6)(x-9)%2B27
May be we can solve this equation with some substitution?


linear algebra
matrix
linear algebra unsolved

Hi, I'm new here. I have a question about a problem I've been working on.

Let A be an n by n real, symmetric positive definite matrix where all it's elements are positive. Matrix A is the product of a binomial matrix B such that A = BB^T where T denotes the transpose operator. In the binomial matrix, note that x+y=1. The structure of the binomial matrix is given by the image attachment.

Let C be an n by n real, diagonal positive definite matrix. What bounds are there on the values of the diagonal elements of C so that the matrix A - C is positive semidefinite?

I thought of using eigenvalues, but it would be difficult to find the eigenvalues of the matrix A. Thanks for your help.

For those who don't want to download the attachment: the structure of a binomial matrix is

$ B(x,y)=\begin{bmatrix}1&0&0&\cdots&0\\ x&y&0&\cdots&0\\ x^{2}&2xy&y^{2}&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ x^{n}&\binom{n}{1}x^{n-1}y&\binom{n}{2}x^{n-2}y^{2}&\cdots&y^{n}\end{bmatrix}$


geometry
number theory unsolved
number theory

Having fun:
1. Grab a calculator. (you won't be able to do this one in your head)
2. Key in the first three digits of your phone number (NOT the area code)
3. Multiply by 80
4. Add 1
5. Multiply by 250
6. Add the last 4 digits of your phone number
7. Add the last 4 digits of your phone number again.
8. Subtract 250
9. Divide number by 2

This was sent to me by a friend on msn. Its kinda fun =)

Let $ x$ be the first three digits and $ y$ be the last four digits of my phone number. After performing all the steps we end up with
\[ \frac{(80x\plus{}1)\times 250\plus{}2y\minus{}250}{2}\equal{}10000x\plus{}y,\]
which is my phone number again! And guess what, I didn't use a calculator! ;)

Ah, no calculator indeed. It was just copy and pasted :P


algebra
function
domain
number theory
greatest common divisor
least common multiple
algorithm

Let A the group of positive integers representing of the form a^2+2(b^2). "a" are integer, and "b" anyone integer not equal zero.
For example, the number 6 is in the group A (6= 2^2+2(1^2))
Prove that if p^2 is in A, p is in A too (p=prime).

Sorry but i dont know Evclidov!! Do you can explain me please??

Maybe he means that $\mathbb Z[\sqrt 2]$ is an [url=http://en.wikipedia.org/wiki/Euclidean_domain]Euclidean domain[/url].

No, for this case used, that $Z[\sqrt{-2}$ Euclidean domain, were $norm(x+y\sqrt{-2})=x^{2}+2y^{2}$.

I explain more. We have $gcd$ in Euclidean domains. By problem hypothetis \[p^{2}=(x+y\sqrt{-2})(x-y\sqrt{-2})\] So consider $r+s\sqrt{-2}=gcd(p,\sqrt{-2})$. (If $r+s\sqrt{-2}=1$ then $gcd(x^{2}+2y^{2},p)=1$ and this is contradiction.) Now $\mbox{norm}(r+s\sqrt{-2})=r^{2}+2s^{2}|p^{2}$ and $<1r^{2}+2s^{2}<p^{2}$. So $r^{2}+2s^{2}=p$ :)

Excuse me, but i dont understand your solution!!

Can you explain me a dominian euclidian!!
Or find another solution!!!

Thanks :maybe:

Euclidean domain is a commutative ring that there exist a function $f: R\longrightarrow\mathbb N$ that for each $a,b\in R$ there exist $q,r\in R$ that $a=bq+r$ and $r=0$ or $f(r)<f(b)$. Euclidean rings have good properties like $\mathbb Z$. We have $gcd$ and $lcm$ with properties similiar to $\mathbb Z$.

Oh, is interesting, thanks!!!

Have you other problems with euclidian domains?'??

In general, in a $\mathbb{Z}[\sqrt{-d}]$ with unique prime factorization, greatest common divisor, and division algorithm, we can do the following.

We know that we can do the following as long as $b \neq 0$.  But we removed that case anyways.  Then, b is a unit in $\mathbb{Z}_{p}$ \[a^{2}+db^{2}= p^{2}\]  \[\implies a^{2}+db^{2}\equiv 0\mod{p}\]  \[\implies \left(\frac{a}{b}\right)^{2}\equiv-d \mod{p}\]  \[\implies \left(\frac{-d}{p}\right)=1\] Then, for some integer $x$, $p|(x^{2}+d)$ \[p|(x+\sqrt{-d})(x-\sqrt{-d})\] Were $p$ prime in $\mathbb{Z}[\sqrt{-d}]$, then $p$ would divide one of $(x+\sqrt{-d})$ and $(x-\sqrt{-d})$.  But clearly, it can't, because it doesn't divide the coefficient of $\sqrt{-d}$ in either.  Therefore, $p$ cannot be prime in $\mathbb{Z}[\sqrt{-d}]$.  Then \[p=(m+n\sqrt{-d})(m-n\sqrt{-d})\] where $m\pm n\sqrt{-d}$ are not units (ie don't have norm 1, have no multiplicative inverse).

Then, because both side have norm $p^{2}$, and norm is always an integers, both of the factors on the right side have norm $p$ (because neither has norm 1).  Thus, \[m^{2}+dn^{2}=N(m \pm n\sqrt{d})=N(p)=p^{2}\] as desired.

You can check yourself, but, for example, $\mathbb{Z}[\sqrt{2}]$ is a euclidean ring, but $\mathbb{Z}[\sqrt{-5}]$ is not.  Another possible method is to construct a lattice of points $(x, y)$ s.t. $p|x^{2}+y^{2}$ such that the area of the fundamental parallelogram is $p$, and then use Minkowski's theorem on a suitable circle.  This can be extended to higher dimensions as well.


calculus
integration
calculus computations

1:  A10 - B100 = 0

2: integral(0 to 10) (Ax - Bx^2)dx = 1

What is A and B?

Continue here please. And still: what have you done so far¿

Do you know how to integrate, theepan? It is just a simple system of equations... :huh:

I don't know, that is the prob.  I know how to fix the prob. with to unknow.

:|  $%Error. "intAx" is a bad command.
-Bx^{2}dx=A\frac{x^{2}}{2}-B\frac{x^{3}}{3}|_{0}^{10}=1$ theepan007 can you solve this system?


How many ways are there to get from A to B by following the segments and moving only up or to the right?
[asy]pair A,B; path p = A -- (1,0)--(1,1)--(0,1)--A; B = (2,3);
for(int i = 0; i < 2; ++i){
for(int k = 0; k < 3; ++k){
draw(shift((i,k))*p);}}
label("A",A,SW);label("B",B,NE);[/asy]

Note that there is only one way to get to any of the sides directly in front of A. There is also only one way to get to any of the sides directly east of A. The rest is just arithmetic.

answer: 10

hello
any random path can be written as Right Up Right Up Up
therefore no. of possible paths =$ \frac{(3\plus{}2)!}{2!3!}\equal{}10$
thank u

[quote="mathemagician1729"]Thats an interesting approach, can you explain how it works?[/quote]

LOL $5C2=\frac{5!}{3!2!}=\frac{5\cdot4}{2}=\boxed{10}$.


Evaluate   $ \int_{0}^{1}\frac{\arctan\frac{x}{x\plus{}1}}{\arctan\frac{1\plus{}2x\minus{}2x^2}{2}}$    :maybe:

_____________________________
Azerbaijan Land of the Fire   :thumbup:  :play_ball:

We can substitute x with 1-x.
Then we can add new integral to the origin and get as final result 1/2

hello, i think this integral is not so easy.
Sonnhard.

After addition the numerator becomes equal to denominator with help of identity. So  2*integral=1


inequalities
superior algebra
superior algebra unsolved

let $ G$ be a finite group, and let $ S,T$ be nonempty subsets of $ G$ such that $ G\not \equal{}ST$ Prove that $ |G|\ge |S|\plus{}|T|$.

if you know about the group action the you can solve this problem. One of the good book talking about this problem is : Algebra, a graduate course by Isaacs Martin I

Suppose that $ \vert G \vert < \vert S \vert \plus{} \vert T \vert$. We'll prove that the latter inequality implies $ G \equal{} ST$.

Clearly enough, we only need to show that $ G \subseteq ST.$

Take any $ g \in G$ and consider the following set

$ \mathfrak{S} \equal{} \{s^{ \minus{} 1}g: s \in S\}.$

It is not difficult to demonstrate that $ \vert \mathfrak{S} \vert \equal{} \vert S \vert$.

Hence,

$ \vert G \vert < \vert S \vert \plus{} \vert T \vert \equal{} \vert \mathfrak{S} \vert \plus{} \vert T \vert.$

From the previous line we conclude that the intersection of $ \mathfrak{S}$ and $ T$ is nonempty. [b]Therefore[/b], there exist $ s \in S$ and $ t \in T$ such that $ s^{ \minus{} 1}g \equal{} t$. This equation implies in turn that $ g \equal{} st \in ST$ and the proof [b][color=red]terminates[/color][/b].

Wasn't the proof supposed to make use of group actions?  :maybe:

Well, at least that's what I'd inferred from [b]1234567a[/b]'s post...

[quote="1234567a"] If you know about the group action then you can solve this problem...[/quote]

A sufficient condition, I know... However, after reading such a disheartening asseveration one would not expect a proof in the vein of [b][color=blue]coquitao[/color][/b]'s one.  :wink:


inequalities
LaTeX
inequalities unsolved

sqr(a)+sqr(b)+sqr(c)=9,a<=b<=c.a,b and c are positive reals.how to show that abc+1>3a

[quote="chengyuLi"]$ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\equal{}9$ and $ a\leq b \leq c$. $ a,b$ and $ c$ are positive reals.how to show that $ abc\plus{}1>3a$[/quote]
hello, please the next time try to write with $ Latex$ for this use this symbol $ \$$ before and after the code,

By the Way look at this

$ abc\plus{}1\ge{a^3}\plus{}1\equal{}(a^3\plus{}1\plus{}1)\minus{}1\ge3a\minus{}1$ by $ AM\minus{}GM$

but the problem tells that $ abc\plus{}1>3a$
thats funny.... :rotfl:

a2+b2+c2=9,a<=b<=c,a,b and c are positive reals,how to show that abc+1>3a.
but how to use latex,i used microsoft word.
thanks in advance.

can't anyone give a correct answer,it couldn't be difficult,just a competetion in sichuan,china.
i want to see your proof,thanks,xiexie,danke sehr.

for more information about latex see here,
http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

thanks,but i'm sorry my english is poor.i downloaded it,but i don't know how to use.sorry
i think this problem can change to 2*sqr(a)*c+sqr(c)=9,we just have to show sqr(a)*c+1>3*a.


function
calculus
integration
derivative
probability
geometric series
calculus computations

Consider the series

$\sum\limits_{k=1}^{\infty} {k^a b^k}$   with  $a>0, a \in R$

The series converges for $b<1$ (it can be prooved with the root test)

Is there an expression for the sum of the finite version of the series? (say:   $\sum\limits_{k=1}^N$)

I mean, something like $\dfrac{N(N+1)}{2}$ for the sum of first $N$ integers..

Thank you

Don't expect any unified formula. There are closed form expressions whenever $a$ is a positive integer, but the complexity grows rapidly.
There are strong relations to the gamma function and the zeta function. I would be surprised by any closed form when $a$ is not an integer.

Can you give me the closed expressions for $a=1$ and $a=2$?

And a reference (book, table, etc) would be useful.

Thanks

I don't know them off the top of my head, and I don't have a reference. Here is a method for finding the closed forms:
Abel's formula (Summation by parts): $\sum\limits_{k=1}^na_kb_k=\sum\limits_{k=1}^n(\Sigma a_k)\cdot(\delta b_k)+(\Sigma a_n)\cdot b_{n+1}-(\Sigma a_0)\cdot b_1$, where $\Sigma a_k=c+\sum\limits_{i=1}^ka_i$ and $\delta b_k=b_k-b_{k+1}$.

This is easy enough to verify (note the analogy with integration by parts). To use it for this problem, sum the geometric series and difference the polynomial. With repeated applications, you eventually reach a closed form for the finite sum, and can take limits to get the infinite sum. It is best in this case to set $c=-\sum\limits_{k=1}^\infty a_k$ in the definiton of $\Sigma$ so that the boundary term at $\infty$ is easy to work with and the series is still geometric for the next step.
This idea can also give recursive formulas when $a$ is not a nonnegative integer.

Another method for the infinite sums: treat $b$ as a variable, and the sum as a function ($f_a$) of $b$. Multiplying by $k$ on the series side is taking a derivative and multiplying by $b$ on the function side. Then $f_0(b)=\dfrac1{1-b}$, $f_1(b)=\dfrac{b}{(1-b)^2}$, and $f_2(b)=\dfrac{b}{(1-b)^2}+\dfrac{2b^2}{(1-b)^3}=\dfrac{b+b^2}{(1-b)^3}$

The infinite sums (for $a$ a nonnegative integer) are essentially the moments of the geometric probability distribution.

Searching for Abel's formula, I found the Abel's theorem:

http://en.wikipedia.org/wiki/Abel's_theorem

Consider the general version.. it seems I have to find the Generating Function of $a_k=k$ and $a_k=k^2$ (it's easy!), to have the sum of the infinite series. I will try to find the GF for $a_k$ stopped at $N$, too.

It's probably easier to find by looking for the other name, or trying a math-specific site.

By the way, I gave the answers for $n=1$ and $n=2$ in my post.

Yes, the GF method confirms that the sums of the infinite series are:

$\dfrac{b}{(1-b)^2}$ for $a=1$

$\dfrac{b(b+1)}{(1-b)^3}$ for $a=2$


Let $a>-1$ and $r \in (0,1)$ be reals. Prove that: \[(1+a)^{r}\leq 1+ra.\]

That's Bernouilli's inequality. You can find proofs of it all over the web.

Is it? I don't think so.

Proof is the same though (the second derivative is negative this time instead of positive).

[quote="MysticTerminator"]Is it? I don't think so.
[/quote]
[url]http://www.imo.org.yu/index.php?options=mbb|Bfa|f|Alg|Ine&p=02lLw[/url]

[quote="Farenhajt"]Posted and proven [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106614]here[/url][/quote]

That's not the same problem; you proved it for natural numbers $n$, while the problem posted here is the reverse inequality for reals $r \in (0,1)$.

Sorry, haven't even noticed that...  :maybe:

Let $b = ar, s = \frac1r$.  Then

$(1+a)^{r}\leq 1+ra \Longleftrightarrow 1+sb \leq (1+b)^{s}$.  So the question is not substantively different then the case $r > 1$.  You do still need to do it for all reals, (not just positive integers) though.

We'll prove the equivalent statement $(1+a)^{r}\geqslant 1+ar$ for $a>-1$ and $\mathbb{R}\owns r>1$.

Put $c=1+a$ and $s=r-1$. Then the inequality becomes

\begin{eqnarray}c^{s+1}\geqslant 1+(c-1)(1+s) &\iff& c^{s+1}\geqslant c+(c-1)s\nonumber\\ &\iff& c(c^{s}-1)\geqslant (c-1)s\end{eqnarray}

for $c>0, s>0$

For $c=1$ $(1)$ is trivially satisfied, hence we'll deal with $c\neq 1$.

[b]Part 1.[/b] $\bold{s\in\mathbb{Q}}$ Put $s={p\over q}, p,q\in\mathbb{N}$

[b]Case 1.1.[/b] $\bold{c>1}$. Then we can write

\begin{eqnarray*}c(c^{p\over q}-1)\geqslant (c-1){p\over q}&\iff& \frac{c(c^{p\over q}-1)}{c-1}\geqslant{p\over q}\\ &\iff& \frac{c(c^{1\over q}-1)(c^{{p-1}\over q}+c^{{p-2}\over q}+\dots+1)}{(c^{1\over q}-1)(c^{{q-1}\over q}+c^{{q-2}\over q}+\dots+1)}\geqslant{p\over q}\\ &\iff& \frac{c(c^{{p-1}\over q}+c^{{p-2}\over q}+\dots+1)}{c^{{q-1}\over q}+c^{{q-2}\over q}+\dots+1}\geqslant{p\over q}\end{eqnarray*}

The numerator on the LHS is not less than $c(\underbrace{1+1+\dots+1}_{p})=pc$, and the denominator is not greater than $\underbrace{c^{{q-1}\over q}+c^{{q-1}\over q}+\dots+c^{{q-1}\over q}}_{q}=qc^{{q-1}\over q}$, hence we have

$\frac{c(c^{{p-1}\over q}+c^{{p-2}\over q}+\dots+1)}{c^{{q-1}\over q}+c^{{q-2}\over q}+\dots+1}\geqslant\frac{pc}{qc^{{q-1}\over q}}={p\over q}c^{1\over q}>{p\over q}$

[b]Case 1.2.[/b] $\bold{c<1}$. Then we can write

\begin{eqnarray*}c(c^{p\over q}-1)\geqslant (c-1){p\over q}&\iff& \frac{c(c^{p\over q}-1)}{c-1}\leqslant{p\over q}\\ &\iff& \frac{c(c^{1\over q}-1)(c^{{p-1}\over q}+c^{{p-2}\over q}+\dots+1)}{(c^{1\over q}-1)(c^{{q-1}\over q}+c^{{q-2}\over q}+\dots+1)}\leqslant{p\over q}\\ &\iff& \frac{c(c^{{p-1}\over q}+c^{{p-2}\over q}+\dots+1)}{c^{{q-1}\over q}+c^{{q-2}\over q}+\dots+1}\leqslant{p\over q}\end{eqnarray*}

The numerator on the LHS is not greater than $c(\underbrace{1+1+\dots+1}_{p})=pc$, and the denominator is not less than $\underbrace{c^{{q-1}\over q}+c^{{q-1}\over q}+\dots+c^{{q-1}\over q}}_{q}=qc^{{q-1}\over q}$, hence we have

$\frac{c(c^{{p-1}\over q}+c^{{p-2}\over q}+\dots+1)}{c^{{q-1}\over q}+c^{{q-2}\over q}+\dots+1}\leqslant\frac{pc}{qc^{{q-1}\over q}}={p\over q}c^{1\over q}<{p\over q}$

[b]Conclusion 1.[/b] $(1+a)^{r}\geqslant 1+ar$ is satisfied for $a>-1$ and $\mathbb{Q}\owns r>1$

[b]Part 2.[/b] $\bold{s\in\mathbb{I}}$. Then we can generalize Conclusion 1 by using Dedekind cuts.

[b]Conclusion.[/b] $(1+a)^{r}\geqslant 1+ar$ is satisfied for $a>-1$ and $\mathbb{R}\owns r>1$


I don't know if it is a silly question or an obvious one!

I want to find matrix (or matrices) which I can multiply them by one matrix and get the transpose of that one.

[hide]I know that transpose of a matrix is similar to it, but I don't know the proof. But it results that these matrices exist, so how to find them?[/hide]


About the equation $ ax^2 \minus{} 2x\sqrt {2} \plus{} c \equal{} 0$, with $ a$ and $ c$ real constants, we are told that the discriminant is zero. The roots are necessarily:

$ \textbf{(A)}\ \text{equal and integral} \qquad\textbf{(B)}\ \text{equal and rational} \qquad\textbf{(C)}\ \text{equal and real}$
$ \textbf{(D)}\ \text{equal and irrational} \qquad\textbf{(E)}\ \text{equal and imaginary}$

[hide="Solution"]The roots of the equation are: $ x\equal{}\frac{2\sqrt{2}\pm 0}{2a}$.

And so they are necessarily $ \boxed{\textbf{(C)}\ \text{equal and real}}$.[/hide]


Let $ r>1$ and $ a,b,c\geq0$ s.t. $ ab\plus{}bc\plus{}ca\equal{}3$.Prove that:
$ a^r(b\plus{}c)\plus{}b^r(c\plus{}a)\plus{}c^r(a\plus{}b)\geq6$

[quote="Inequalities Master"]Let $ r > 1$ and $ a,b,c\geq0$ s.t. $ ab \plus{} bc \plus{} ca \equal{} 3$.Prove that:
$ a^r(b \plus{} c) \plus{} b^r(c \plus{} a) \plus{} c^r(a \plus{} b)\geq6$[/quote]
$ LHS\equal{} ab(a^{r\minus{}1}\plus{}b^{r\minus{}1})\plus{}bc(b^{r\minus{}1}\plus{}c^{r\minus{}1})\plus{}ca(c^{r\minus{}1}\plus{}a^{r\minus{}1})\ge 2((ab)^{\frac{r\plus{}1}{2}} \plus{} (bc)^{\frac{r\plus{}1}{2}}\plus{}(ca)^{\frac{r\plus{}1}{2}}) \ge 6(\frac{ab\plus{}bc\plus{}ca}{3})^{\frac{r\plus{}1}{2}}\equal{}6$; :)


Which of the following conjecture is true? Justify

[latex]10n  = O(n)[/latex]

[latex]10n^2  = O(n)[/latex]

[latex]10n^{55}  = O(n^2 )[/latex]

1) [latex]10n \leq cn[/latex]  => [latex]n=1[/latex]  [latex]c=10[/latex]

[latex]f(n) \leq 10g(n)[/latex] for all [latex]n \geq 1[/latex]. [latex]f(n) \in O(g(n))[/latex]

How can I solve the other two?

I found this problem on the Internet:
It is true that [latex]n^2 + 200n + 300  =   O(n^2[/latex]) ?

And [latex]n^2 -200n -300  =   O(n)[/latex]

@Apprentice123: to use LaTeX formulas, just take two dollar sings instead of [latex] and [/latex].  :wink:

All of these should be taken as $ n\to\infty$; there is an implicit "as $ x\to ?$ in any "$ f(g)$ is $ O(g(x))$" statement, but we should clarify what we expect the variables to tend to.
There's a sticky in the calculus forum you may find useful, and this thread really belongs there- because big-O notation is about limits.

[color=green][Moved by moderator.][/color]


Let $ k$ be a positive integer. Find all functions $ f: \mathbb  N \to \mathbb N$ satisfying
\[f(f(n))\plus{}f(n)\equal{}2n\plus{}3k\] for all positive integers $n$.

[quote="KDS"]Let $ k$ be a positive integer.Find all functions $ f: N \to N$ satisfying $ f(f(n)) \plus{} f(n) \equal{} 2n \plus{} 3k$ $ \forall n \in N$.[/quote]

Let $ f^{[p]}(n) \equal{} a_pf(n) \plus{} b_pn \plus{} c_pk$ $ \forall n > 0$
We get $ f^{[p \plus{} 1]}(n) \equal{} f^{[p]}(f(n)) \equal{} a_pf(f(n)) \plus{} b_pf(n) \plus{} c_pk$ $ \equal{} a_p( \minus{} f(n) \plus{} 2n \plus{} 3k) \plus{} b_pf(n) \plus{} c_pk$ $ \equal{} (b_p \minus{} a_p)f(n) \plus{} 2a_pn \plus{} (c_p \plus{} 3a_p)k$
 hence the equations :
$ a_{p \plus{} 1} \equal{} b_p \minus{} a_p$
$ b_{p \plus{} 1} \equal{} 2a_p$
$ c_{p \plus{} 1} \equal{} c_p \plus{} 3a_p$

And the result : $ f^{[p]}(n) \equal{} \frac {1 \minus{} ( \minus{} 2)^p}3f(n) \plus{} \frac {2 \plus{} ( \minus{} 2)^p}3n \plus{} (p \minus{} \frac {1 \minus{} ( \minus{} 2)^p}3)k$ $ \forall n > 0$, $ \forall p\geq 0$

Setting $ p \equal{} 2q$ and writing $ f^{[p]}(n) > 0$, we get $ f(n) < \frac {4^q \plus{} 2}{4^q \minus{} 1}n \plus{} (1 \plus{} \frac {6q}{4^q \minus{} 1})k$
Using $ q\to \plus{} \infty$ in this line, we get $ f(n)\leq n \plus{} k$

Setting $ p \equal{} 2q \plus{} 1$ and writing $ f^{[p]}(n) > 0$, we get $ f(n) > \frac {2\cdot 4^q \minus{} 2}{2\cdot 4^q \plus{} 1}n \plus{} (1 \minus{} \frac {6q \plus{} 3}{2\cdot 4^q \plus{} 1})k$
Using $ q\to \plus{} \infty$ in this line, we get $ f(n)\geq n \plus{} k$

So $ f(n) \equal{} n \plus{} k$ which, indeed, is a solution.

So the unique solution : $ \boxed{f(n) \equal{} n \plus{} k}$


During a severe thunderstorm, Tim saw a flash of lightning and then six seconds later heard a clap of thunder.  The speed of sound is 1088 ft/sec.  How far is Tim from the lightning?  Express your answer to the nearest thousand feet.

1088*6=6528.

express 6528 nearest thousand feet = 7000

answer : 7000


Ascertain the integer numbers $a,b$ so that the equations
$x^4+x^3+3x^2+2x+a=0$ and $x^3-x^2-x+b=0$
have at least two common roots.[hide="Answer."]$a=2$ and $b=-2\ .$[/hide]

[quote="chess64"]Do you want $a\ne0$?[/quote]

Oh nevermind, I didn't pay attention to the signs.


Find the largest constant $ t$ for which the following inequality holds for $ a,b,c\geq 0$:

$ \frac {bc}{b \plus{} c} \plus{} \frac {ca}{a \plus{} c} \plus{} \frac {ab}{a \plus{} b}\geq t(a \plus{} b \plus{} c)$

(inspired by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=148725]this one[/url])

Set $x=1/a$, $y=1/b$ and $z=1/c$ (assuming $a,b,c > 0$)

(if one of $a,b,c$ is $0$, then it is easy to see that $t \leq \frac{1}{4}$)

So we have

$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\geq t(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ but

we have that

$\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}$

Thus we have that

$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 2(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})$

Thus $t \le \frac{1}{2}$

The value $t = \frac{1}{2}$ is achieved when $a=b=c$


inequalities
geometry
rhombus
trigonometry
circumcircle
geometric transformation
geometry proposed

Let $ ABCD$ be a rhombus with $ \angle BAD \equal{} 60^{\circ}$. Points $ S$ and $ R$ are chosen inside the triangles $ ABD$ and $ DBC$, respectively, such that $ \angle SBR \equal{} \angle RDS \equal{} 60^{\circ}$. Prove that $ SR^2\geq AS\cdot CR$.

op02 project, Yugoslavia problem 3.

[hide="Solution by Myth and me"]
[color=blue][b]Problem.[/b] Let ABCD be a rhombus with < BAD = 60°. The points S and R lie inside the triangles ABD and DBC respectively such that < SBR = < RDS = 60°. Prove that $SR^2 \geq AS \cdot CR$.[/color]

[i]Solution by Myth and me.[/i] We start with a simple lemma:

[b]Lemma 1.[/b] For any angle t, we have $\sin\left(  60^{\circ}-t\right)+\sin\left(  60^{\circ}+t\right)  \leq2\sin60^{\circ}$.

[i]Proof.[/i] Using the addition formulas for sin, we get:

$\sin\left(  60^{\circ}-t\right)  +\sin\left(  60^{\circ}+t\right)$
$=\left(  \sin60^{\circ}\cos t-\cos60^{\circ}\sin t\right)  +\left(\sin60^{\circ}\cos t+\cos60^{\circ}\sin t\right)$
$=2\sin60^{\circ}\cos t\leq2\sin60^{\circ}$        (since $\sin 60^{\circ}=\frac{\sqrt3}{2}\geq 0$ and $\cos t\leq 1$).

Now let's solve the problem:

Since the quadrilateral ABCD is a rhombus, we have DA = AB. Since we also have < BAD = 60°, the triangle BAD is equilateral; since a rhombus is symmetric with respect to each of its diagonals, the triangles BAD and BCD are congruent, so that the triangle BCD is also equilateral.

WLOG assume that $DS\leq DR$. Let X be a point on the ray DS such that DX = DR, and let Y be a point on the ray DR such that DY = DS. Then, since DX = DR and < RDX = 60° (this follows from < RDS = 60°), the triangle XDR is equilateral. Similarly, the triangle YDS is equilateral.

Since the triangle XDR is equilateral, we have < DXR = 60°. In other words, < SXR = 60°. The equilateral triangle YDS yields < DYS = 60°, so that < RYS = 180° - < DYS = 180° - 60° = 180° - < SXR. This implies that the quadrilateral RYSX is cyclic. In other words, the points R, Y, S and X lie on one circle. The point B also lies on this circle, since < SBR = 60° and < SXR = 60° imply < SBR = < SXR. Hence, all five points R, Y, S, X and B lie on one circle. Let r be the radius of this circle. Then, by the fact that the length of a chord in a circle equals the double circumradius times the sine of the chordal angle of this chord, we have $SR=2r\sin\measuredangle SBR$, $BX=2r\sin\measuredangle BSX$ and $BY=2r\sin\measuredangle BSY$.

Since the triangle YDS is equilateral, we have < DSY = 60°, and thus < BSX + < BSY = 180° - < DSY = 180° - 60° = 120°. Hence, there exists an angle t such that < BSX = 60° - t and < BSY = 60° + t (this angle t can be positive or negative or 0°).

By Lemma 1, we thus have

$\sin\measuredangle BSX+\sin\measuredangle BSY=\sin\left(  60^{\circ}-t\right)+\sin\left(  60^{\circ}+t\right)  \leq2\sin60^{\circ}=2\sin\measuredangle SBR$

(since < SBR = 60°). Hence,

$2r\sin\measuredangle BSX+2r\sin\measuredangle BSY\leq2\cdot2r\sin \measuredangle SBR$,

or, in other words, $BX+BY\leq2\cdot SR$.

Since the triangle BAD is equilateral, we have < ADB = 60°, what, combined with < RDS = 60°, yields < RDS = < ADB, so that < RDB = < RDS - < SDB = < ADB - < SDB = < SDA. In other words, < YDB = < SDA. Since we also have BD = AD (this is again because the triangle BAD is equilateral) and DY = DS, it follows that the triangles YDB and SDA are congruent, and thus BY = AS. Similarly, BX = CR. Thus, the inequality $BX+BY\leq2\cdot SR$ rewrites as $CR+AS\leq2\cdot SR$. In other words, we have $2\cdot SR\geq AS+CR$. Thus,

$SR\geq\frac{AS+CR}{2}$.

But by the AM-GM inequality, $\frac{AS+CR}{2}\geq\sqrt{AS\cdot CR}$. Hence, $SR\geq\sqrt{AS\cdot CR}$, and $SR^2\geq AS\cdot CR$. And we are done.

Equality holds if and only if < BSD = 120°. The proof is left to the reader (hint: < BSD = 120° is equivalent to t = 0° and to AS = CR = SR).[/hide]

  Darij

Funny! I used EXACTLY the same construction!!

I found it in a very short time, not even 10 minutes, but it might not be that obvious if you don't see it immediately. Do you think this is a hard problem Darij?

Well, I don't know, but it took me way longer. Maybe it's easier to give a straightforward sine law solution which does not show the stronger inequality $SR\geq\frac{AS+CR}{2}$. Anyway, it was more than half a year ago when we - Myth and me - were solving the problem, and Myth was pretty busy while I was not too experienced with geometric inequalities. Still I find it quite a difficult problem.

  darij

My write-up of the solution was much shorter.

I assumed that $DR \geq DS$. Pick points $R'$ and $S'$ on $BS$ and $DR$ respectively such that triangles $BRR'$ and $DSS'$ are equilateral. It is easy to see, by for example spiral similarity, that $CR = DR'$ and $AS = BS'$. Also, $DSR'R$ and $BRS'S$ are cyclic. So, we need to prove $\frac{SR}{DR'}\frac{SR}{BS'} \geq 1$, or $\frac{\sin 60}{\sin \alpha}\frac{\sin 60}{\sin \beta} \geq 1$, where $\alpha = \angle{S'RR'}$ and $\beta = \angle{R'SS'}$ and $\alpha + \beta = 120$. But that is obvious from Jensen, for example.


geometry
perimeter
rectangle
perpendicular bisector

[color=indigo][hide]x=width of lot
3x*x = 3x2 =1200
x=20
She could divide the lot on the diangonal, the perpendicular bisector of the long side or the perpendicular bisector of the short side.
diangonal :approx: 63.2
per. bisector of short side=60
per. bisector of long side=30
The per. bisector of the short side is the best so she has to get 180 feet of fence.[/color][/hide]

[hide]
this is how i did it!
A=lw
1200=(3w)(w)
w=20
so, the length is
60 and the width
is 20.
than u split the rectangle
verticly and u ll get to two
rectangles and than
the measurements would be
30, 30, 30, 30, 20, 20, 20
so the answer is 180! :D[/hide]

[quote="MCrawford"]Ms. Salman purchases a rectangular lot with area 1,200 sq ft which has a length equal to 3 times its width.  She wishes to divide the lot into exactly 2 plots having equal area and install fencing along the perimeters of both lots.  What is the least number of feet of fencing she must buy?[/quote]
[hide="solution"]
3x:^2: = 1200
x:^2: = 400
x = +/- 20

20 by 60. Divide it widthwise - 20 feet plus the perimeter.

20 + 2(20 + 60)
20 + 2(80)
20 + 160
[b]180 feet[/b]
[/hide]


We should go back and dig up the old test and put the solutions here...

For educational purposes... ;)

Unless Code like them to be kept secret in order to sell his solution book later... :D

actually i was looking for "solutions"

yea, i been to ur site a couple of times before too...

thanks

Since we're on the topic of solutions to problems -

I'd like to see a solution to Week 7 nr. 9 that does not involve brute force, which is how I attempted it (and missed.)

Oh sorry, you are correct.  :blush:  I'll tell him.

Thanks.

Oops.  Actually, Samson, I thought you had posted those answers (it was actually DJ or Nathan, I can't tell which)...don't worry about telling him (them?), they'll see it.

No, you caught the culprit. I told thoobik (who in this case, is DJ) that the answer was 462.  :blush:

For number 9, you just have to use a recursion.  First, say that you won't use any 0's.  After that, you just figure out what the distribution is in an efficient manner.  And then add up.  Signposts along the way: (5, 4), (20, 21) and (105, 104).


hi, I'm sort new to form, and I just want to ask people what they think about this site. I started creating this site a couple days ago.

My goal is to create simple math notes that people can easily understand. I sort of dislike those other sites since it takes a bit too much effort to find what I'm looking for. Even though they have a lot of content.

What I'm planning to add are flash cards that contain formulas that a person should know. Questions and how a person should analyse these questions and how they should answer to get full marks.

Even though there is hardly anything on the site. What do you guys think about the format and the direction of the site?

ezmathnotes.hosthelpers.net

I like the layout, I really do :) But, I would consider using LaTeX for the math notation -- it looks so much prettier than formatted HTML! There's tons of guides around AoPS for getting LaTeX up and running -- give it a shot! (Oh, and this thread doesn't really fit in the problem solving forum :()

Sorry about that, this was sort of the closest thing I could find to what I was posting, where should I have posted it.

Thanks for the advice too. Ill add it soon.

This thread doesn't really fit anywhere else, and this forum [i]is [/i]about math resources, so this is probably the best place to put this thread.


function
algebra
domain
calculus
calculus computations

Let $ a$ be a positive constant. Find the maximum and minimum values of $ y\equal{}a\sqrt{1\plus{}x}\plus{}\sqrt{a^2\minus{}x}$.

First, write:

$ y'(x) \equal{} \frac {1}{2}\left(\frac {a}{\sqrt {1 \plus{} x}} \minus{} \frac {1}{\sqrt {a^2 \minus{} x}}\right)$

and note that the domain of $ y$ is $ [ \minus{} 1,a^2]$.

Now, if $ a \leq 0$ we see that $ y' < 0$ and therefore the minimum is:

$ y_{max} \equal{} y( \minus{} 1) \equal{} \sqrt {1 \plus{} a^2}$

$ y_{min} \equal{} y(a^2) \equal{} a\sqrt {1 \plus{} a^2}$

On the other hand, if $ a > 0$ we get

$ x_{0} \equal{} a^2 \minus{} 1$

that turns out to be a maximum. So, since $ x_0 \geq \minus{} 1$ we see that the minimum is either on the left side or on the right side of the domain. But, as we can see from before, the right value is bigger that the left, so we have:

$ y_{max} \equal{} y(a^2 \minus{} 1) \equal{} a^2 \plus{} 1$

$ y_{min} \equal{} y( \minus{} 1) \equal{} \sqrt {1 \plus{} a^2}$

And that's it! :)

First of all $ a>0!$ the max is correct, but the min is not.

Agreed. The minimum varies for $ a < 1$ and $ a > 1$. For $ a < 1$, $ y_{min} \equal{} y(a^2)$ and for $ a > 1$ it is $ y_{min} \equal{} y( \minus{} 1)$. For $ a \equal{} 1$ we have $ y_{min} \equal{} y( \minus{} 1) \equal{} y(a^2)$.

That's right. Let me summarize the answer,

$ y_{max}=y(a^2-1)=a^2+1$.

$ y_{min}=\left\{
\begin{array}{ll}
y(a^2)=a\sqrt{a^2+1}\ (0<a<1)  &\quad  \\
y(-1)=\sqrt{a^2+1}\ (a\geqq 1)  &\quad
\end{array}
\right.$

Here is my solution:

Let ${ a\sqrt{1+x}}=X,\ \sqrt{1+x}=Y$, the problem is restated that

If $ X^2+Y^2=a^2+1\ (X\geq 0,\ Y\geq 0)$, find the maximum and minimum values of $ aX+Y$ for a positive constant $ a$.


This might be too easy or too hard, I'm not sure.

Given a quadrilateral find, with proof, what point minimizes the sum of the distances from that point to the vertices.

[hide="ummm...."]
Isn't it just the intersection of the diagonals?
Lable the vertices a,b,c,d counterclockwise. Then to minimize $AX+XC$ we must have $X\in AC$ and likewise $X\in BD$ so that $X_{min}=AC\cap BD$
[/hide]

The fermat point is the solution for a [b]triangle[/b]


$S(N)$ denote the sum of the digits of the positive integer $N$.
Prove that $S(9^n)>9$ for $n>2$.

hé, hé  :D  you still like this one...

Pierre.

A related problem was solved in the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8590]exponents problem[/url].

As far as I remember, it is a difficult problem.

Pierre.

You're right Pierre. As you know, I'm eating digits for my breakfast :wacko:
Thanks to Ravi B for the link but I think it's more complicated in this case...

Yes, I agree this problem is more difficult.


(a) Show that, for all non-zero values of the constant $ k$, the curve

$ y\equal{}\frac{kx^{2}\minus{}1}{kx^{2} \plus{} 1}$

has exactly one stationary point.

(b) Show that, for all non-zero values of the constant $ m$, the curve

$ y \equal{} e^{mx}(x^{2} \plus{} mx)$

has exactly two stationary points.

$ y \equal{} \frac {kx^{2} \minus{} 1}{kx^{2} \plus{} 1} \equal{} \allowbreak 1 \minus{} \frac {2}{kx^{2} \plus{} 1}$

set $ \frac {dy}{dx} \equal{} \allowbreak \frac {4kx}{\left( kx^{2} \plus{} 1\right) ^{2}} \equal{} 0$ to locate stationary points.

Obviously, $ (0, \minus{} 1)$ is the only one for all $ k\neq 0$

$ y \equal{} e^{mx}\left( x^{2} \plus{} mx\right)$

$ \frac {dy}{dx} \equal{} \allowbreak \allowbreak e^{mx}\left( mx^{2} \plus{} (m^{2} \plus{} 2)x \plus{} m\right) \equal{} 0$

Since the discriminant $ (m^{2} \plus{} 2)^{2} \minus{} 4m^{2} \equal{} \allowbreak m^{4} \plus{} 4 > 0,$

there are always $ 2$ stationary points for all non zero $ m$.


Let $x,y$ be positive integers. Can $x^4+x^2y^2+y^4$ be a perfect square?

[quote="seamusoboyle"]That only works if the two brackets are relatively prime. eg:

$(2)(18)=6^2$[/quote]
You're right, my mistake.

[quote="Jimmy"]No, $x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)$.
So for this to be a perfect square, we must have $x^2+xy+y^2=x^2-xy+y^2\Rightarrow xy=-xy$, which cannot happen in positive integers.[/quote]

But what if x^2 + xy + y^2 and x^2 - xy + y^2 are both square?

[hide]Jimmy's decomposition doesn't help much here. It's better off to use this one:
\[ x^4+ x^2y^2 + y^4 = (x^2+y^2)^2 - x^2y^2 = k^2 \ \Rightarrow \ xy = x^2-y^2, k = 2xy\] (from the Pytagorean triplets) therefore looking at the latter as a quadratic in $x$ we have $\Delta = 4y^2 + y^2 = 5y^2 $ which cannot be a perfect square if $y$ is a positive integer. [/hide]


trigonometry
algebra
system of equations
algebra proposed

Solve the following system of equations:
$\cos x+\cos y=1$,
$\sin x \cdot \sin y =-\frac34$
if $0 \leq x \leq 2\pi$ and $0 \leq y \leq 2 \pi$

[hide="Change variables"]
$x+y=2u, \ \ x-y=2v$

Obtain easy system
$\cos u \cos v = \frac{1}{2}$
$\cos^{2}u-\cos^{2}v = \frac{3}{4}$[/hide]


[quote="raebre"]The length of a median is 36 units. Describe the location of the centroid on that median. :?:[/quote]

http://en.wikipedia.org/wiki/Centroid

This question is too weak for a society like this....

I am sorry that it is too weak for this society. I am in college and thought I would see if I could get some help here. I am not a math wizard or genius. Iam though studying to become a teacher and I have not had this kind of math in years and found this site and thought I would see if this forum would be nice to try and help me figure it out. If not that is alright.

It's O.K. But it'll be more appropriate to post them in Lower level places like "High School Basics" or "Intermediate Topics" in this site.

I just thought since it was geometry that I would put in the geometry section and this is my first time using this site so I had no clue as to how it works. I will make sure to do that next time, I actually think I have them all figured out now but I do appreciate your help and kindness. If I can not get any answers in the other forums you suggested would it be alright to repost them here. I mean since you all seem to be pretty good with math and all.


Let us consider $25^{\frac{1}{2}}$

On page 4 of AoPS 1 it asserts $(5^2)^{\frac{1}{2}} = 25^{\frac{1}{2}}$
and therefore $25^{\frac{1}{2}} = 5$

But how about this?  $-5^1 = (-5^2)^{\frac{1}{2}} = 25^{\frac{1}{2}}$

So does $25^{\frac{1}{2}} = \pm{5}$

Or have I made a mistake? If so point it out for me!

When you take the square and then the square root, you are actually finding the absolute value, which is why you're gettting $-5=25^{\frac{1}{2}}=5$. And when taking the square root of a positive real number, it's implied that you get the positive square root. Of course, this isn't the case when you have something like $x^2=25$, which would have $-5$ and $5$ as its answers.


I go to an American International school in India, and was trying to get my school to register for the AMC. When trying to convince the school to register for it, what websites should I point them towards?

I'm particulary looking for sites that point out the prestige of this competition (This is the biggest in the US, leading to the IMO...)

Thanks

Start with our basic webpage
http://www.unl.edu/amc

On it you will find our Frequently Asked Questions, the Sliffe Awards for teachers, pictures of the USAMO winners, pictures of the USA IMO teams, statistics and much more.  Point out the number of participating schools listed among the statistics.
Good luck.

Steve Dunbar
MAA Director, American Mathematics Competitions


[quote="xdanielxedgex"]If x,y,z > 0 show that
$ \frac {{4x \minus{} 1}} {{2x^2 \plus{} 1}} \plus{} \frac {{4y \minus{} 1}} {{2y^2 \plus{} 1}} \plus{} \frac {{4z \minus{} 1}} {{2z^2 \plus{} 1}} \leqslant 2$[/quote]

Try $ x\equal{}y\equal{}z\equal{}1$.


geometry
3D geometry
sphere
calculus
calculus computations

A solid hemisphere of variable radius x (0<x<a) with its base lowermost and horizontal is placed symmetrically into a hemispherical bowl of radius a,filled with water and resting on a horizontal surface.
Find the greatest volume of water displaced.

[quote="benorin"]The maximum volume of water displaced is $\frac{4\pi \sqrt{10}a^{3}}{15}$, will post solution later.  --Ben[/quote]

The volume of a sphereical cap of height h (if the radius of the sphere is r) is

$S(r,h)=\frac{\pi h^{2}}{3}(3r-h)$

and the volume of a sphere is $V(r)=\frac{4\pi r^{3}}{3}$

Let h denote the distance from the water line to the top of the inner-sphere, then the volume of water displaced, say w(x), is given by

$w(x)=V(x)-S(h,x)$

some simple calculations give $h=x-\sqrt{a^{2}-x^{2}}$ so that we have

$w(x)=V(x)-S(x-\sqrt{a^{2}-x^{2}},x)$
$=\frac{4\pi x^{3}}{3}-\frac{\pi(x-\sqrt{a^{2}-x^{2}})^{2}}{3}(3x-\sqrt{a^{2}-x^{2}})$
$=\frac{\pi}{3}\left[ 2x^{3}+(4x^{2}-a^{2})\sqrt{a^{2}-x^{2}}\right]$

to maximazed the volume of water displaced, differentiate w(x) w.r.t. x and set =0, i.e.

$w^{\prime}(x) = \frac{\pi x\left( 2x\sqrt{a^{2}-x^{2}}-4x^{2}+3a^{2}\right)}{\sqrt{a^{2}-x^{2}}}=0$
$\Rightarrow x=0,-\frac{a}{\sqrt{2}},\frac{3a}{\sqrt{10}}$

of which the last value of x gives the maximum, namely

$w\left( \frac{3a}{\sqrt{10}}\right) = \frac{4\pi \sqrt{10}a^{3}}{15}$