Baby Steven weighed six pounds at birth and gained two pounds a week. Mother Ilene weighed 159 pounds when Steven was born. Ilene started exercising and eating low-fat foods and lost one pound per week. How many weeks past his birth did Baby Steven weigh one-fourth as much as Mother Ilene?

So, you can create a chart:
Baby Steven|6|8|...
Mother Ilene|159|158|...
 and so on, until you get to |36|
                                        |144|
You can stop here, because 36 * 4 = 144.
So, to find the number of weeks, subtract 6 from Baby Steven's weight, since that is what he weighed at birth, and divide by 2 (the number of pounds he gained per week), to get 30/2 = [b]15[/b].
This solution is rather long, and I barely managed to finish in time. If anyone has a better solution, could they please share it.
Thanks,
phr33k

Yes, charts and tables take time that may take longer than the given time.  Using an equation is a lot faster in my opinion.  Let's call the number of weeks $ w$.  We can create the equation
\[ 6\plus{}2w\equal{}\frac{1}{4}(159\minus{}w)\implies4(6\plus{}2w)\equal{}159\minus{}w\implies\] \[ 24\plus{}8w\equal{}159\minus{}w\implies 9w\equal{}135\implies w\equal{}\boxed{15}\]


quadrilateral
geometry
circumcircle
area
david monk
IMO
IMO 1993

Let $A$, $B$, $C$, $D$ be four points in the plane, with $C$ and $D$ on the same side of the line $AB$, such that $AC \cdot BD = AD \cdot BC$ and $\angle ADB = 90^{\circ}+\angle ACB$. Find the ratio
\[\frac{AB \cdot CD}{AC \cdot BD}, \]
and prove that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles $ACD$ and $BCD$ at the point $C$ are perpendicular.)

There is an easy way to calculate the ratio AB.CD/AC.BD.
Applying rule of cosines for nonconvex quadrilateral ADBC we have
(AB.CD)^2=(AC.BD)^2+(BC.AD)^2-2.AC.BD.BC.AD.cos(AB.CD)^2=
=(AC.BD)^2+(BC.AD)^2-2.AC.BD.BC.AD.cos(<ADB-<ACB)=(AC.BD)^2+(BC.AD)^2-2.AC.BD.BC.AD.cos90=
=(AC.BD)^2+(BC.AD)^2=2.(AC.BD)^2 therefore the ratio is equa to   :rt2:

my solution:

see one inversion $(D,R^2)$
let $X'$ the inverse of $X$.

we have $A'C' = \frac{R^2.AC}{AD.DC} = \frac{R^2.BC}{BD.CD} = B'C'$. (by our condition)
so, $\angle{DCB} = \angle{ABD}$.

now, with simple angle chasing, we get $\angle{A'C'B'} = 90$.
(this implies that the circuncircles of $ADC$ and $BDC$ are orthogonal).

so, $A'C'B'$ is one isoseles right angled triangle.

now,

$\sqrt{2} = \frac{A'B'}{B'C'}$

$= \frac{\frac{R^2.AB}{AD.BD}}{\frac{R^2.BC}{BD.CD}} = \frac{AB.CD}{AD.BC}$.  :)

[quote="armpist"]Hard to belive it was a problem for IMO, but back in '93  we did not know that much

   mainly because of infancy of The Internet, Data Bases were not  properly

   loaded, I conjecture.

   Here is my solution:

   Similar triangles in circles ACD  and BCD are 90deg rotated because of

   condition angle ADB = 90 + ACB.

   So we have two vectors that are equal and at 90 deg.   The connecting vector

   is sqrt(2) times longer.

   Thank you.

   M.T.[/quote]

I think this is the same idea as my solution, but it seems like you skipped a lot of steps. More detailed proof:

Construct E such that a spiral similarity takes ACE to ADB. CE = BD*AC/AD = BC, and by the angle condition, ECB is an isosceles right triangle. Then

$\frac{AB \cdot CD}{AC \cdot BD}= \frac{EB \cdot AD}{EC \cdot AD}= \sqrt{2}$

and the angle condition easily gives the two circumcircles orthogonal.

First of all, the correct statement is "Let $ D$ be a point inside acute triangle $ ABC$ such that $ \angle ADB\equal{}\angle ACB\plus{}\pi/2$ and $ AC\cdot BD\equal{}AD\cdot BC$. (...)"
I'll solve the second part first, because it can be solved without making reference to $ \overline{AC}\cdot\overline{BD} \equal{} \overline{AD}\cdot\overline{BC}$.
Let $ E\ne C$ be the intersection of the circumcircle of $ [ACD]$ with the tangent to $ [BCD]$ at $ C$.
Let $ F\ne C$ be the intersection of the circumcircle of $ [BCD]$ with the tangent to $ [ACD]$ at $ C$.
It's easy to find out $ C\hat{A}D \plus{} C\hat{B}D \equal{} 90^{\circ}$ and therefore, using the Half-Inscribed Angle Theorem, it's easy to find out $ E\hat{C}F \equal{} 90^{\circ}$, and the result follows.
For the first part, invert through $ C$. Points $ A$, $ B$ and $ D$ are sent to points $ A'$, $ B'$ and $ D'$, respectively. Then $ \overline{AC}\cdot\overline{BD} \equal{} \overline{AD}\cdot\overline{BC}$ is equivalent to $ \overline{A'D'} \equal{} \overline{B'D'}$ and $ \frac {\overline{AB}\cdot\overline{CD}}{\overline{AC}\cdot\overline{BD}}$ is equal to $ \frac {\overline{A'B'}}{\overline{B'D'}}$.
Since $ C\hat{A}D \plus{} C\hat{B}D \equal{} 90^{\circ}$, it's easy to find out $ A'\hat{D'}B' \equal{} 90^{\circ}$, so, using Pythagorean Theorem, we find out $ \frac {\overline{A'B'}}{\overline{B'D'}} \equal{} \sqrt {2}$.

Let $X$ and $Y$ respectively denote the circumcenters of $\triangle{ACD}$ and $\triangle{BCD}$. Construct point $P$ such that $\triangle{BDP} \sim \triangle{BCA}$. Combining the ratio of similarity and the condition, we now have that

\[DP \cdot BC = AC \cdot BD = AD \cdot BC\]
Hence $DP=BC$. Also, we have that $\angle{ADP}=\angle{ADB}-\angle{PDB}=\angle{ADB}-\angle{ACB}=90$. Hence $\triangle{ADP}$ is an isosceles right triangle.

Now note that since $\triangle{BDP} \sim \triangle{BCA}$, spiral similarity implies that $\triangle{BAP} \sim \triangle{BCD}$. The ratio of similarity implies that

\[\frac{AP}{CD}=\frac{AB}{BD} = \frac{AC\sqrt{2}}{CD} \quad \Rightarrow \quad \frac{AB\cdot CD}{AC\cdot BD}= \sqrt{2}\]
Also, we have the following

\[\angle{CBD} = \angle{ABP} = 180-\angle{PAB} - \angle{APB}=135-\angle{CAP}=90-\angle{CAD}\]
Therefore we have that $\angle{CXD}=180-2\angle{CBD}=180-\angle{CYD}$. Therefore $XCYD$ is cyclic. Since $XCYD$ is symmetrical around line $XY$, we have that $\angle{XCY}=\angle{XDY}=90$. Hence circles $ACD$ and $BCD$ are orthogonal.

Weird. When I first saw this problem, it seemed very projective [due to Apollonian circle and the ratio condition], but none of the other sols took that approach. But then I am not good at projective, so I do not know if such methods trivialize this. Is this the case?

[asy] unitsize(1.2inches); pair A=(9/13)*dir(0); pair B=(13/9)*dir(0); pair CC=dir(37); pair DD=CC*dir(90); pair C=2*foot(0,B,CC)-CC; pair D=2*foot(0,B,DD)-DD; pair Y=dir(0); pair X=dir(180);

draw(circumcircle(X,Y,CC)); draw(B--C); draw(B--DD); draw(X--B); draw(A--D); draw(A--C); draw(0*dir(0)--CC); draw(0*dir(0)--DD);

dot("$A$",A,dir(-90)); dot("$B$",B,dir(0)); dot("$C$",C,dir(C)); dot("$D$",D,dir(40)); dot("$C'$",CC,dir(CC)); dot("$D'$",DD,dir(DD)); dot("$X$",X,dir(180)); dot("$Y$",Y,dir(-45)); dot("$O$",0,dir(-90)); [/asy]

The condition $AC\cdot BD=AD\cdot BC$ implies that $C$ and $D$ are on an Appolonius circle with foci $A$ and $B$. This means that there are points $X,Y$ on line $AB$ such that $(AB;XY)=-1$ and $C,D\in(XY)$. If we let $O$ be the midpoint of $XY$, then the condition $(AB;XY)=-1$ just tells us that $A$ and $B$ are inverses with respect to the Appolonius circle $(XY)$.

Let $\phi_B$ be inversion at $B$ with power $BX\cdot BY$. Note that $\phi_B$ fixes $(XY)$ and swaps $O$ and $A$. Let $D'=\phi_B(D)$ and $C'=\phi_B(C)$. Note that by the inversion, we have \[\angle ADB=\angle \phi_B(D)\phi_B(A)B=\angle D'OB\] and similarly $\angle ACB=\angle C'OB$. Thus, the angle condition in the problem simply tells us that $OC'\perp OD'$.

We see that $\phi_B((ACD))=(OC'D')$ and $\phi_B((BCD))=C'D'$. Since $(OC'D')=(C'D')$, we have $(OC'D')\perp C'D'$. Inverting back tells us that $(ACD)\perp(BCD)$.

To find the ratio, we will resort to a complex bash. Let $(XY)$ be the unit circle with $y=1$ and $x=-1$. Furthermore, suppose $C'=z$, so $D'=iz$. Finally, suppose that $B=\beta$ and $A=1/\beta$ where $\beta\in\mathbb{R}$.

We see that $B$ lies on the chord $CC'$, so \[c+c'-cc'\overline{b}=b\implies c=\frac{c'-b}{c'\overline{b}-1},\] and similarly $d=\frac{d'-b}{d'\overline{b}-1}$. Thus, \[\boxed{c=\frac{z-\beta}{z\beta-1}}\quad\text{and}\quad \boxed{d=\frac{iz-\beta}{iz\beta-1}}.\] We now compute \begin{align*} AC\cdot BD &= \left|\frac{1}{\beta}-\frac{z-\beta}{z\beta-1}\right|\cdot\left|\beta-\frac{iz-\beta}{iz\beta-1}\right| \\ &= \frac{|\beta^2-1|^2}{|\beta|\cdot|z\beta-1|\cdot|iz\beta-1|} \end{align*} and \begin{align*} AB\cdot CD &= \frac{|\beta-1/\beta|}{|z\beta-1|\cdot|iz\beta-1|}\cdot|-iz\beta^2-z+z\beta^2+iz| \\ &= AC\cdot BD\cdot|1-i| \\ &=\sqrt{2}\cdot AC\cdot BD. \end{align*} Thus, $\boxed{\frac{AB\cdot CD}{AC\cdot BD}=\sqrt{2}}$.

Let $O_1$ and $O_2$ be the centers of $(ACD)$ and $(BCD).$ Notice $$\angle O_1DO_2=\angle O_2DC+\angle O_1DC=\angle DAC+\angle CBD=360-\angle ACB-(360-\angle ADB)=90$$ so $(ACD)$ and $(BCD)$ are orthogonal. Invert about $A$ with radius $1.$ Then, $$1=\frac{AC\cdot BD}{AD\cdot BC}=\frac{AC\cdot\frac{B'D'}{AB'\cdot AD'}}{AD\cdot\frac{B'C'}{AB'\cdot AC'}}=\frac{B'D'}{B'C'}$$ and $$\angle AD'B'=90+\angle AB'C'$$ so $B'D'=B'C'$ and $\angle C'B'D'=90.$ Hence, $$\frac{AB\cdot CD}{AC\cdot BD}=\frac{AB\cdot\frac{C'D'}{AC'\cdot AD'}}{AC\cdot\frac{B'D'}{AB'\cdot AD'}}=\frac{C'D'}{B'D'}=\sqrt{2}.$$ $\square$

Apparently complex numbers works for the first part.

The first condition means that $\frac{(b-d)(a-c)}{(a-d)(b-c)}$ is pure imaginary by dividing the rotations between $\overline{AD}$ and $\overline{BD}$ and between $\overline{AC}$ and $\overline{BC}$. However, because $|a-c||b-d| = |a-d||b-c|$, the magnitude of this number is $1$, so it is necessarily $i$. Then,
\begin{align*}
(a-b)(c-d) &= ((a-c) - (b-c))((b-d)-(b-c)) \\
&= (b-c)[i(a-d) - (b-d) - (a-c) + (b-c)] \\
&= (1-i)(b-c)(a-d).
\end{align*}So$$\frac{AB \cdot CD}{AC \cdot BD} = \frac{|a-b||c-d|}{|b-c||a-d|} = |1-i| = \boxed{\sqrt 2}.$$

Invert about $D$ with radius $1$; denote images with $\bullet'$. The condition $\angle ADB=90^\circ+\angle ACB$ translates to $\angle A'C'B'=90^\circ$. Furthermore, by applying the inversion distance formula the condition $AC\cdot BD=AD\cdot BC$ is equivalent to $A'C'=B'C'$. Then $\frac{AB\cdot CD}{AC\cdot BD}=\frac{A'B'}{A'C'}=\sqrt{2}$. Furthermore, it is clear that perpendicular lines, when inverted around a point not lying on them, become orthogonal circles, so since $\overline{A'C'} \perp \overline{B'C'}$ we have $(DAC)$ and $(DBC)$ orthogonal. $\blacksquare$

Invert at $A$ with radius 1. Our angle condition is transformed to $\angle C^*B^*D^* = 90$, so the orthogonality is clear. Our length condition is
\[\frac{1}{AC^*} \cdot B^*D^* \cdot \frac{1}{AB^* \cdot AD^*} = \frac{1}{AD^*} \cdot B^*C^* \cdot \frac{1}{AB^* \cdot AC^*} \implies B^*D^* = B^*C^*.\]

Thus our desired ratio can be written as
\[\frac{\frac{1}{AB^*} \cdot C^*D^* \cdot \frac{1}{AC^* \cdot AD^*}}{\frac{1}{AC^*} \cdot B^*D^* \cdot \frac{1}{AB^* \cdot AD^*}} = \frac{C^*D^*}{B^*D^*} = \boxed{\sqrt 2}.\]

Invert about $D$ with radius $1.$ Note that by angle chasing with the $\angle ADB=\angle ACB+90$ we have $\angle A'C'B'=90.$ As the inversion of two perpendicular lines that don't go through the center of inversion is two orthogonal circles, we have proved the second part of the problem. For the first part, using the inversion distance formula we have
\begin{align*}
A'C' &= AC \cdot \frac{1}{CD \cdot AD} \\
B'C' &= BC \cdot \frac{1}{CD \cdot BD} \\
A'B' &= AB \cdot \frac{1}{AD \cdot BD}. \\
\end{align*}
Using $\frac{BC}{BD}=\frac{AC}{AD},$ we have that $A'C'=B'C$ by dividing the the first equation by the second. Dividing the third equation by the first we get $$\frac{AB \cdot CD}{AC \cdot BD}=\frac{A'B'}{A'C'}=\sqrt{2},$$ and we're done.  You could also spiral the length condition quite easily.

Upon inverting at $A$ we get that $B'C'D'$ is an isosceles right triangle with right angle at $B',$ so the ratio is $\frac{C'D'}{B'D'}=\sqrt2,$ and the orthogonality is because $C'D'$ is a diameter of $(B'C'D').$


Its a beautifull problem :
we are going to extend the set of real numbers, $R$, with a new creature with the name of $\delta$
and we will show the new set with $R[\delta]$ and its elements are creatures with the  form of $ a +b\delta$
such that $a , b \in R$

we have these rules in our new set
1.$a+b\delta=a^`+b^`\delta$ if and only if $a=a^`$ and $ b=b^`$
2.$\delta$ is a very small creature such that it is not zero but $\delta^2=0$
3.$ (a+b\delta)+(a^`+b^`\delta)=(a+a^`)+(b+b^`)\delta $
4.$ (a+b\delta)(a^`+b^`\delta)=aa^`+(ab^`+ba^`)\delta$

let $P(x)$ be a polynomial with real coefficients

show that there exist a number in $R$ like $a$ such that $P(x)$ is divisable by $(x-a)^2$ if and only if $P(x)$
has a root in $R[\delta]$( a root in $R[\delta]$ means a root with the form of $a+b\delta$ such that $b$is not zero)

It does look beautiful, but unless I'm making some stupid mistake, it also looks realy easy :?.

Assume $P$ has a double real root $a$. In this case, $P(x)=(x-a)^2Q(x)$, and we get $P(a+\delta)=0$.

Conversely, assume $P(a+b\delta)=0$. Denote the polynomial $P(a+bx)$ by $\tilde P(x)$. We have $\tilde P(\delta)=0$. If we write $\tilde P(x)=a_nx^n+\ldots+a_1x+a_0$, we get $a_1\delta+a_0=0$, which is equivalent to $a_1=a_0=0$, so $x^2|\tilde P(x)=P(a+bx)$, which means that $(b^{-1}x-b^{-1}a)^2|P(x)\Rightarrow (x-a)^2|P(x)$, and we are done.

no mr grobber
as i see you are not making any mistake thanks for the solotion


analytic geometry
conics
ellipse
calculus
geometry
inequalities
ratio

Find the locus of points $ (x,y)$ in the Cartesian coordinate system for which $ x^{2}\minus{}xy\plus{}y^{2}\leq{9}$.

[quote="lasha"]Find the locus of points $ (x,y)$ in the Cartesian coordinate system for which $ x^{2} \minus{} xy \plus{} y^{2}\leq{9}$.[/quote]

This is obviously the inside part of an ellipse whose characteristics are easy to find thru some minor calculus :

Axis are : $ y \equal{} x$ and $ y \equal{} \minus{} x$

Focus points are : $ (\sqrt 6,\sqrt 6)$ and $ ( \minus{} \sqrt 6, \minus{} \sqrt 6)$

Sum of distances to focus points : $ 6\sqrt 2$

semimajor axis : $ 3\sqrt 2$

semiminor axis : $ \sqrt 6$

excentricity : $ \frac{\sqrt 6}3$

[quote="lasha"]Thank you, pco. I needed it for the Romanian Masters 2008 problem 1, where I denote the distances from point $ P$ to the sides of equilateral triangle by $ a^{2},b^{2},c^{2}$. We are asked to find the locus of points $ P$, for which $ a,b,c$ are the sides of a triangle. It's obvious that if WLOG we assume that the sidelength of a given triangle is $ 1$, then the area is $ \frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{2}$, but on the other hand it's $ \frac {\sqrt {3}}{4}$. So, $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} \frac {\sqrt {3}}{2}$ is fixed and as $ a > |b \minus{} c|$, we get $ b^{2} \minus{} bc \plus{} c^{2}\leq{\frac {\sqrt {3}}{4}}$. Analogeous inequality must be true for any permutation of $ {a,b,c}$. As I've read in the official solution, answer is the inside part of an inscribed circle. Is it in any relation to my proposed problem?
 P.S. Sorry for talking about that problem in other topic.[/quote]

I dont know what kind of response you are waiting for.

I've not checked your text but I dont think there is a direct relation between the locus of points $ (x,y)$ such that $ x^2\minus{}xy\plus{}y^2\le u$ and the locus of points $ P$ such that $ PA^2\minus{}PA\cdot PB\plus{}PB^2\le v$ where $ PA$ and $ PB$ are the distances between $ P$ and two given points $ A$ and $ B$.

You should notice too that $ x^2\minus{}xy\plus{}y^2\equal{}u$ is never a circle since ratio between semi major axis and semi minor axis is always $ 3$.


Let a,b,c such that $ a,b,c<2$ and $ abc\equal{}1$.Prove that:
$ \frac{1}{(2\minus{}a)^3}\plus{}\frac{1}{(2\minus{}b)^3}\plus{}\frac{1}{(2\minus{}c)^3}\minus{}\frac{2}{(2\minus{}a)(2\minus{}b)(2\minus{}c)}\geq 1$

[quote="natural"]Let a,b,c such that $ a,b,c < 2$ and $ abc \equal{} 1$.Prove that:
$ \frac {1}{(2 \minus{} a)^3} \plus{} \frac {1}{(2 \minus{} b)^3} \plus{} \frac {1}{(2 \minus{} c)^3} \minus{} \frac {2}{(2 \minus{} a)(2 \minus{} b)(2 \minus{} c)}\geq 1$[/quote]

Denote $ x \equal{} \frac {1}{{2 \minus{} a}},y \equal{} \frac {1}{{2 \minus{} b}},z \equal{} \frac {1}{{2 \minus{} c}}$ then the inequality becomes
\[ x^3 \plus{} y^3 \plus{} z^3 \minus{} 2xyz \ge 1
\]
But we always have
\[ x^3 \plus{} y^3 \plus{} z^3 \minus{} 2xyz \ge xyz
\]
Therefor it's enough if we can show that $ xyz \ge 1$. But by AM-GM we have
\[ xyz \equal{} \frac {1}{{\left( {2 \minus{} a} \right)\left( {2 \minus{} b} \right)\left( {2 \minus{} c} \right)}} \ge \frac {1}{{\frac {{\left( {2 \minus{} a \plus{} 2 \minus{} b \plus{} 2 \minus{} c} \right)^3 }}{{27}}}} \equal{} \frac {{27}}{{\left( {6 \minus{} a \minus{} b \minus{} c} \right)^3 }} \ge 1
\]
So the inequality is proved.
----------------------
Edit. Sorry my above solution is only true if $ a \plus{} b \plus{} c \ge 3$. But if $ a \plus{} b \plus{} c \le 0$ try your inequality with $ a \equal{} b \minus{} \frac{3}{2},c \equal{} \frac{1}{{a^2 }}$.

[quote="Honey_S"]
\[ x^3 \plus{} y^3 \plus{} z^3 \minus{} 2xyz \ge xyz
\]
[/quote]
Try $ x \equal{} \minus{}100$, $ y \equal{} \minus{}100$, $ z \equal{} \frac{1}{10^4}$

[quote="Honey_S"][quote="Mathias_DK"][quote="Honey_S"]
\[ x^3 \plus{} y^3 \plus{} z^3 \minus{} 2xyz \ge xyz
\]
[/quote]
Try $ x \equal{} \minus{} 100$, $ y \equal{} \minus{} 100$, $ z \equal{} \frac {1}{10^4}$[/quote]
Are you sure?  :wink:[/quote]
$ x^3 \equal{} \minus{} 10^6$ $ y^3 \equal{} \minus{} 10^6$ $ z^3 \equal{} 10^{ \minus{} 12}$ and $ xyz \equal{} 1$.
So $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 2xyz \equal{} \minus{} 2*10^6 \plus{} 10^{ \minus{} 12} \minus{} 2 < 1$. Do you see it now?

[quote="karis"]$ a;b;c < 2\Rightarrow x;y;z > 0$[/quote]
What is $ x,y,z$?? And btw. $ a,b,c < 2$ does not imply $ a,b,c > 0$

[quote="Mathias_DK"][/quote][quote="Honey_S"][quote="Mathias_DK"][quote="Honey_S"]
\[ x^3 + y^3 + z^3 - 2xyz \ge xyz
\]
[/quote]
Try $ x = - 100$, $ y = - 100$, $ z = \frac {1}{10^4}$[/quote]
Are you sure?  :wink:[/quote]
$ x^3 = - 10^6$ $ y^3 = - 10^6$ $ z^3 = 10^{ - 12}$ and $ xyz = 1$.
So $ x^3 + y^3 + z^3 - 2xyz = - 2*10^6 + 10^{ - 12} - 2 < 1$. Do you see it now?

[quote="karis"]$ a;b;c < 2\Rightarrow x;y;z > 0$[/quote][quote="Mathias_DK"]
What is $ x,y,z$?? And btw. $ a,b,c < 2$ does not imply $ a,b,c > 0$[/quote]

Don't you see this?

[quote="Honey_S"]
Denote $ x = \frac {1}{{2 - a}},y = \frac {1}{{2 - b}},z = \frac {1}{{2 - c}}$ then the inequality becomes
[/quote]

And this

[quote="karis"]$ a;b;c < 2\Rightarrow x;y;z > 0$[/quote]


calculus
integration
function
trigonometry
inequalities
real analysis
real analysis unsolved

Let $ f : [a,b]\to \mathbb{R}$ be a continuous function such that $ 0\leqslant f(x) \leqslant M$, and $ f$ is not the $ 0$ function. Show that \[ 0< \left( \int_a^b f(t)\ dt \right)^2 \minus{} \left( \int_a^b f(t)\cos t\ dt \right)^2 \minus{} \left( \int_a^b f(t)\sin t\ dt \right)^2 \leqslant \frac{M^2(b\minus{}a)^4}{12}\]

The function in the middle can be rewritten as
\[ \int_a^b\int_a^b f(x)f(y)(1 \minus{} \cos(x \minus{} y))dxdy\]
So the left inequality follows. because $ \int \int 1 \minus{} cos(x \minus{} y) \equal{} (b \minus{} a)^2 \minus{} 2 \plus{} 2\cos(a \minus{} b)$
the right inequality follows from $ 1 \minus{} \frac {x^2}{2} \plus{} \frac {x^4}{24}\geq \cos x$


Let $ a, b, c$ be non-negative numbers, no two of then are zero. Then

$ \frac{a^2}{a^2 \plus{} a^2b^2 \plus{} b^2} \plus{} \frac{a^2}{a^2 \plus{} a^2c^2 \plus{} c^2} \plus{} \frac{b^2}{b^2 \plus{} b^2c^2 \plus{} c^2} \ge 1$.

[quote="lipako"]Let $ a, b, c$ be non-negative numbers, no two of then are zero. Then

$ \frac {a^2}{a^2 \plus{} a^2b^2 \plus{} b^2} \plus{} \frac {a^2}{a^2 \plus{} a^2c^2 \plus{} c^2} \plus{} \frac {b^2}{b^2 \plus{} b^2c^2 \plus{} c^2} \ge 1$.[/quote]
$ a\equal{}9;b\equal{}99999;c\equal{}2$
;)


[quote="ashwinrk_jain"]the question is not clear... can you please explain the question.[/quote]  we have to prove that: $ p^3$ divide $ a{}^2{}^P + b{}^2{}^P$

[quote="ashwinrk_jain"]but my friend the thing which we have to prove is same that is given..... please check it...[/quote] i've made one change now it's ok:[b]p[/b] it's an odd, prime number from the set $ \mathbb{N}$ . we have the numbers [b]a[/b] and [b]b[/b] from the set $ \mathbb{Z}$, thus $ a^p^ + ^1 + b^p^ + ^1$. Knowing that $ p^2$ devide $ a{}^2{}^p + b{}^2{}^p$ than we have to  prove that $ p^3$ divide $ a{}^2{}^P + b{}^2{}^P$.


ratio
algebra
polynomial
function
geometric sequence
algebra unsolved

I have a question about recurrency. We have to find a sequence satisfying:
$a_0=a$
$a_1=b$
$a_{n+2}=Aa_{n+1}+Ba_n$

I've found out, that we can find formula of this sequence by using Euler's Method, but I'm not sure if I called it correctly.
Please explain to me how we do it if you know something about that.

Do you know about the characteristic equation etc?

We look for  solutions in the form $\alpha^n$
$\alpha^2-A\alpha-B=0$ this is a characteristic equation.
I want to know is it called Euler's Method, I've tried to look in Google, but there are many things called Euler's Method.
I just want to find out more about this

And I have an additional question:
Can we find [b]all[/b] sequences satisfying aforementioned condition by using characteristic equation or maybe we'll find one among group of sequences satisfying that?

I guess it's clear that only one such sequence can exist.

Hummm...

The proposed method to solve recurrences says us to look for sequences of the form $a_n=\lambda q^n$ which satisfy the recurrence, which leads us to the characterist equation. Actually, I really wanted to know why geometric progressions always works, and is a suitable guess for solutions...it's an old doubt of mine, the same of yours, and I've never given my mind enough time to think about the answer. But I would be glad to know. =]

For me it's not so obvious, that there is [b]only one[/b] solution.
By the way what is geometric progressions Aodeath?

Neither for me =]

A geometric progression is a sequence of the form

$a_n=bq^n$

where you obtain the next term of the sequence by multiplying the previous one by a constant term, named the ratio, in our case $q$. I remember I asked one friend of mine why do the geometric progressions worked as though, and he explained me something about the Jordan Canonic form; by that time I couldn't understand nothing. And when I learned to solve these recurrence relations [the techniques], I remember my teacher told us he didn't know how to explain why the geometric progressions worked well, which for me was kind of saying that it wasn't so obvious, since this question remains "unsolvable" for me. I hope someone could help! =]

[quote="Arne"][quote="dondigo"]For me it's not so obvious, that there is [b]only one[/b] solution.[/quote]

The first two terms are given (and fixed). Each term can be determined uniquely from the two previous terms (because of the recursion). Hence such a sequence must be unique.[/quote]

Yes, but there is another interpretation of the question: does the sequence have a unique expression as a sum of (polynomials multiplied by) exponentials?    That is, are such functions linearly independent.


I don't have the answer for this question:

$ \text{How many integers are solutions to the equation: }\\
\\
\left( x \minus{} 2 \right) ^{\left( 25 \minus{} x^2 \right)} \equal{} 1 ?$

Clearly, for integers $ x^y\equal{}1$, then either $ x\equal{}1$, $ y\equal{}0$, or $ x\equal{}\minus{}1$ and $ y\equiv 0\pmod 2$

So, $ x\equal{}3$

$ 25\equal{}x^2$ so $ x\equal{}\pm 5$

And since $ 25\minus{}1\equal{}24\equiv 0\pmod 2$, $ x\equal{}\minus{}1$ works.

Four integers.


Daniel writes over a board, from top to down, a list of positive integer numbers less or equal to 10. Next to each number of Daniel's list, Martin writes the number of times exists this number into the Daniel's list making a list with the same length. If we read the Martin's list from down to top, we get the same
list of numbers that Daniel wrote from top to down. Find the greatest length of the Daniel's list can have.

The answer is 39. we first show the list must be similar, then we prove that there must be only one odd number, and because we're searching for the largest number, we have that the numbers are <= 10 so we consider that the odd number is 9 wich lead us to the sum: 2+4+6+8+10+9 = 39
1)suposse daniel's list's first number is x and martin's is x+y, wich means there are x+y x+y's in martin's lists and it means that in daniel's list's also according to the fact that they must have the same terms, but daniel's last must be x+y and martin's x, this means there are x x+y wich lead us to a contradiction.
2) because the list is simetrical, there must be only one odd number wich may be put in the center of the list and between separating different pairs also. we use 9 so we can reach the largest amount of numbers as possible.
3) the arrangement is like this:
2 9 4 4 9 6 6 6 9 8 8 8 8 9 10 10 10 10 10 9 10 10 10 10 10 9 8 8 8 8 9 6 6 6 9 4 4 9 2

@rafaelguedez

Your Proof of part 1) wasnt very convincing. to show you why, let me give the following argument -

im rewriting the case which you have taken -

let daniels list = $D$
Martins list = $M$ let $D_i$ and $M_i$ reresent the ith element of each list

suppose daniel's list's first number is $D_1 = x$ and martin's is $M_1 = x+y$, wich means there are $x+y$  $x$'s. but for each $D_i = x$ we know that $M_i = x+y$ hence there must be [u][b]AT LEAST[/b][/u] $(x+y)$ number of $i$ for which $M_i = x+y$
Also, since i have assumed that they are not equal hence $y \neq0$

let $D_n$ and $M_n$ be the last elements of respective lists. from condition of equaity of reverse of $M$ to $D$ we have

$M_n = D_1 = x$ and $D_n = M_1 = x+y$

$\implies$ number of $(x+y)'s$ in Daniels list $=x$
since lists are indentical if one is inverted then number of each element in each list must be equal
hence, number of $(x+y)'s$ in Martins list $=x$

thus from our previous condition we have
$x > x+y \implies y < 0$  you havent really shown that this is impossible

ill post the rectification a little later. if you are not convinced by my argument, feel free to discuss. i always love a little discussion.ill let u try till then

My argument goes as follows:

If some integer $k$ is on David's list, then it is also on Martin's list. This means that another (or maybe the same one) integer $k'$ occurs $k$ times on David's list.
So then, the maximum is the sum of the distinct integers on the list. i.e. $1+2+3+...+10=55$. And this is the answer.
Note that a simple one of these list is David's list:
$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ... ,10$
With Martin's being:
$10, 10, 10, 10, 10, 10, 10, 10, 10, 10,..., 1$
Where each $k$ occurs $k$ times.


Let $a, b, c\in \mathbb R^{+}$. Prove that $\frac{a+b+c}{\sqrt[3]{abc}}-1\le \frac{max(a,b,c)}{min(a,b,c)}+\frac{min(a,b,c)}{max(a,b,c)}$

someone..........?
the problem isn't that hard but you may think its quite hard at the first glimpse.

Firs let's prove the following inequality.

If $a,b,c>0$=>
\[3\frac{a+b+c}{\sqrt[3]{abc}}\leq (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3\]

Let's assume that $abc=1$

then
\[3\frac{a+b+c}{\sqrt[3]{abc}}\leq (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3\]
<=>
\[3(a+b+c)\leq (a^{2}b+a^{2}c+1)+(b^{2}a+b^{2}c+1)+(c^{2}a+c^{2}b+1)\]

From AM-GM we have that $a^{2}b+a^{2}c+1\geq 3\sqrt[3]{a^{4}bc}=3a$=>
\[(a^{2}b+a^{2}c+1)+(b^{2}a+b^{2}c+1)+(c^{2}a+c^{2}b+1)\geq 3a+3b+3c=3(a+b+c)\]

Now to complete the solution you only need to notice that when $a\geq b\geq c$=>$\frac{a}{c}+\frac{c}{a}\geq \frac{b}{c}+\frac{c}{b}$ and $\frac{a}{c}+\frac{c}{a}\geq \frac{a}{b}+\frac{b}{a}$
=>
\[3\frac{a+b+c}{\sqrt[3]{abc}}\leq (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3\leq 3(\frac{max(a,b,c)}{min(a,b,c)}+\frac{min(a,b,c)}{max(a,b,c)}+1)\]=>

\[\frac{a+b+c}{\sqrt[3]{abc}}\leq \frac{max(a,b,c)}{min(a,b,c)}+\frac{min(a,b,c)}{max(a,b,c)}+1\] :wink:

[quote="Tiks"]Firs let's prove the following inequality.

If $a,b,c>0$=>
\[3\frac{a+b+c}{\sqrt[3]{abc}}\leq (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3 \]
Let's assume that $abc=1$

then
\[3\frac{a+b+c}{\sqrt[3]{abc}}\leq (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3 \]
<=>
\[3(a+b+c)\leq (a^{2}b+a^{2}c+1)+(b^{2}a+b^{2}c+1)+(c^{2}a+c^{2}b+1) \]
From AM-GM we have that $a^{2}b+a^{2}c+1\geq 3\sqrt[3]{a^{4}bc}=3a$=>
\[(a^{2}b+a^{2}c+1)+(b^{2}a+b^{2}c+1)+(c^{2}a+c^{2}b+1)\geq 3a+3b+3c=3(a+b+c) \]
Now to complete the solution you only need to notice that when $a\geq b\geq c$=>$\frac{a}{c}+\frac{c}{a}\geq \frac{b}{c}+\frac{c}{b}$ and $\frac{a}{c}+\frac{c}{a}\geq \frac{a}{b}+\frac{b}{a}$
=>
\[3\frac{a+b+c}{\sqrt[3]{abc}}\leq (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3\leq 3(\frac{max(a,b,c)}{min(a,b,c)}+\frac{min(a,b,c)}{max(a,b,c)}+1) \]
=>
\[\frac{a+b+c}{\sqrt[3]{abc}}\leq \frac{max(a,b,c)}{min(a,b,c)}+\frac{min(a,b,c)}{max(a,b,c)}+1 \]
:wink:[/quote]

i have another solution.
whats the meaning of wlog please?

[quote="nayel"]i have another solution.
whats the meaning of wlog please?[/quote]

Aren't you gonna post it?

[quote="Tiks"]

Aren't you gonna post it?[/quote]

okey here's my solution:

first we prove a lemma.

lemma: Nayel's Inequality: for $a,b,c,x,y,z\in \mathbb R^{+}$ and $a\ge b\ge c$ and $x\le y\le z$ we have
\[\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\ge \frac{a+b+c}{\sqrt[3]{xyz}}\]
set $p=\frac{1}{x},q=\frac{1}{y},r=\frac{1}{z}$. we have to prove that
\[ap+bq+cr\ge (a+b+c)(\sqrt[3]{pqr}) \]
for $a\ge b\ge c$ and $p\ge q\ge r$. which follows from Tchebyshef and AM-GM. so the lemma is proved.

now for the inequality assume that $a\ge b\ge c$ and set $x=c,y=b,z=a$ and by the lemma we are done.


There are many types of spirals. There's an equiangular spiral, an Archimedian spiral, a logarithmic spiral (or is that the same as an equiangular one?) and probably many others?

Yes i'm sure most people here have messed around with the polar graphing capabilities of a calculator, with my ti-89, ive done sprials, squares, stars, and even a nice looking butterfly once, but it gets old very quickly :)

Try this equation:

r = e^cos(:theta:) - 2cos(4:theta:) + sin^5(:theta:/12)

note: I forget what the interval required is, and i'm currently too lazy (and hungry) to test it, try 0 = 2pi and im sure it will work.

Speaking of being hungry, try graphing this one:

r=tan(1+:theta:/8).

[quote="TheSouthpaw"]I was just screwing around with my TI - 85. and I thought this was amazing. Well, not amazing, just pretty darn cool.

Anybody know the general equation for a spiral?  :D[/quote]

ive never heard of a ti-85 before. did you mean 86? or am i just ignant (i know i know, ignant is not a word, but it will be one)

the ti-85 is an older version calculator.

ah, cool beans. but when did it come out? like before the 82?

don't quote me on this, i'm too lazy to look it up... but I think the 79 came out, then the 82, then the 85, then the 92.. years are before my time, couldn't tell ya

lol. do you know of any new calculators coming out from ti?

the 85 is just the older, classic version of the 86, just like the 82 is the predecessor of the 83, which came before 83+ and 83+. similarly, the 92 came before the 89 and 92+. im not sure when the 79 came in, but ive seen it before. yes there is a new calc coming out, though i have no details. go to tis website.

Texas Instruments website: [url=http://www.ti.com]http://www.ti.com[/url]

I have a TI-55.  I'm sure it predates all those others.  It has a display a lot like a LCD clock (except smaller and it flickers) and it plugs into the wall.  There was also a battery pack, but the batteries are obviously dead.  My dad bought the calculator 20 years ago or something insane like that.

WOW. :shock: I wonder if there's TI-15 :)

Lower numbers don't always mean older or worse, they sometimes jump around...

I have a ti-36x Solor, it's a very nice scientific calculator that I use when a graphing one isnt allowed.

Oh, yeah. Lower numbers are for scientific/normal calculators...
Isn't the 30s scientific calculators?

Interesting Fact: Technically it's "30s's" which is the only word (Well, numbers) which can have "s's"


July 4, 1903, was a Thursday.  On what day of the week was July 4, 1904?

Saturday, b/c you add 365+1 (you went past feb 29 on the leap year of 1904) so you get 366, which yields a remainder of 2 when divided by seven (i just know that there are 52 and 1/7 weeks in a year) and so you add one. Two days from Thursday is [b]Saturday[/b], the answer.

[hide]fisrt see if 1904 is a leap year...it is(1904/4 has remainder of 0)!

so there are 396 days

366/7=__remainde 2

2 days afer thursday is Saturday[/hide]


inequalities
function
logarithms
inequalities proposed

If $ a,b,c$ are nonnegative real numbers, then

(a) $ 4(a^3\plus{}b^3\plus{}c^3)\plus{}3abc \le \sum a\sqrt{16a^4\plus{}9b^2c^2}$;

(b) $ 5(a^3\plus{}b^3\plus{}c^3)\plus{}24abc \ge \sum a\sqrt{25a^4\plus{}144b^2c^2}$.

[quote="auj"]M.M., I'm new here.
Sorry for bothering you with a question probably very plain for you and many other users of this forum.

But what does  [i]RCF-Corollary[/i]  stand for?

Please give an explanation or post a link.
Thanks![/quote]

  [b]Right Convex Function Corollary[/b](RCF-Corollary).

 ,,Let f be a function defined on $ (0,\infty)$, and let $ r > 0$.  If the function $ f_1(u) \equal{} f(e^u)$ is convex for $ u\geq\ln r$,and
\[ f(a_1) \plus{} f(a_2) \plus{} ... \plus{} f(a_n)\geq nf(\sqrt [n]{a_1a_2...a_n}) (*)
\]
for all $ a_1,a_2,....a_n > 0$ such that
\[ \sqrt [n]{a_1a_2...a_n} \equal{} r, and , a_2 \equal{} a_3 \equal{} .... \equal{} a_n\geq r,
\]
then (*) holds for all $ a_1,a_2,...a_n > 0$  such that $ \sqrt [n]{a_1a_2...a_n}\geq r$''

[quote="Marius Mainea"]a)  Presume $ abc\neq 0$
Let
\[ f(t) \equal{} \sqrt {16t^2 \plus{} 9} \minus{} 4t ,
\]

\[ t > 0
\]
Inequality is equivalent with
\[ f(x) \plus{} f(y) \plus{} f(z)\geq 3,
\]

\[ xyz \equal{} 1
\]
But $ f_1(u) \equal{} f(e^u)$ is convex for $ u\geq\ln1$

 According to RCF-Corollary , we need to show that
\[ f(x) \plus{} 2f(y)\geq 3f(1)
\]
for $ 0 < x\leq 1\leq y$ and $ xy^2 \equal{} 1$

 This inequality is equivalent with
\[ \sqrt {16x^2 \plus{} 9} \plus{} 2\sqrt {\frac {16}{x} \plus{} 9} \minus{} 4(x \plus{} \frac {2}{\sqrt {x}})\geq 1.
\]

\[ 0 < x\leq1
\]
wich is true after squaring.

Cosequently, apply RCF-Corollary for r=1 , inequality is prove.

 b) Similarly.[/quote]
You are right.  :lol:
In the second part of the proof, we can also prove the original inequality for $ b\equal{}c\equal{}1$.


function
Ring Theory
topology
superior algebra
superior algebra unsolved

suppose R=C[0,1]={all of the continuous function from [0,1] to $ %Error. "display" is a bad command.
\Re$ },show every maximal ideal in R is in this form:$ %Error. "display" is a bad command.
\ M_r$={f$ %Error. "display" is a bad command.
\in$ R$ %Error. "display" is a bad command.
\mid$           f(r)=0} .r is a constant number and r$ %Error. "display" is a bad command.
\in$ [0,1]

First, we prove that the $ M_r$ are maximal ideals. Clearly, they form proper ideals and for any $ u\not\in M_r$ we have $ u(r)\not= 0$. Let $ u'\in R$ be the constant function equal to $ \frac{1}{u(r)}$, then $ 0=u(r)u'(r)-1(r)\Longrightarrow uu'\in 1+M_r\Longrightarrow 1\in M_r(u)$ and hence, $ M_r$ must be a maximal ideal, since $ u$ was arbitrary. Conversely, let a maximal ideal $ M$ be given. If $ M=M_r$ for some $ r$, we are finished. Therefore, we can assume for all $ r\in [0,1]$ the existence of some $ m_r\in M$ with $ m_r(r)\not=0$. Since $ m_r$ is a continuous function, there exist an open set $ r\in U(r)$ such that $ m_r$ has no zero in $ U(r)$. Since $ [0,1]$ is compact, we can determine a finite cover $ \bigcup_{i=1}^nU(r_i)$ of $ [0,1]$. Then $ m=\sum_{j=1}^nm_{r_j}^2\in M$ and $ m$ is strictly positive on $ [0,1]$, since the $ m_{r_j}^2$ are all non-negative and $ m_{r_i}^2$ is strictly positive on $ U(r_i)$ for $ i=1,\ldots,n$. Hence, ${ \frac{1}{m}\in R}$ and $ 1=m\cdot\frac{1}{m}\in M$, a contradiction.


Prove that $ \forall\,x,y,z\in\mathbb R^\plus{},$
\[ (x^2\plus{}y^2\plus{}z^2)^2\ \geqslant\ x^3y\plus{}y^3z\plus{}z^3x\plus{}2(xy^3\plus{}yz^3\plus{}zx^3)\]

That's a weaker version of the so called "Vasc's Inequality".

[quote="itiselizabeth"]Prove that $ \forall\,x,y,z\in\mathbb R^ \plus{} ,$
\[ (x^2 \plus{} y^2 \plus{} z^2)^2\ \geqslant\ x^3y \plus{} y^3z \plus{} z^3x \plus{} 2(xy^3 \plus{} yz^3 \plus{} zx^3)\]
[/quote]

without loss of generality, we may assume that $ x\ge y\ge z$ or $ x\le y\le z$.
if $ x\ge y\ge z$,
$ (x^2 \plus{} y^2 \plus{} z^2)^2 \ge x^3y \plus{} y^3z \plus{} z^3x \plus{} 2(xy^3 \plus{} yz^3 \plus{} zx^3)$
$ \Longleftrightarrow (x^2  \minus{} xy \plus{} y^2)(x \minus{} y)^2 \plus{} (2x(x\minus{}y)\plus{}y(y\minus{}z) \plus{} z^2)(x\minus{}z)(y\minus{}z) \ge 0$
obvious ture;
if $ x\le y\le z$,
$ (x^2 \plus{} y^2 \plus{} z^2)^2 \ge x^3y \plus{} y^3z \plus{} z^3x \plus{} 2(xy^3 \plus{} yz^3 \plus{} zx^3)$
$ \Longleftrightarrow (x^2  \minus{} xy \plus{} y^2)(x \minus{} y)^2 \plus{} (x^2\plus{}(x\minus{}y)^2\plus{}z(z\minus{}y))(z\minus{}x)(z\minus{}y) \ge 0$
obvious ture.
so...:o


Prove that sufficient condition for $n$ and $n+2$ to be simultanously primes is:

$4((n-1)!+1)\equiv -n \mod n(n+2)$

i.e. if above congruence is true, $n$ and $n+2$ are primes

By Wilson's Theorem, $(n-1)! \equiv -1 \bmod n$ if and only if $n$ is prime.

Suppose that $4((n-1)!+1)\equiv -n \bmod n(n+2)$ and that $n$ is odd.  Then $4((n-1)!+1)\equiv 0 \bmod n$ implies that $(n-1)! \equiv -1 \bmod n$ and so $n$ is prime.  Further, $(n+1)! \equiv (n+1)n.(n-1)! \equiv (n+1)n (-n/4) \bmod (n+2)$.  But $n^3 + n^2 \equiv -4 \bmod (n+2)$, so $(n+1)! \equiv -1 \bmod (n+2)$ and so $n+2$ is prime.

If now $4((n-1)!+1)\equiv -n \mod n(n+2)$ and $n=2k$ is even then $2k$ divides $4((2k-1)!+1)$, which implies $k$ divides 4.  A simple check shows that in fact the condition never holds for even $n$.


Can someone explain to me about hybrid orbitals?
How do we know that an element (e.g. C) will form sp3, sp2, or sp orbitals?
Thx

In the first place, you have to know the number of atoms to which your atom is bonded to. For example, carbon 1 in 1-butene is bonded to three atoms, so we will explain bonding in that carbon by considering an hibridization scheme that will produce three orbitals: in this case, sp2. Do you follow?

Carcul

Oh I see.. Now I understand..
So the C in methane will produce 4 sp3 orbitals.
Then Al in AlCl3 will produce 3 sp2 orbitals.
O in water will produce 2 sp orbitals.
Am I right?

Thx.

[quote="boonsp"]O in water will produce 2 sp orbitals.[/quote]

This is incorrect because one also has to know how many nonbonding pairs of electrons there are.  The O in water is bonded to two atoms, but it also has two unshared pairs of electrons.  Thus, that O forms 4 $\text{sp}^{3}$ orbitals.

Oh.. So it depends on the number of regions of electon density, not on number of atoms bonded, doesnt it?

Isnt it the same with predicting molecular shape?

It depends both on number of atoms bonded and on number of unshaired pairs.


LaTeX
algebra
polynomial
calculus
integration
national olympiad

please some one help me in typing in latex for this paper

Problem 2:
Let $ a,b,c$ and  $ d$ be any four real numbers, not all equal to zero. Prove that the roots of the polynomial $ f(x)\equal{}x^{6} \plus{} ax^{3} \plus{} bx^{2}\plus{} cx \plus{} d$ can't all be real.

[quote="makar"]Problem 1
Prove that the Polynomial $ f(x) \equal{} x^{4} \plus{} 26x^{3} \plus{} 56x^{2} \plus{} 78x \plus{} 1989$ can't be expressed as a product $ f(x) \equal{} p(x)q(x)$ , where $ p(x)$ and $ q(x)$ are both polynomial with integral coefficients and with degree at least 1[/quote]

I'm not sure, but...

consider $ f(x \plus{} 1) \equal{} x^{4} \plus{} 30x^{3} \plus{} 140x^{2} \plus{} 272x \plus{} 2150$, which is irreducible by Eisenstein's Irreducibility Criterion. :thumbup:

Problem 3:
Let $ A$ denote a subset of the set $ \{ 1,11,21,31, \dots ,541,551 \}$ having the property that no two elements of $ A$ add up to 552. Prove that $ A$ can't have more than 28 elements.

Problem 4:
Determine with proof, all the positive integers $ n$ for which:
[b](a)[/b]  $ n$ is not the square of any integer ; and
[b](b)[/b]  $ \left[\sqrt{n} \right ]$ divides $ n^{2}$.
( Notation : $ [x]$ denotes the largest integer that is less than or equal to $ x$ )

Problem 5:
For positive integers $ n$, define $ A(n)$ to be $ \frac{(2n)!}{(n!)^{2}}$.
Determine the sets of positive integers $ n$ for which
[b](a)[/b] $ A(n)$ is an even number,
[b](b)[/b] $ A(n)$ is a multiple of 4.

Post each problem in it's own thread. Just the problem statement in the body text.


You are printing from your computer onto paper that measures $ 8\frac12$'' by 11''. Each margin on the printed page is 1''. How many inches are in the perimeter of the rectangular region available for printing?

I have no idea what this means, I guess $ 2(\frac{13}{2}\plus{}9)\equal{}31$.

same here i have no idea wut they r askin.

If you have a piece of paper, the printer will not print out until the edges. It leaves a little blank area called the margin. If the margin was 1 inch, how much could we use? Make a sketch and then draw the margins.

O, ok. I misunderstood the question. Thanks.

Wouldn't it be 35 though?

[hide=(How?)]
So you could do it like $8.5*11-6.5*9=\boxed{35}$ or $2(8.5+11)+4=\boxed{35}$ Where we add $4$ for the corners.[/hide]
Edit: Or maybe the question is weird and you don't add $4$ for the corners...

@above The problem is asking for perimeter not area.


geometry
3D geometry
modular arithmetic
Diophantine Equations
pen

Does there exist an integer such that its cube is equal to $3n^2 +3n+7$, where $n$ is integer?

Consider $ 3n^2 \plus{} 3n \plus{} 7$ $ \pmod 9$. If $ n\equiv 0\pmod 3$ or $ n\equiv 2\pmod 3$ then $ n^2 \plus{} n \equal{} n(n \plus{} 1)\equiv 0\pmod 3$; if $ n\equiv 1 \pmod 3$ then $ n^2 \plus{} n\equiv 2\pmod 3$. Therefore, $ n^2 \plus{} n\equiv 0$ or $ 2 \pmod 9$ hence $ 3n^3 \plus{} 3n\equiv 0$ or $ 6\pmod 9$, so $ 3n^2 \plus{} 3n \plus{} 7\equiv 7$ or $ 4\pmod 9$. However, because any cube of an integer is congruent to $ \minus{} 1,0,$ or $ 1$ $ \pmod 9$ then it is impossible for a cube of an ineteger to be congruent to $ 4$ or $ 7\pmod 9$, so no cube of an integer can be equal to $ 3n^3 \plus{} 3n \plus{} 7$ where $ n$ is an integer, QED.

@rem

$ n^2 + n\equiv 0$ or $ 2 \pmod 9$

This conclusion is wrong , such as $ n \equiv 2 \pmod{9} $   then $ n^2 + n\equiv 6 \pmod{9}$

The elliptic curve $y^2 + 9y = x^3 - 189$ has Mordell-Weil group $E(\mathbf{Q}) = 0$ (no non-trivial torsion and trivial $2$-Selmer group).

[hide=Solution:]
Taking $\pmod{9}$, we have,
$$3n^2+3n+7 \equiv 4,7 \pmod{9}$$
However,
$$x^3 \equiv 0,1, 8 \pmod{9}$$
Hence, there exit no such integer.


Let $ a_0,a_1,..,a_n$ be integers not all zero and all at least $ \minus{}1$.Given $ a_0\plus{}2a_1\plus{}....\plus{}2^na_n\equal{}0$.Prove that: $ a_0\plus{}a_1\plus{}..\plus{}a_n>0$

[quote="math10"]Let $ a_0,a_1,..,a_n$ be integers not all zero and all at least $ \minus{} 1$.Given $ a_0 \plus{} 2a_1 \plus{} .... \plus{} 2^na_n \equal{} 0$.Prove that: $ a_0 \plus{} a_1 \plus{} .. \plus{} a_n > 0$[/quote]

For $ a_i\equal{}2k_i\plus{}1$, where $ k_i$ is integer ,  we have

$ 2^i(2k_i\plus{}1)\plus{}2^{i\plus{}1}a_{i\plus{}1}\equal{}2^i(1)\plus{}2^{i\plus{}1}(a_{i\plus{}1}\plus{}k_i)\equal{}0$

And $ (2k_i\plus{}1)\plus{}a_{i\plus{}1}>1\plus{}(a_{i\plus{}1}\plus{}k_i)$ for $ k_i \ge 1$

For $ a_i\equal{}2k_i$ , where $ k_i$ is integer , we have

$ 2^i(2k_i)\plus{}2^{i\plus{}1}a_{i\plus{}1}\equal{}2^i(0)\plus{}2^{i\plus{}1}(a_{i\plus{}1}\plus{}k_i)$

And $ 2k_i\plus{}a_{i\plus{}1}>a_{i\plus{}1}\plus{}k_i$ for $ k_i \ge 1$

So we only need to prove the case that $ a_i \in \{\minus{}1,0,1 \}$, where $ a_i$ may all be zero.
[b]Proof:[/b]
For $ n \ge 1$, then we have$ |2^n|>|a_0\plus{}2a_1\plus{}...\plus{}2^{n\minus{}1}a_{n\minus{}1}| \Rightarrow a_n\equal{}0 \Rightarrow a_i\equal{}0$ for all $ i$

So $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}0$ for this case.

$ \Rightarrow$ in the original case ,  $ a_1\plus{}a_2\plus{}...\plus{}a_n \ge 0$,

equality holds $ \Leftrightarrow$ no modification needed , i.e. $ k_i \le 0$ for all $ i$
$ \Leftrightarrow a_i \in \{\minus{}1,0,1 \}$ for all $ i$
$ \Leftrightarrow a_i\equal{}0$ for all $ i$, impossible as it contradicts the condition that $ a_i$ are not all zero in the original question.

Done.

[quote="stephencheng"][quote="math10"]Let $ a_0,a_1,..,a_n$ be integers not all zero and all at least $ \minus{} 1$.Given $ a_0 \plus{} 2a_1 \plus{} .... \plus{} 2^na_n \equal{} 0$.Prove that: $ a_0 \plus{} a_1 \plus{} .. \plus{} a_n > 0$[/quote]

For $ a_i \equal{} 2k_i \plus{} 1$, where $ k_i$ is integer ,  we have

$ 2^i(2k_i \plus{} 1) \plus{} 2^{i \plus{} 1}a_{i \plus{} 1} \equal{} 2^i(1) \plus{} 2^{i \plus{} 1}(a_{i \plus{} 1} \plus{} k_i) \equal{} 0$

And $ (2k_i \plus{} 1) \plus{} a_{i \plus{} 1} > 1 \plus{} (a_{i \plus{} 1} \plus{} k_i)$ for $ k_i \ge 1$

For $ a_i \equal{} 2k_i$ , where $ k_i$ is integer , we have

$ 2^i(2k_i) \plus{} 2^{i \plus{} 1}a_{i \plus{} 1} \equal{} 2^i(0) \plus{} 2^{i \plus{} 1}(a_{i \plus{} 1} \plus{} k_i)$

And $ 2k_i \plus{} a_{i \plus{} 1} > a_{i \plus{} 1} \plus{} k_i$ for $ k_i \ge 1$

So we only need to prove the case that $ a_i \in \{ \minus{} 1,0,1 \}$, where $ a_i$ may all be zero.
[b]Proof:[/b]
For $ n \ge 1$, then we have$ |2^n| > |a_0 \plus{} 2a_1 \plus{} ... \plus{} 2^{n \minus{} 1}a_{n \minus{} 1}| \Rightarrow a_n \equal{} 0 \Rightarrow a_i \equal{} 0$ for all $ i$

So $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 0$ for this case.

$ \Rightarrow$ in the original case ,  $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \ge 0$,

equality holds $ \Leftrightarrow$ no modification needed , i.e. $ k_i \le 0$ for all $ i$
$ \Leftrightarrow a_i \in \{ \minus{} 1,0,1 \}$ for all $ i$
$ \Leftrightarrow a_i \equal{} 0$ for all $ i$, impossible as it contradicts the condition that $ a_i$ are not all zero in the original question.

Done.[/quote]
I can't understand your solution.ou can write clear solution??
My solution using induction(with condition $ a_0$ is even so $ a_0\equal{}0$ or $ a_0\plus{}2a_1\geq 0$)


Daca $1 \geq |{z}^{n}+1|$ pentru orice numar $n$ natural atunci $z=0$ .
BAFTA MAINE TUTUROR!!

Fie $z = r \left( \cos \theta+i \sin \theta \right)$. Rezulta ca $z^{n}= r^{n}\left( \cos n \theta+i \sin n \theta \right)$.

Acuma observam ca daca $\text{Re}\ w > 0$, atunci $|1+w| > 1$, deci $z^{n}$ nu are niciodata partea reala mai mare ca $0$.

Daca $r > 0$, asta inseamna ca $n \alpha \, \bmod \, 2 \pi$ nu e niciodata in $\left(-\pi, \pi \right)$. Se poate obtine o contradictie in felul urmator: dacă $\frac{\theta}{2 \pi}$ e raţional, atunci pentru un $m$ potrivit avem $\text{arg}\, z^{m}= 0$. Dacă nu, folosind teorema lui Kronecker putem găsi $k$ astfel incat $\text{arg}\, z^{k}$ sa fie oricât de aproape de $0$.

Mai rămâne doar cazul $z = 0$, care se vede că verifică inegalitatea, $\forall n \geq 1$.

[quote="posabogdan"]$\left(\forall\right) n\in\overline{1,3}\ ,\ \left|1+{z}^{n}\right|\le 1\Longleftrightarrow z=0$.[/quote]
[color=darkblue][b]Demonstratie.[/b] Presupunem prin absurd ca $r>0$, unde $z=r(\cos x+i\sin x)$.
Se arata usor ca $\left|1+z^{n}\right|\le 1$ $\Longleftrightarrow$ $\cos nx\le-\frac{r^{n}}{2}$, unde $n\in\overline{1,3}$. Asadar,
$\left\{\begin{array}{c}\cos x<0\\\ \cos 2x<0\end{array}\right\|\Longrightarrow \boxed{\cos x\cos 2x>0}$. Insa $\cos 3x=2\cos x\cos 2x-\cos x$.
Deci $\cos 3x<0\Longrightarrow$ $2\cos x\cos 2x<\cos x<0$ $\Longrightarrow$ $\boxed{\cos x\cos 2x <0}$.

Cele doua relatii impachetate conduc la absurd. In concluzie, $r=0$, adica $z=0$.

[b][u]Observatie[/u][/b] ([u]L. Panaitopol, etapa locala[/u], nu-mi amintesc anul).

$\boxed{z^{3}\ne 1\ ,\ |1+z|\le 1\ ,\ \left|1+z^{2}\right|\le 1\Longrightarrow |z|<1}$.

[u]Demonstratie.[/u] $2|z|=\left|\left(1+z\right)^{2}-\left(1+z^{2}\right)\right|\le \left|1+z\right|^{2}+\left|1+z^{2}\right|\le 2$ $\Longrightarrow$

$|z|\le 1$. Presupunem prin absurd ca $|z|=1$, adica $z\cdot\overline z=1$. Asadar,

$\left\{\begin{array}{c}|1+z|\le 1\\\ \left|1+z^{2}\right|\le 1\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}(1+z)(1+\overline z)\le 1\\\ \left(1+z^{2}\right)\left(1+\overline z^{2}\right)\le 1\end{array}\right\|$ $\Longrightarrow$

$\left\{\begin{array}{c}z+\overline z\le-1\\\ z^{2}+\overline z^{2}\le-1\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}z+\overline z\le-1\\\ \left(z+\overline z\right)^{2}\le 1\end{array}\right\|$ $\Longrightarrow$

$\left\{\begin{array}{c}z+\overline z\le-1\\\-1\le z+\overline z\le 1\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}z+\overline z=-1\\\ z\cdot\overline z=1\end{array}\right\|$ $\Longrightarrow$

$z^{2}+z+1=0$ $\Longrightarrow$  $z^{3}=1$, absurd. In concluzie, $|z|<1$.[/color]


Let $AD$ be the common chord of two circles $\Gamma_1$ and $\Gamma_2$.  A line through $D$ intersects $\Gamma_1$ at $B$ and $\Gamma_2$ at $C$.  Let $E$ be a point on the segment $AD$, different from $A$ and $D$.  The line $CE$ intersects $\Gamma_1$ at $P$ and $Q$.  The line $BE$ intersects $\Gamma_2$ at $M$ and $N$.

(i) Prove that $P, Q, M, N$ lie on the circumference of a circle $\Gamma_3$. (This is obvious; it's there because of the second part.)

(ii) If the centre of $\Gamma_3$ is $O$, prove that $OD$ is perpendicular to $BC$.

N.B. Out of the 4 questions which we were given, this is the only question worth posting.  If Valentin ever gets around to making an olympiads 2004 book, he's only going to have 3 problems inside, because I am way too ashamed of the 4th one to let anybody see it :blush:.

This one in my opinion is easy.
 It is sufficient to prove that $ OB^2-OC^2=DB^2-DC^2$.
  We have $ OB^2-OC^2 = BM.BN - CP.CQ=BD.BC-CD.CB=BD^2-CD^2$ :D

Unfortunately enough, the other 3 problems were easier :(


Hi.
Here is problem from CNMO 2005.
Some Mathlinkers solved it. But my solution is coming out little different. Where is my mistake?

Question:
Define sequence of natural numbers $ a_n$ as $ a_1  \equal{} 1,\,a_n  \equal{} a_1  \cdot a_2  \cdot ... \cdot a_{n \minus{} 1}  \plus{} 1$, for $ n \ge 2$. Find smallest $ M$ such that $ \sum\limits_{n \equal{} 1}^m {{1 \over {a_n }} < M\,,\,(m \in N)}$.

Comment:
Many of Mathlinkers claiming that $ M\equal{}2$, but my sol is giving me $ M\equal{}5/2$.

My solution:
See that $ {1 \over {a_n }} \equal{} {1 \over {a_1  \cdot a_2  \cdot ... \cdot a_{n \minus{} 1}  \plus{} 1}} < {1 \over {a_{n \minus{} 2}  \cdot a_{n \minus{} 1} }} < {{a_{n \minus{} 1}  \minus{} a_{n \minus{} 2} } \over {a_{n \minus{} 2}  \cdot a_{n \minus{} 1} }} \equal{} {1 \over {a_{n \minus{} 2} }} \minus{} {1 \over {a_{n \minus{} 1} }}$.
$ \sum\limits_{n \equal{} 1}^m {{1 \over {a_n }}}  \equal{} {1 \over {a_1 }} \plus{} {1 \over {a_2 }} \plus{} \left( {\sum\limits_{n \equal{} 3}^m {{1 \over {a_n }}} } \right) \equal{}{3 \over 2}\plus{} \left( {\left( {{1 \over {a_1 }} \minus{} {1 \over {a_2 }}} \right) \plus{} \left( {{1 \over {a_2 }} \minus{} {1 \over {a_3 }}} \right) \plus{} ... \plus{} \left( {{1 \over {a_{m \minus{} 2} }} \minus{} {1 \over {a_{m \minus{} 1} }}} \right)} \right)\equal{}{3\over 2} \plus{} {1 \over {a_1 }} \minus{} {1 \over {a_{m \minus{} 1} }} \equal{} {5 \over 2} \minus{} {1 \over {a_{m \minus{} 1} }} < {5 \over 2} \equal{} M.$

Please help where is the mistake???
 :blush:
p.s: Sorry for bad typing.... :oops:

I fear you should work it out by yourself where your mistake is. The correct answer is $ 2$, and comes from the classical way to express $ 1$ as sum of Egyptian fractions.

Using the identity $ \frac {1} {a} \equal{} \frac {1} {a\plus{}1} \plus{} \frac {1} {a(a\plus{}1)}$, notice that
$ \frac {1} {a_n} \equal{} \frac {1} {\prod_{i\equal{}1}^{n\minus{}1} a_i} \minus{} \frac {1} {\prod_{i\equal{}1}^n a_i}$.
Therefore, by telescoping, $ \sum_{i\equal{}1}^m \frac {1} {a_m} \equal{} 1 \plus{} 1 \minus{} \frac {1} {\prod_{i\equal{}1}^m a_i} < 2$, and $ 2$ is approached as closely as wanted when $ m \to \infty$.

Well thanx Mavropnevma, I think I find the mistake.
I think I did a mistake by inserting $ {1 \over {a_{n \minus{} 2}  \cdot a_{n \minus{} 1} }} < {{a_{n \minus{} 1}  \minus{} a_{n \minus{} 2} } \over {a_{n \minus{} 2}  \cdot a_{n \minus{} 1} }}$  which make my total sum bigger...

I think... :?:

Your mistake is that you only have proved that sum $ < \frac{5}{2}$ and assumed $ \frac{5}{2}$ is smallest $ m$ without proving it .


conics
ellipse
parabola
analytic geometry
algebra
polynomial

Fine the points on the ellipse $ 4x^2\plus{}y^2\equal{}4$ that are the farthest away from the point $ (1,0)$.

[hide="Hint"]It suffices to maximize the square of the distance[/hide]

[hide="Solution"]We have that $ D^2 \equal{} (1 \minus{} x)^2 \plus{} y^2 \equal{} (1 \minus{} x)^2 \plus{} (4 \minus{} 4x^2) \equal{} 5 \minus{} 2x \minus{} 3 x^2$

This is maximized at the vertex of the above parabola, when $ x \equal{} \frac { \minus{} ( \minus{} 2)}{2 ( \minus{} 3)} \equal{} \minus{} \frac {1}{3}$

Thus, the maximum distance is produced by $ \boxed{(x,y) \equal{} \left( \minus{} \frac {1}{3} , \; \pm \frac {4 \sqrt {2}}{3} \right)}$

At these points, we have $ D \equal{} \sqrt {5 \minus{} 2 \left( \minus{} \frac {1}{3} \right) \minus{} 3 \left( \minus{} \frac {1}{3} \right)^2} \equal{} \frac {4}{\sqrt {3}}$[/hide]

JBL, can you explain better what you're saying? If the distance between them is 0, they're the same point, obviously not true.

Right -- TZF showed that the square of the distance between (1, 0) and the point on the elipse with $ x$-coordinate $ a$ is $ 5 \minus{} 2a \minus{} 3a^2$.  When $ a \equal{} \minus{}5/3$, this gives a distance-squared of 0, and so a distance of zero.  But this is, as you note, obviously false -- so why doesn't this argument work?

because there is no point on the ellipse with an x coordinate of $ \minus{}5/3$.  The points on the ellipse must have $ x \in [\minus{}1,1]$ and $ y \in [\minus{}2,2]$

[quote="JBL"][quote="TZF"]We have that $ D^2 \equal{} (1 \minus{} x)^2 \plus{} y^2 \equal{} (1 \minus{} x)^2 \plus{} (4 \minus{} 4x^2) \equal{} 5 \minus{} 2x \minus{} 3 x^2$[/quote]  For extra credit, explain why this doesn't imply that the distance between the point (1, 0) and the point on the elipse with $ x$-coordinate -5/3 is 0.[/quote]

What would bring up the zero of the polynomial?


linear algebra
matrix
linear algebra unsolved

Let $ A$ and $ B$ be a matrices of size $ m \times n$ and $ k \times n$, respectively, such that if $ AX\equal{}0$ for a certain column $ X$, then $ BX\equal{}0$. Prove that $ B\equal{}CA$, where $ C$ is a matrix of size $ k \times m$.

Define two $ (m\plus{}k)\times n$ partitioned matrices, $ D_1, D_2$, as follows $ D_1\equal{}\left(\begin{array}{cc}A\\B\end{array}\right), D_2\equal{}\left(\begin{array}{cc}A\\0\end{array}\right)$. Clearly, $ \ker(D_1)\equal{}\ker(A)\equal{}\ker(D_2)$. Now, according to [url=http://www.mathlinks.ro/viewtopic.php?t=173786]this post[/url] of mine, $ D_1$ and $ D_2$ are row equivalent, and hence $ B\equal{}CA$ for some matrix $ C$.

Or we can proceed by letting $ W$ to be a row space of $ A$, by condition we have  row of $ B$ is in $ (W^{\bot})^{\bot}$ and thus in $ W$ since $ W$ is finite dimensional space. so $ A$ and $ B$ is row equivalent.


I have two and a half dozen coins that total 32 cents. What are the coins?

They are [hide]a quarter and seven pennies.  The sentence is slightly ambiguous; it can be read "(two and a half) dozen" or "two and (a half dozen)."  I think actually there might need to be punctuation in order to make the first interpretation operative, meaning that the second interpretation is "better" in some sense.[/hide]

[quote="Geehoon"][hide="in other case"]4 nickels and 12 pennies[/hide][/quote]

Whooooooaaaaaa... That would make 16 coins, which is not 2 and a half dozen no matter how you interpret it.

In response to jbl, I was intentionally ambiguous in the problem. I was thinking of 6 nickels and 2 pennies, but 1 quarter and 7 pennies are equally good.

Guess nobody got tripped up!!!  :lol: [/quote]


again,
More text for more learners.

In how many ways can I choose 131 items from a set of 131 items..

"think about it"

- if there is a bunch of stuff, how many ways can take all of it?

"simpler problem"

how many ways can you take 3 cookies from 3 cookies? just ONE! you take them all!

[hide="Solution"]Once again, in general, $ \binom{n}{n} \equal{} 1$.[/hide]

HIDE YOUR SOLUTIONS!!!
[hide="solution#26"]Any combination(n, n) is equal to 1.[/hide]

Yeah. Seriously, just think about it. More broadly:

For any combination $\dbinom{x}{x}$ or $\dbinom{x}{0}$, the answer is 1.


I created a table that looks like this:

[code]
Table 1

A B
1 9
1 8
1 7
2 9
2 3
3 5
[/code]

How do you return max(B) and min(B) value of A when A=1? But the problem is that you do not know A=1. Is there a way to create a view that will look like this? Such that it will look for similar values in A and just put them in the view.

[code]
A B
1 9
1 8
1 7
[/code]

[quote="Ravi B"]Regarding your first question, does the following SQL statement do what you want?

[code]
SELECT A, MAX(B), MIN(B)
FROM <table name>
GROUP BY A
[/code][/quote]

Yes that works, thank you. However, do you know if it is possible to have both MAX(B) and MIN(B) in the same column?

ie. for A=1

[code]

Result

A B
1 9
1 7

[/code]

You can use a UNION operation:

[code]
SELECT A, MAX(B)
FROM <table name>
GROUP BY A
UNION
SELECT A, MIN(B)
FROM <table name>
GROUP BY A
[/code]

If the minimum and maximum are the same, then this statement will generate only one record.  If you wanted to generate two (duplicate) records, then change "UNION" to "UNION ALL".


Does anyone know how to convert numbers from base to base?
such as the base 4 representation of 11011000(base 2)??

Yes I do.
Usually the best way is to convert to base 10, then to the other base.
However for this problem,

Note that the base two representation is

11011000

$2^3+2^4+2^6+2^7$

If we halve all of the exponents, we get it in base 5.

$2(4^1)+4^2+4^3+2(4^3)$

We take out a factor of 2 in the  $2^3$ and the $2^7$ so that you can divide the exponent by 2.

Thus we have
$2(4^1)+(4^2)+3(4^3)=3120$

EDIT: im a retard

These problems are rarely used on MATHCOUNTS, but I would know them anyway (especially since there not that hard)

[quote="biffanddoc"]$2^3+2^4+2^6+2^7$

If we halve all of the exponents, we get it in base 5.

$2(4^2)+4^2+4^3+(4^3)$

We take out a factor of 2 in the  $2^3$ and the $2^7$ so that you can divide the exponent by 2.

Thus we have
$3(4^2)+3(4^3)=3300$[/quote]

Be careful, where did the other $4^2$ come from?

[hide]It should be $3(4^3)+1(4^2)+2(4^1)$ because $2^3=2(4^1)$ and $2^4=1(4^2)$...[/hide]


geometry
circumcircle
angle bisector
perpendicular bisector

In $ \triangle ABC$,if the angle bisector of angle A and perpendicular bisector of BC intersect,prove that they intersect on the circumcircle of the triangle ABC.

Suppose that the bisector of $ A$ meet the circumcircle at $ P$, we have that $ <CBP\equal{}<CAP\equal{}<BCP$ . the triangle PBC is isosceles , then $ PB\equal{}PC$ ---> $ P$ lies on the perpendicular bisector of $ BC$

Label the intersection of the angle bisector of $ A$ with the circumcircle centered at $ O$ as $ D$. Because of the angle bisection, we get that Arc $ CD\equal{}BD$. Because if two arcs on a circle are equal, the chords with their endpoints will be equal, lines $ CD\equal{}BD$. Since $ \triangle DBC$ is isosceles, $ D$ lies on the perpendicular bisector of $ BC$ and therefore the circle around $ O$, the angle bisector of $ A$ and the perpendicular bisector of $ BC$ are concurrent at $ D$.


geometry
trigonometry
calculus
AMC 10
complex numbers
AMC

I think you can order past competitions online and take them. Or you can print out practice problems from the teacher manuals or brochures

Preparing from past test questions is probably the best way. I know from other people's experiences that if the AMC 10 is very easy for you and you can correctly answer all but the last few, then the AMC 12 is better practice...

Do some past contests (AMC 12's too - they share many common problems). For the questions you miss, look at the solution, make sure you understand it, and then either [i]a) let the problem sit for a little while and then do it again, or b) explain the solution to someone else.[/i] If you don't understand the solution, get help (that's what AoPS is here for).

Since you're only in Alg II (and AMC can test stuff up to pre-calc and stuff that aren't normally taught in school) you may run into topics that you have very little experience in (combinatorics, geometry, etc. though I can't really say for you which ones are your strengths and which ones are your weaknesses). For these topics, you can either do more problems or pick up a book that describes the topic in better depth than your textbook (for example: AoPS (the book)).

The biggest difference between the AMC 10 and 12 is the curriculum background required. Most takers of the AMC 10 are 9th and 10th graders in geometry or Alg. II classes, and the test is designed with that in mind. While some of the questions will involve material you probably haven't seen in the classroom (particularly discrete math), there won't be questions that you need next year's class for.

You probably also need to master precalculus to do well on the AMCs (in general)

I found that very few, if any, precalc questions come up on the test. I thought the AMC 10 doesn't include trig,complex numbers, and logs?

Oh.

But however, you need to master precalculus for the AMC 12, but even if you do, you'll still succeed on the AMC 10.

Eh, it depends. I hadn't done precalc when I got a 112.5; sure, it's not great, but I still count it as succeeding somewhat.

I don't know about other schools, but in my school's precalc class we covered combinatorics/probability and analytical geometry in addition to the standard trig/limits/etc., which I found to be helpful for the AMC's.

[quote="tridecalogism_lover"]This will be my first time doing it; how do I prepare?

( currently taking Alg 2)[/quote]

Just take a tour on Mathlinks  :D

I don't know how much precalc is needed for AMC 12 vs AMC 10 but I heard that they have like 2/3 of the problems the same. Also can't tell how much having had precalc vs not helps- already had like 6 months of precalc before AMCs last year. got 121.5 on the AMC 10 that way, doing AoPS problems this year to prepare.

i don't think that you need to have mastered any particularly high-leveled math class (e.g. above geometry/alg 2) to succeed in math competitions. i have a friend that's taking geometry in school, and this guy could easily pass AMC 10/12.
likewise, i have a friend that's taking calculus AB, but he couldn't even get to state mathcounts.


abstract algebra
group theory
superior algebra
superior algebra unsolved

1. If N is an R-module and there is a family ( )  of epimorphisms,  :N  such that  = {0}, prove that N is isomorphic to a subdirect product of

2. Show that Z is isomorphic to a subdirect product of (Z/nZ)n>1.

3. If M, N are R-modules. Let  (M, N) be the set of R-morphisms : M N. Show that  (M, N) is an abelian group which is an R-module whenever R is commutative.
Suppose now that R is commutative and that  , ... ,  are sub-modules of M such that M= …….. . For j= 1, 2,…,n, let   denote the set of R- morphisms  g: M N such that  . Prove that each  is an R-module and that  .

4. Let  : M M be an R-morphism. Show that if   is a monomorphism, then   not a zero divisor in the ring  M. if M is free, establish the converse: that if  M is not a left zero divisor then   is a monomorphism.

5. Let p be a prime. Show that the Z-module  /Z is not free. (Hint: show that the endomorphism on  /Z described by   is neither a left zero divisor nor a monomorphism and use question 4.

6. Let M be an R-module of finite height. If N is a sub-module of M. prove that there is a Jordan Hӧlder tower,
M=
With, for some index k,  = N. (Hint: consider a tower of submodules of N)

I have a hunch that this is a homework problem set. If it is, please do not post homework on this forum.

At the very least, do you have any ideas? Where are you stuck?


Nobody like it or it was not new? :maybe:

[quote="NextPeace"]It is true problem?
if x=1, y=z=0, LHS<RHS.[/quote]
Yes,It is true. :)

I really have no clue solving in this kind of 3 variable inequalities... but i will try. Please post solution if nobody can find it for a while.
but i have question. is it possible or is there a way, strategy to prove inequalities in the form of :

$ \sum \frac {ma^2 \plus{} nb^2}{pa^2 \plus{} qb^2 \plus{} rab} \geq  k$ where variables $ a, b, c, m, n, p, q, r, k \geq 0$

Please suggest ways to prove such inequalities. Thanks.

I obtained by RLCF-Theorem the following related statement:

Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. If $ p$ and $ q$ are nonnegative real numbers such that $ p \plus{} 4q\ge 2$, then

$ \frac {1 \minus{} a}{1 \plus{} pa \plus{} qa^2} \plus{} \frac {1 \minus{} b}{1 \plus{} pb \plus{} qb^2} \plus{} \frac {1 \minus{} c}{1 \plus{} pc \plus{} qc^2}\ge 0$.

This inequality is equivalent to

$ \sum \frac {(1 \plus{} p)y^2 \plus{} qx^2}{qx^2 \plus{} y^2 \plus{} pxy} \ge 3$.

I conjecture that the inequality holds for $ n$ numbers if $ p \plus{} 4q \ge n \minus{} 1$. I proved this for $ n \equal{} 3,4,5$.

[b]Remark.[/b] The following particular cases are nice (for $ abc\equal{}1$):

$ \frac {1\minus{}a}{2\plus{}a^2}\plus{}\frac {1\minus{}b}{2\plus{}b^2}\plus{}\frac {1\minus{}c}{2\plus{}c^2} \ge 0$

and

$ \frac {1\minus{}a}{(2\plus{}a)^2}\plus{}\frac {1\minus{}b}{(2\plus{}b)^2}\plus{}\frac {1\minus{}c}{(2\plus{}c)^2} \ge 0$.


Let f be differentiable on (0,inf), if lim x-->inf f'(x) and lim x--> inf f(x) both exist and are finite, prove that lim x-->inf f'(x) = 0.

I know that i want to prove the conclusion false and say that lim x--->inf f'(x) is not = 0 and us the mean value theorem to see what happens as x increased in integer steps but i have been trying and trying to come up with this but keep banging my head against the wall can anyone please help me?!
i really need some good help on this one.
thanks
john

So.. why did you post it here? There is a calculus/analysis forum where this belongs, and it gets much more traffic.

Try arguing by contradiction: if $\lim_{x\to\infty}f'(x)=L\neq0$, show that $\lim_{x\to\infty}f(x)=\infty$ if $L>0$ and $\lim_{x\to\infty}f(x)=-\infty$ if $L<0$. You still want to use the mean value theorem.

I see you also posted it in the proper forum, and it was answered there as well.
Don't post the same thing more than once; it annoys people.


geometry
geometric transformation
homothety
Euler
incenter
symmetry
reflection

Let Na be the Nagel point of a triangle, G the triangles centroid, and I the incenter. Prove that Na, G, and I are always collinear and that GNa=2*IG.

you can be lazy and use barycentric coordinates...also i think a homothety argument works here [it works for the euler line...i am not sure if it works here too]

[hide="purely geometric but a bit beastly"]
The problem is equivalent to showing that the incenter of a triangle is the Nagel point of its midpoint triangle.

Suppose $\triangle ABC$ is our original triangle with incenter $I$ and midpoint triangle $\triangle DEF$ ($D$ opposite $A$, etc.) By symmetry, it suffices to prove that if $P$ is the point where the excircle of $\triangle DEF$ opposite $D$ touches $\overline{EF}$, then $D,I,P$ collinear.

It is a well-known fact that in $\triangle DEF$, if you take the point where the incircle touches $\overline{EF}$ and reflect it over the midpoint of $\overline{EF}$, you get the point where the excircle opposite $D$ touches $\overline{EF}$. (You can compute the lengths of all of the tangent segments yourself if you doubt this.) Using this fact, we see that $P$ is in fact touched by the incircle of $\triangle AEF$ because that circle is the reflection of the incircle of $\triangle DEF$ over the midpoint of $\overline{EF}$. Therefore, if $T$ is the point where the incircle of $\triangle ABC$ touches $\overline{BC}$, then $P$ is the midpoint of $\overline{AT}$.

After dilation centered at $T$ with factor 2, showing that $D,I,P$  collinear is equivalent to showing that $Q,R,A$ collinear, where $R$ is the point opposite $T$ on the incircle of $\triangle ABC$ and $Q$ is the point where the excircle of $\triangle ABC$ opposite $A$ touches $\overline{BC}$. (Again, this relies on the well-known fact.) But this is a much nicer statement. If we let the tangent line to the incircle of $\triangle ABC$ at $D$ intersect $\overline{AB}$ at $X$ and $\overline{AC}$ at $Y$, then $\overline{XY}\perp \overline{RT}\perp \overline{BC}$, so $\overline{XY}\parallel \overline{BC}$. Also, the incircle of $\triangle ABC$ is the excircle of $\triangle AXY$ opposite $A$, so the dilation centered at $A$ that takes $X\rightarrow B$ takes $Y\rightarrow C$ and $R\rightarrow Q$. It follows that $Q,R,A$ are collinear.
[/hide]

Dear Mathlinkers,
five theorems (including the present question) of Nagel has been put on my website
http://perso.orange.fr/jl.ayme
Sincerely
Jean-Louis


It isn't hard. Let's put $x=a^{\frac{n}{4}}, y=b^{\frac{n}{4}}, z=c^{\frac{n}{4}}, t=d^{\frac{n}{4}}$. Then the inequality rewrites as: \[\frac{\sum{\frac{a^{3}}{bcd}}}{4}\geq 1.\] Now it is simple $\mathrm{AM-GM}$.

[quote="lotrgreengrapes7926"][quote="SBHD"]If a,b,c and d are positive real numbers and n is a positive integer, prove that

$\frac{a^\frac{3n}{4}}{(bcd)^\frac{n}{4}}+\frac{b^\frac{3n}{4}}{(acd)^\frac{n}{4}}+\frac{c^\frac{3n}{4}}{(abd)^\frac{n}{4}}+\frac{d^\frac{3n}{4}}{(abc)^\frac{n}{4}}\geq 4$

When does the equality hold?[/quote]
Just AM-GM:
$\frac{a^\frac{3n}{4}}{(bcd)^\frac{n}{4}}+\frac{b^\frac{3n}{4}}{(acd)^\frac{n}{4}}+\frac{c^\frac{3n}{4}}{(abd)^\frac{n}{4}}+\frac{d^\frac{3n}{4}}{(abc)^\frac{n}{4}}\geq 4(\frac{a^{3}b^{3}c^{3}d^{3}}{a^{3}b^{3}c^{3}d^{3}})^\frac{n}{4}=4$[/quote]

It's in fact the same solution as I wrote.  :maybe:  :mad:


function
limit
calculus
calculus computations

$f(x)=1-x^p+o(x^p)$ when $x$ is near $0$ where $p>0$ is fixed integer

$g_n(x)=\frac{n}{\frac{1}{f(x)}+...+\frac{1}{f(nx)}}$ and $n\geq 1$ fixed integer

Compute $\lim_{ x \rightarrow 0}\frac{1-g_n(x)}{x^p}$

--------------------------------------------------------------------------------

For those who are not familiar with small o

$f(x)=1-x^p+o(x^p)$ when $x$ is near $0$

means

$f(x)=1-x^p+x^p\epsilon (x)$ where $\lim_{x\rightarrow 0}\epsilon(x)=0$

I was thinking of writing out an explanation of big-O and little-o notation and how it's used, and posting that explanation in this forum. I haven't done it yet but should have it up some time next week. This, of course, is [i]exactly[/i] the kind of problem that language and mindset is most useful for.

I'll post an answer, but hide it to let others try for it.

[hide="The answer"]$\frac1n\sum_{k=1}^nk^p$[/hide]
[hide="The method"]For future reference, let $S=\sum_{k=1}^nk^p.$

$f(kx)=1-k^px^p+o(x^p)$

$\frac1{f(kx)}=1+k^px^p+o(x^p)$

$\sum_{k=1}^n\frac1{f(kx)}=n+\sum_{k=1}^nk^px^p+o(x^p)=n+Sx^p+o(x^p)$

$g_n(x)=\frac{n}{n+Sx^p+o(x^p)}=\frac1{1+\frac{S}{n}\,x^p+o(x^p)} = 1-\frac{S}{n}\,x^p+o(x^p)$

$1-g_n(x)=\frac{S}{n}\,x^p+o(x^p)$

$\frac{1-g_n(x)}{x^p}=\frac Sn+o(1).$ Take the limit and we are done.[/hide]
[hide="Extra comment"]For large $n$, this can be written as $\frac{n^{p}}{p+1}+o(n^{p}).$[/hide]


Is it just me or does the link not work?

sorry...

the link is: [url=http://hseagle.sas.edu.sg/hscounseling/Tests/NMSCindex.htm]hseagle.sas.edu.sg/hscounseling/Tests/NMSCindex.htm[/url]

if only i lived in arkansas, south dakota, west virginia, or wyoming  :D

they removed it from the site -_-"

Anyone remember what Florida's was? Or about..?

[quote="ffdbzathf"]if you take the psat in 10th grade and make the cutoff, can you get national merit?  what if you are graduating after 11th grade?[/quote]

I'm fairly certain that only the 11th grade testing counts. I'm not sure about people who graduate early...

If you are graduating early, your 10th grade score can count. Investigage at the NMS website.

In any case, call up the College Board and see what they say. Unless you're just doing the standard Take PSAT in junior year -> see if score qualifies method, chances are it depends on your circumstances. Like for example, I'm pretty sure you can also qualify without taking the PSAT (substitute SAT score).


geometry
3D geometry
tetrahedron
vector
symmetry
ratio
calculus

I don't really understand how you got your result, but it is the correct answer for the COM of a tetrahedron as you described it.

I multiplied each vertex (which I treated as a vector) by a mass "m."  I added those four results together and divided by the total mass, which is "4m."  The "m's" cancel, and I am left with 1/4 the sum of the vertices.

for $n$ points with weights $w_i$, the center of mass (barycenter) is the weighted arithmetic mean of the points (well, of the vectors)
this is true in any $n$-D space, right?

[quote="Fermat2"]Capisco!!

That's neat.
I must remember that method.

So, what you did is:

You put a point mass at each of the 4 vertices and and took the sum of the vector moments of each mass about the origin and equated that with the vector moment of the total mass about its COM (a vector position)
The basic assumption being: The COM of the 4 point masses is the same as the COM of the homogemeous solid (tetrahedron).

I did it with vectors using the fact that: If you draw a line (vector) from a vertex of a tetrahedron to the centroid of the opposite face, then the intersection of all such lines (vectors) is at the COM of the object.
Of course you would only need two such lines to find the point of intersection.

However, your method is much shorter.

I got the same result as yourself which verifies your assumption  :)[/quote]
OK thanks a lot, I appreciate the confirmation and alternate method :)

Caveat: This applies in general only to simplices; $n$-dimensional figures with $n+1$ vertices. Why does it work for simplices? It clearly works for regular simplices by symmetry, and every simplex is an affine image of a regular simplex. Affine transformations preserve ratios of mass densities, so the centers of mass are moved by the same transformation.

[quote="amcavoy"]I need to find the center of mass of the tetrahedron with vertices (0,0,0), (1,0,0), (0,1,0), and (0,0,1) assuming uniform density.  At first, I thought I would need to integrate, but then I thought of this:

$\left(x,y,z\right)_{\textrm{CM}}=\frac{1}{4}\sum_{n}\left(x_n,y_n,z_n\right)=\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)$

Is this the right answer or did I try to make it easier than it is?[/quote]

How can take masses to be point masses at the given points when you have mentioned the tetrahedron to be of uniform density.
I tried it using integration but got the coordinates $({\frac 1{2}},{\frac 1{2}},{\frac 1{2})}$ for com. Which really seems to be wrong.

[quote="anjani_kumarsingh"]How can take masses to be point masses at the given points when you have mentioned the tetrahedron to be of uniform density.[/quote]

Consider the solid tetrahedron to be infinite shells of tetrahedra, each smaller than the previous, filling the entire volume. They're all similar, and they're all scaled but not translated, so they should all have the same center of mass. Thus, the center of mass of a solid tetrahedron is the same as the center of mass of the surface of a tetrahedron. You can repeat this argument with equilateral triangles to reduce it to the edges, and then with lines to reduce it to the vertices.

[quote="Fermat2"][quote="anjani_kumarsingh"]Fermat 2
Can you show the steps how did you do it using vectors.[/quote]

Hope you can read this all right.

Fermat[/quote]

What you wrote was clearly visible.
Thanks for taking so much pain.
I will have to say that your handwriting is better than mine.
 :roll:
Thanks Again.
Bye.  :lol:


Be $ n\in \mathbb{N},\: n\ge2$
Prove that $ (n\minus{}\varphi(n))|n$ if and only if $ n$ is a power of a prime number.

(where $ \varphi$ is the Euler's indicator.)

But if $ n \equal{} \prod p_i^{a_i}$ is the prime decomposition of $ n$ then $ n \minus{} \varphi (n) \equal{} \prod p_i^{a_i \minus{} 1}$ so $ n \minus{} \varphi(n)$ divides $ n$...
Am I missing something?

Pierre.

yes!
$ n\minus{}\phi(n)\equal{}\prod p_i^{a_i\minus{}1}\left(p_1p_2\cdots p_k\minus{}(p_1\minus{}1)(p_2\minus{}1)\cdots(p_k\minus{}1)\right)$
and notice that the other factor doesn't divide $ n$ unless k=1
 :)

OOps...I forgot the parenthesis in the product... :blush:

Pierre.


Determine the value of  k so that x^2 –3x – 4k divides x^3 –37x –12k

[hide] We have $ (x^2\minus{}3x\minus{}4k)(ax\plus{}b)\equal{}(x^3\minus{}37x\minus{}12k)$ for some real numbers $ a$ and $ b$. Clearly, $ a\equal{}1$ and $ b\equal{}3$ (consider the third degree terms and the $ k$ terms). Thus, by comparing $ x$-values, we have $ \minus{}9x\minus{}4kx\equal{}\minus{}37x \implies k\equal{}\boxed{7}$.[/hide]


A relatively prime day is one where the month number (e.g. January = $ 1$, February = $ 2$, etc.) and date of the month $ (1, 2, \ldots, 31)$ have no common factor other than $ 1$. Which month has the fewest relatively prime days?

Someone correct me if I'm wrong, but I believe the answer is $ \boxed{\text{February}}$, because only half of the days in the month are relatively prime with it, and as there are only 28(or 29) days, only $ 14$ days are relatively prime.

I got November because 11 is the greatest prime from 1 to 12. The only two days that aren't prime days are the eleventh (11) and the twenty-second (22).

umm...
dragon
u found the month with the [b]most[/b] relatively prime dates
the question asks for which month has the fewest
i would probably say february

The same number of days are relatively prime, but december also has 31, which june does not

i think its [b]june[/b]... cuz december has the same as june but one more cuz of dec 31st


It is given a real number sequence $x_1,x_2,...,x_{40}$ with $x_1,x_2,...,x_{40}$ are all different. Then, we compare $x_n$ and $x_{n+1}$ for $n=1,2,...,39$. If $x_{n+1}<x_n$, we switch the order such that the smaller number is at the left of the bigger number. Do the procedure until $x_{40}$.
If the sequence is formed randomly, what is the probability that after doing the procedure, $x_{20}$ will become the 30th term ?

After some steps, the new x_19 is the biggest of the old $x_1 \dots x_{19}$.  Now we must have $x_{20} > x_{19}$, etc and $x_{20 + k} < x_{20}$ for k = 1...10.  Then, we have $x_{20} < x_{31}$.  So we have 12 conditions.  so 1/ 2^12.


I can not edit my post  in http://www.mathlinks.ro/Forum/viewtopic.php?t=128305 why?  :maybe:

i don't know but if you want to edit your post you must ask a moderator to do it

[quote="jmerry"]I think there's a 24-hour limit on edits for non-moderators. You were just past that.[/quote]

I think so.


Let $ H_{a}, H_{b}, \ldots$ be an ascending chain of subgroups of  a finitely generated abelian group.  Does it necessarily hold that any two (different) subgroups of the chain have different ranks?

Any insights you can provide me with are thanked in advance.

Best of Luck.

No, not in general. Imagine that the chain starts with the trivial subgroup. But eventually the chain stabilizes since the group is a Noetherian $ Z$-module, so that not only the rank, but also the groups are eventually the same.

In $ \mathbb{Z}$, we can have something like $ 60\mathbb{Z},12\mathbb{Z},6\mathbb{Z},3\mathbb{Z},\mathbb{Z}$. We can have chains of arbitrary finite size at the same rank, but any ascending chain must be finite overall since $ \mathbb{Z}$ (and thus any finitely generated $ \mathbb{Z}$-module) is Noetherian.

Thanks for taking the time to reply...

Your posts have proved to be very useful to me.


analytic geometry
conics
ellipse
calculus
integration
trigonometry

If you have two objects in space, then the gravitational force between them is $\mathrm{F}_{g}=\frac{\mathrm{G}m_{1}m_{2}}{r^{2}}$. Therefore, the acceleration of Object 1 towards Object 2 is $a_{1}= \frac{\mathrm{G}m_{2}}{r^{2}}$ and $a_{2}= \frac{\mathrm{G}m_{1}}{r^{2}}$ for the acceleration of Object 2. Since $r$ is decreasing because of the acceleration of the objects, acceleration of the objects toward each other also changes. If both objects are at rest, how long would it take them to hit each other?

Note that you cannot integrate $\frac{\mathrm{G}m_{2}}{r^{2}}$ to get the velocity, because we do not have time in the equation. We need $t$ in the equation to integrate it with respect to time.

to cuong:
can you show that method with polar coordinates?

$a = a_{1}-a_{2}= \frac{F_{12}}{m_{1}}-\frac{F_{21}}{m_{2}}= \left(\frac{1}{m_{1}}+\frac{1}{m_{2}}\right) F_{12}=$

$=-\frac{m_{1}+m_{2}}{m_{1}m_{2}}\cdot \frac{G m_{1}m_{2}}{r^{2}}=-\frac{GM}{r^{2}}$

because $F_{21}=-F_{12}$ and $F_{12}=-\frac{Gm_{1}m_{2}}{r^{2}},$ where $M = M_{1}+M_{2},\ r = r_{1}-r_{2}.$ Since $a = \ddot r,$ this is a 2nd order diff. equation $\ddot r =-\frac{GM}{r^{2}}$ not containing the independent variable t, hence its order can be reduced by the standard substitution $v = \dot r,$ yielding a separable 1st order diff. equation:

$\ddot r = a = \dot v = \frac{\text dv}{\text dt}= \frac{\text dv}{\text dr}\cdot \frac{\text dr}{\text dt}= \frac{\text dv}{\text dr}\ v =-\frac{GM}{r^{2}}$

$v\ \text dv =-GM\ \frac{\text dr}{r^{2}},\ \ \ \frac{v^{2}}{2}= \frac{GM}{r},\ \ \ \frac{\text dr}{\text dt}= v = \sqrt{\frac{2GM}{r}}$

This is another separable 1st order diff. equation.

$\text dt = \sqrt{\frac{r}{2GM}}\ \text dr,\ \ \ t = \frac{2}{3}\sqrt{\frac{r^{3}}{2GM}}$

BTW, if the collision is ellastic, the bodies rebounce, the oscillation period is $T = 2t,$ the "ellipse" semimajor axis $a = \frac{r}{2},$ hence $T = \frac{8}{3}\ \sqrt{\frac{a^{3}}{GM}},\ \ \frac{a^{3}}{T^{2}}= \frac{9GM}{64},$ which is something like Kepler's 3rd law. I wonder why did not I get $\frac{a^{3}}{T^{2}}= \frac{GM}{4 \pi^{2}},$ because the problem solution should be based on Kepler's 3rd law and the result should be $t = \frac{\pi}{2}\sqrt{\frac{r^{3}}{2GM}}.$

Yetti you forgot the angle diffrential, there is no collision here

If you know much about Differential Equation,you can solve it, dun try elementary way.

First
Consider the Polar coordinate :in which the Newton 2 second law can express as:

$m(r''-r(phi')^{2})=F_{r}$
$m(r(phi')+2r'phi')=F_{p}hi$

so use can solve it using some The center of mass method and conservation of angular momentum, which is the so_called Keppler 1 and 2 St law

[b](1)[/b] There will be a collision, because the initial velocity is zero, hence the orbital momentum is zero.
[b](2)[/b] For the same reason, polar coordinates are not necessary, because $\phi = \text{const}= 0$ at all times.
[b](3)[/b] The elementary solution of the problem is to consider the path as a degenerate ellipse with the semimajor axis $a = \frac{r_{0}}{2},$ where $r_{0}$ is the initial distance, the orbital period $T = 2t,$ where t is the time to the collision, and apply Kepler's 3rd law $\frac{a^{3}}{T^{2}}= \frac{GM}{4\pi^{2}},$ yielding $t = \frac{\pi}{2}\sqrt{\frac{r_{0}^{3}}{2GM}}.$
[b](4)[/b] In the 1st method, I did not forget any differential, but an integration constant.

You are wrong on all counts and you did not even try to solve the problem.
_______________________________

Integrating $v\ \text dv =-GM\ \frac{\text dr}{r^{2}},$ we get

$\frac{v^{2}}{2}= \frac{GM}{r}+\text{const},\ \ \ v = \sqrt{\frac{2GM}{r}+\text{const}}$

At t = 0, we have v = 0 and $r = r_{0},$ hence the integration constant in the last equation is $-\frac{2GM}{r_{0}}.$ Then

$\frac{\text dr}{\text dt}= v = \sqrt{2GM}\sqrt{\frac{1}{r}-\frac{1}{r_{0}}},\ \ \ \text dt = \frac{1}{\sqrt{2GM}}\frac{\text dr}{\sqrt{\frac{1}{r}-\frac{1}{r_{0}}}}$

Integrating and substituting $\rho = \frac{r}{r_{0}},$

$t = \frac{1}{\sqrt{2GM}}\int_{0}^{r_{0}}\frac{\text dr}{\sqrt{\frac{1}{r}-\frac{1}{r_{0}}}}= \sqrt{\frac{r_{0}^{3}}{2GM}}\int_{0}^{1}\frac{\text d\rho}{\sqrt{\frac{1}{\rho}-1}}$

Now we have to show that the last integral is equal to $\frac{\pi}{2},$ which is elementary. Substituting $\rho = \sin^{2}\frac{\theta}{2},$

$\int_{0}^{1}\frac{\text d\rho}{\sqrt{\frac{1}{\rho}-1}}= \int_{0}^{\pi}\sin^{2}\frac{\theta}{2}\ \text d\theta = \int_{0}^{\pi}\frac{1-\cos \theta}{2}\ \text d\theta = \left[\frac{\theta-\sin \theta}{2}\right]_{0}^\pi = \frac{\pi}{2}$

Thanks, yetti.

[quote="cuong"]hahahahaaha :rotfl:
if two object (with different mass) attract each other with centripital forces(dun care what kind of force), it will not hit ,but run in an ellipsoide or paroboloide path, depend on velocity

It is easy to prove with the polar coordinate .[/quote]

Did you even read the question? Both objects are at rest. But can you prove what you thought I asked using polar coordinates here? Thanks.


limit
ratio
real analysis
real analysis unsolved

Let $(x_{n})_{n \geq 1}$ be a sequence so that $x_{1}=1$, $x_{2} = 2$, $\ds x_{n+1} = \frac{x_{n}+x_{n-1}^{2}}{n}$. Prove that the limit $\ds \lim_{n \to \infty} (x_{1} + \ldots + x_{n})$ is finite.

By explicit computation, $x_6=\frac{631}{720}<1$ and also $x_7<1.$ Once that happens, for all $n\ge 7,\,x_{n+1}<\frac{1+1}n<1.$

Then we have $0<x_{n+1}=\frac{x_n+x_{n-1}^2}n<\frac{2x_n}n.$

Thus $0<\frac{x_{n+1}}{x_n}<\frac2n$ and $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=0.$

By the ratio test, $\sum_{n=1}^{\infty}x_n$ converges.


Let $ f\colon [0,1]\to \mathbb{R}^2$ be a continuous function. Define $ g\colon [0,1]\to\mathbb{R}$ as $ g(t) \equal{} diam(f([0,t]))$, $ t \in [0,1]$, where $ diam(S)$ stands for the diameter of the set $ S$, $ diam(S) \equal{} \sup(\{d(a,b) \colon a,b\in S\})$. Must $ g$ be continuous?

Yes, even with the same modulus of continuity. Hint: if $ A$ and $ B$ are two sets with nonempty intersection, then $ \mathrm{diam}\, (A\cup B)\le \mathrm{diam}\, (A)\plus{}\mathrm{diam}\, (B)$.

Thanks mlok for the hint. :) Could somebody please check the following:

Since $ f$ is continuous and $ [0,1]$ is compact, it follows that $ f$ is uniformly continuous so for each $ \varepsilon > 0$ there is a $ \delta_\varepsilon > 0$ such that for all $ t,\, t_0 \in [0,1]$ such that $ |t \minus{} t_0| < \delta_\varepsilon$, $ d(f(t),f(t_0)) < \varepsilon$ holds.

Let $ t_0 \in [0,1]$, $ \varepsilon > 0$ and $ \delta \equal{} \delta_\varepsilon > 0$. Suppose $ t \in [0,1]$ is such that $ |t \minus{} t_0| < \delta_\varepsilon$. Suppose $ t_0 < t$ (case for $ t_0 \geq t$ should be similar). Then $ \mathrm{diam}\, (f([0,t]))\geq \mathrm{diam}\, (f[0,t_0])$ and
\[ |g(t) \minus{} g(t_0)| \equal{} |\mathrm{diam}\, (f([0,t])) \minus{} \mathrm{diam}\, (f[0,t_0])| \equal{}
\]

\[ \equal{} \mathrm{diam}\, (f([0,t])) \minus{} \mathrm{diam}\, (f[0,t_0]) \equal{} \mathrm{diam}\, (f([0,t_0] \cup [t_0,t])) \minus{} \mathrm{diam}\, (f[0,t_0])\leq
\]

\[ \mathrm{diam}\, (f([0,t_0])) \plus{} \mathrm{diam}\,(f([t_0,t])) \minus{} \mathrm{diam}\,(f([0,t_0])) \equal{} \mathrm{diam}\, (f([t_0,t])) \equal{}
\]

\[ \equal{} \sup{\{d(f(x),f(y)) \, | \, x,y\in [t_0,t]\}}.
\]
Since for any $ x,\, y \in [t_0,t]$, $ |x \minus{} y| < \delta_\varepsilon$ holds, uniform continuity of $ f$ does not allow $ \mathrm{diam}\, (f([t_0,t]))$ to exceed $ \varepsilon$, so $ |g(t) \minus{} g(t_0)| < \varepsilon$ and therefore $ g$ is continuous.


algebra
polynomial
induction
number theory unsolved
number theory

We say that a polynomial $ P(x)$ with integer coefficients is divisible by a natural number $ m$ if $ P(a)$ is divisible by $ m$ for every integer $ a$. Prove that if a polynomial $ P(x)\equal{}a_0x^n\plus{}...\plus{}a_{n\minus{}1}x\plus{}a_n$ is divisible by $ m$, then $ n!a_0$ is also divisible by $ m$.

[hide="A hint"]Induction on $ n$ . Note that if $ P$ has that property then $ P(x\plus{}1)\minus{}P(x)$ too. [/hide]


geometry rectangle calculus integration perimeter

We have two rectangles $ A$ and $ B$ with integral sides. Can you find
the dimensions of the two rectangles based on the following facts?

1) Area of rectangle $ A$ = $ 3$ times the area of rectangle $ B$

2) Perimeter of rectangle $ B$ = $ 3$ times the perimeter of
rectangle $ A$

[quote="123456789"]0 x 0 rectangles[/quote]

means not possible?


I was looking at some AMC10 practice problems, but I'm having trouble with a few. I would greatly appreciate it if someone could explain how to solve the problem that I am having trouble with.

http://www.unl.edu/amc/b-registration/b1-archive/2003-2004/04bro1012/04bro1012html/04bro1012-que.html

It is number 25.

As for number 15, I do not know a very fast way of solving the problem, aside from seeing which numbers fit. What's a faster method?

Thanks

For #15:

You need to find the number of positive integers from 1 to 100 that are divisible by two, and then subtract from those the number of integers from 1 to 100 that are divisible by both two and three. The number of positive integers from 1 to 100 that are divisible by two is 50. The number of positive integers from 1 to 100 that are divisible by both two and three (which is the same as being divisible by 6) is 16. 50 - 16 = 34, so the answer is 34/100 = 17/50.

For #25:

If n/100 has quotient q and remainder r, then n = 100q + r. We want q + r to be divisible by 11. If 100q + r is divisible by 11, then q + r must be divisible by 11, because since 99q is divisible by 11 and q + r is divisible by 11, their difference must also be divisible by 11. So we need to find all five-digit numbers that are divisible by 11. These range from 10010 to 99990. Now, out of the numbers in this range, exactly one out of every 11 will be divisible by 11 if we ignore the last one for the moment. The number of integers in the range 10010 to 99989 is 99989 - 10010 + 1 = 89980. 89980/11 = 8180, and adding back the one we skipped over we get 8181.

Hope this helps! If I was not clear enough in my solutions, please let me know and I will be glad to go over them again.

thank you both! It seems so simple now that you've explained it.