trigonometry
geometry
circumcircle
perpendicular bisector
geometry proposed

Let ABCD be a convex quadrilateral such that AD = CD, BD = BC, < ADC = 168° and < ABC = 66°. Find < BAD.

I am not trained at such problems at all, so I usually bash them with trig, but for this one I have a trig solution which requires no more trigonometry than the formulas sin (180° - x) = sin x, sin (90° + x) = cos x, sin (2x) = 2 sin x cos x and $\sin 30^{\circ}=\frac12$, each of them applied once. So I didn't even look for a synthetic one...

  darij

Denote $b = BC = BD$ and $c = CD = DA$.

$\measuredangle C = 360^o - (\measuredangle A + \measuredangle B + \measuredangle D) = 360^o - (\measuredangle A + 66^o + 168^o) = 126^o - \measuredangle A$

$\measuredangle ABD = 180^o - (168^o - \measuredangle C + \measuredangle A) = 180^o - (42^o + 2 \measuredangle A)$

Using cosine theorem for the triangle $\triangle BCD$ and sine theorem for the triangle $\triangle ABD$, we have

$ b^2 + c^2 - 2bc \cos{\widehat C} = b^2$

$\frac c b = 2 \cos{(126^o - \widehat A)}$

$\frac c b = \frac{\sin{\widehat ABD}}{\sin{\widehat A}} = \frac{\sin{(2 \widehat A + 42^o)}}{\sin{\widehat A}}$

The angle $\measuredangle A$ has to satisfy the equation

$2 \cos{(126^o - \widehat A)} \sin{\widehat A} = \sin{(2 \widehat A + 42^o)}$

$2 \sin{(\widehat A - 36^o)} \sin{\widehat A} = \cos{36^o} - \cos{(2 \widehat A - 36^o)} = \sin{(2 \widehat A + 42^o)}$

Substitute $\theta = \measuredangle A + 36^o$. The equation becomes

$\cos{36^o} - \cos{(2 \theta - 108^o)} = \sin{(2 \theta - 30^o)}$

$\cos{36^o} + \cos{(2 \theta + 72^o)} = \cos{(2 \theta - 120^o)}$

Denote $G = \frac{1 + \sqrt 5}{2}$ the golden ratio. From a regular pentagon,

$\cos{36^o} = \frac{G + 1}{2G}$, $\cos{72^o} = \frac{1}{2G}$, $\sin{72^o} = \frac{\sqrt{4G^2 - 1}}{2G}$

$\frac{G + 1}{2G} + \frac{1}{2G} \cos{2 \theta} - \frac{\sqrt{4G^2 - 1}}{2G} \sin{2 \theta} = \frac{\sqrt 3}{2} \sin{2 \theta} - \frac 1 2 \cos{2 \theta}$

${(G + 1) (1 + \cos{2 \theta}) = (G \sqrt 3 + \sqrt{4G^2 - 1}}) \sin{2 \theta}$

${(G + 1) \cos^2{\theta} = (G \sqrt 3 + \sqrt{4G^2 - 1}}) \sin{ \theta} \cos{\theta}$

Either $\cos{\theta} = 0$, $\theta = 90^o$, $\measuredangle A = \theta - 36^o = 54^o$, or

${\tan{\theta} = \frac{G + 1}{G \sqrt 3 + \sqrt{4G^2 - 1}}}$, $\theta = 24^o$, $\measuredangle A = -12^o$

The second solution is perfectly valid - it corresponds to the other intersection of the perpendicular bisector of the segment $CD$ with the circumcircle $(O)$ of the triangle $\triangle ABC$. However, this would make the quadrilateral $ABCD$ concave and the the angle $\measuredangle B = 180^o - 66^o = 114^o$, hence it does not correspond to the conditions of the problem. Therefore, $\measuredangle A = 54^o$.

As a bonus, $\measuredangle C = 126^o - \measuredangle A = 72^o$,  $\measuredangle BCA = \measuredangle C - 6^o = 66^o$. Hence, $\measuredangle BCA = \measuredangle ABC$, the triangle $\triangle ABC$ is isosceles and $CA = AB$.

Here another solution (similar to that one of Yetti).

Denoting   $\measuredangle BDC = \vartheta $ we have:

$\measuredangle ABD = 66^\circ  - \left( {180^\circ  - 2\vartheta } \right) = 2\vartheta  - 114^\circ $

$\measuredangle BAD = 360^\circ  - 66^\circ  - 168^\circ  - \vartheta  = 126^\circ  - \vartheta $

From the sine theorem for the triangle $\triangle ABD$ it follows that:

$\frac{{BD}}{{\sin \measuredangle BAD}} = \frac{{AD}}{{\sin \measuredangle ABD}}$

and since  $AD = 2 \cdot BD \cdot \cos \vartheta $  we have:

$BD \cdot \sin \left( {2\vartheta  - 114^\circ } \right) = 2 \cdot BD \cdot \cos \vartheta  \cdot \sin \left( {126^\circ  - \vartheta } \right){\rm{   }} \Rightarrow $

$\sin \left( {2\vartheta  - 114^\circ } \right) = 2 \cdot \frac{1}{2}\left[ {\sin 126^\circ  + \sin \left( {126^\circ  - 2\vartheta } \right)} \right]{\rm{  }} \Rightarrow $

$\sin \left( {2\vartheta  - 114^\circ } \right) - \sin \left( {126^\circ  - 2\vartheta } \right) = \sin 126^\circ {\rm{   }} \Rightarrow $

$2\cos 6^\circ  \cdot \sin \left( {2\vartheta  - 120^\circ } \right) = \sin 126^\circ $

Thus:

$\begin{array}{l}
 \sin \left( {2\vartheta  - 120^\circ } \right) = \frac{{\cos 36^\circ }}{{2\cos 6^\circ }} = \sin 24^\circ  \cdot \frac{{\cos 36^\circ }}{{2\cos 6^\circ  \cdot \sin 24^\circ }} =  \\  \\
  = \sin 24^\circ  \cdot \frac{{1 - 2\sin ^2 18^\circ }}{{\sin 30^\circ  + \sin 18^\circ }} = \sin 24^\circ  \cdot \frac{{1 - 2\left( {\frac{{\sqrt 5  - 1}}{4}} \right)^2 }}{{\frac{1}{2} + \frac{{\sqrt 5  - 1}}{4}}} = \sin 24^\circ  \\
 \end{array}$

Since  $0^\circ  < \vartheta  < 90^\circ $ it follows that

$2\vartheta  - 120^\circ  = 24^\circ {\rm{   }} \Rightarrow {\rm{   }}\vartheta  = 72^\circ {\rm{   }} \Rightarrow {\rm{   }} \measuredangle BAD = 126^\circ  - \vartheta  = 54^\circ $

and we are done.  $\blacksquare$

And here is my solution of the original problem:

[quote="darij grinberg"]Let ABCD be a convex quadrilateral such that AD = CD, BD = BC, < ADC = 168° and < ABC = 66°. Find < BAD.[/quote]

We will prove that < BAD = 54°.

First we introduce a notation: A quadrilateral ABCD is called [i]nice[/i] if and only if this quadrilateral ABCD is convex and the equations AD = CD, BD = BC, < ADC = 168° and < ABC = 66° hold. The problem asks to prove that every nice quadrilateral ABCD satisfies the equation < BAD = 54°. In order to prove this, I will perform two steps:

[b]Step 1:[/b] I will show that all nice quadrilaterals are similar. In other words, I will prove that there exists, up to similitude, only one nice quadrilateral.

[b]Step 2:[/b] I will construct a quadrilateral ABCD which, on one hand, is nice, while, on the other hand, it can be easily seen to satisfy the equation < BAD = 54°.

Combining these two steps, it will become clear that every nice quadrilateral ABCD must satisfy the equation < BAD = 54°. And the problem will be solved.

So it remains to establish the Steps 1 and 2.

Establishing Step 1 is easy:

Let's try to construct a nice quadrilateral ABCD. Such a quadrilateral must satisfy the conditions AD = CD and < ADC = 168°; these two conditions uniquely determine the shape of triangle ACD (namely, the triangle ACD must be an isosceles triangle with the angle 168° at its apex D). In other words, the triangle ACD is determined uniquely up to similitude. Once the triangle ACD is constructed, it remains to find the point B; this can be done as follows:

- the condition that the quadrilateral ABCD must be convex places the point B into the halfplane with respect to the line AC which does not contain the point D;
- the condition BD = BC places this point B on the perpendicular bisector of the segment CD;
- the condition < ABC = 66° places the point B on a given circular arc spanning the chord AC (actually, it would even place the point B on one of [i]two[/i] circular arcs if we hadn't the prior condition that the point B lies on the halfplane with respect to the line AC which does not contain the point D).

Hence, the point B can be localized as the common point of the perpendicular bisector of the segment CD with a given circular arc spanning the chord AC. But it is clear that the perpendicular bisector of the segment CD and this given circular arc have exactly one common point (for a watertight proof, use the fact that the perpendicular bisector of the segment CD intersects the segment AC). Hence, the point B is uniquely determined when the triangle ACD is given. Since the triangle ACD is determined uniquely up to similitude, it follows that the whole nice quadrilateral ABCD is determined uniquely up to similitude. This establishes Step 1.

Now we have to perform Step 2. In other words, we must construct a quadrilateral ABCD which is nice and satisfies the equation < BAD = 54°.

We construct such a quadrilateral as follows: Let B and D be two arbitrary distinct points in the plane. In one half-plane with respect to the line BD, take a point A such that < ABD = 30° and < ADB = 96°; in the other half-plane, take a point C such that < CBD = 36° and < CDB = 72°. Then, we claim that the quadrilateral ABCD is nice and satisfies the equation < BAD = 54°.

Well, showing that the quadrilateral ABCD satisfies the equation < BAD = 54° is simple: The sum of the angles of triangle ABD yields

  < BAD = 180° - < ABD - < ADB = 180° - 30° - 96° = 54°,

and < BAD = 54° is proven. So it remains to show that the quadrilateral ABCD is nice. In other words, we must show that the quadrilateral ABCD is convex and that the equations AD = CD, BD = BC, < ADC = 168° and < ABC = 66° hold.

Well, the equations < ADC = 168° and < ABC = 66° are almost trivial again:

  < ADC = < ADB + < CDB = 96° + 72° = 168°;
  < ABC = < ABD + < CBD = 30° + 36° = 66°.

The equation BD = BC is easy to verify, too: By the sum of angles in triangle CBD, we have

  < BCD = 180° - < CBD - < CDB = 180° - 36° - 72° = 72°,

so that < BCD = < CDB, and thus, the triangle CBD is isosceles, and we obtain BD = BC.

The convexity of the quadrilateral ABCD is also quite clear, since all four interior angles < BAD = 54°, < ABC = 66°, < BCD = 72° and < ADC = 168° are < 180°.

Thus, it remains to prove that AD = CD.

Let W be the point on the ray DA such that < WCD = 6°. We will prove that the lines BA, CW and DA concur. This will be the key to the solution of the problem; in fact, once this will be shown, the problem will become very easy: Since the lines BA, CW and DA concur, the line CW passes through the point of intersection of the lines BA and DA; in other words, the line CW passes through the point A. This is equivalent to saying that the point W lies on the line CA. But we also know that the point W lies on the line DA. Thus, the point W is the point of intersection of the lines CA and DA. In other words, W = A. Hence, the equation < WCD = 6° becomes < ACD = 6°. Together with < ADC = 168°, this yields, after the sum of angles in triangle ADC, that

  < CAD = 180° - < ACD - < ADC = 180° - 6° - 168° = 6°;

in other words, < CAD = < ACD, so that the triangle ADC is isosceles, and AD = CD is proven.

Hence, the problem will be solved as soon as we will have proved that the lines BA, CW and DA concur.

By the trigonometric version of the Ceva theorem, applied to the triangle BCD, in order to prove that the lines BA, CW and DA concur, it is enough to establish the equality

$\left(-\frac{\sin\measuredangle DBA}{\sin\measuredangle ABC}\right)\cdot\frac{\sin\measuredangle BCW}{\sin\measuredangle WCD}\cdot\left(-\frac{\sin\measuredangle CDA}{\sin\measuredangle ADB}\right)=1$.

Hereby, the minus signs are due to the fact that the lines BA and DA divide the angles DBC and CDB externally.

Hence, the problem will be solved once the equality

$\left(-\frac{\sin\measuredangle DBA}{\sin\measuredangle ABC}\right)\cdot\frac{\sin\measuredangle BCW}{\sin\measuredangle WCD}\cdot\left(-\frac{\sin\measuredangle CDA}{\sin\measuredangle ADB}\right)=1$

will be shown.

But we have

< DBA = < ABD = 30°;
< ABC = 66°;
< BCW = < BCD - < WCD = 72° - 6° = 66°;
< WCD = 6°;
< CDA = < ADC = 168°;
< ADB = 96°.

Thus, the above equality, which we have to prove, becomes

$\left(-\frac{\sin 30^{\circ}}{\sin 66^{\circ}}\right)\cdot\frac{\sin 66^{\circ}}{\sin 6^{\circ}}\cdot\left(-\frac{\sin 168^{\circ}}{\sin 96^{\circ}}\right)=1$.

This immediately simplifies to

$\frac{\sin 30^{\circ}}{\sin 6^{\circ}}\cdot\frac{\sin 168^{\circ}}{\sin 96^{\circ}}=1$.

Now, we have sin 96° = sin (90° + 6°) = cos 6°, and we also have $\sin 168^{\circ}=\sin\left(180^{\circ}-12^{\circ}\right)=\sin 12^{\circ}=\sin\left(2\cdot 6^{\circ}\right)=2\cdot\sin 6^{\circ}\cdot\cos 6^{\circ}$. Finally, it is known that $\sin 30^{\circ}=\frac12$. Thus, the above equation becomes

$\frac{\left(\frac12\right)}{\sin 6^{\circ}}\cdot\frac{2\cdot\sin 6^{\circ}\cdot\cos 6^{\circ}}{\cos 6^{\circ}}=1$.

But this is obviously true. Hence, the equality

$\left(-\frac{\sin\measuredangle DBA}{\sin\measuredangle ABC}\right)\cdot\frac{\sin\measuredangle BCW}{\sin\measuredangle WCD}\cdot\left(-\frac{\sin\measuredangle CDA}{\sin\measuredangle ADB}\right)=1$

is established, and the problem is solved.

By the way, Yetti's "bonus" result that CA = AB can be easily seen as a consequence of this solution: Since W = A, we can rewrite the equality < BCW = 66° in the form < BCA = 66°. Comparing this with < ABC = 66°, we get < BCA = < ABC; thus, the triangle BAC is isosceles, and it follows that CA = AB.

  Darij


geometry
3D geometry
number theory unsolved
number theory

Does there exist a positive integer such that both $ n\plus{}3$ and $ n^{2}\plus{}3n\plus{}3$ are perfect cubes?

note that $ x \ge 2$.
if $ n\plus{}3\equal{}x^3$ then we have $ y^3\equal{}n^2\plus{}3n\plus{}3\equal{}(n\plus{}3)^2\minus{}3(n\plus{}2)\equal{}x^6\minus{}3(x^3\minus{}1)\equal{}x^6\minus{}3x^3\plus{}3$.
but we have $ (x^2\minus{}1)^3 < y^3\equal{}x^6\minus{}3x^3\plus{}3 < (x^2)^3$
so we r done!

Or we could just say that $ (n\plus{}3)(n^2\plus{}3n\plus{}3)\equal{}n^3\plus{}6n^2\plus{}12n\plus{}9$ must also be a cube but $ (n\plus{}2)^3\equal{}n^3\plus{}6n^2\plus{}12n\plus{}8$ and no two positive cubes differ by $ 1$.


Solve in positive numbers

$x+y^2+z^3=3$
$x^2+y^3+z=3$
$x^3+y+z^2=3$

Wlog, we may assume that $x \leq y,z.$
Then $x+x^2+x^3 \leq 3$ which leads to $(x-1)(x^2+2x+3) \leq 0$ so that $x \leq 1.$

In another hand, the bigger of the three numbers has to be not less than $1$.
- If $y,z \geq 1$ then :
$3 = x^3 + y + z^2 \leq x + y^2 + z^3 = 3$. Thus we are in the equality case and $x=y=z=1.$
- If $y \leq 1$ and $z \geq 1$ then :
$3 = x^2 + y^3 + z \leq x + y^2 + z^3 = 3$ which also leads to $x=y=z=1.$
- If $z \leq 1$ and $y \geq 1$ then :
$3 = x^3 + y + z^2 \leq x^2 + y^3 + z = 3$ which also leads to $x=y=z=1.$

Threfore, the only solution in positive numbers is $x=y=z=1.$

Pierre.


A person contributes $ Rs. 14$ per month from $ 1.1.1982$  to $ 31.12.1985$ and from $ 1.1.86$ to $ 31.12.89$ at $ Rs.28$ per month and from $ 1.1.90$ to $ 30.9.2008$ at $ Rs. 42$ per month. What will be his total contribution if interest is compounded quarterly at $ 10.8%
$

I do not wish to be sarcastic, but this was given to me by someone.I thought it was more linked to some maths rather than finance.

Thank you in advance.  :?:  :rotfl:

hello,
this should ideally be in other forum  :P

btw,check out for a chapter(on installments),you'll get a formula for such calculations.

I suppose question is more time consuming(taking in acc the 3 different situations)

P.S:In any exam i'ld have left this question  :rotfl:

Just have a look at any ICSE Class 10th book. Its a terribly lame question, which I would always get wrong without a calculator...

[quote]btw,check out for a chapter(on installments),you'll get a formula for such calculations.

I suppose question is more time consuming(taking in acc the 3 different situations)
[/quote]

trues :rotfl:  :huh:  :P  :maybe:

sorry for renewing this thread!  In this problem I have used the $ formula for annuity$ and have calculated the mode of RoI.Can the mode be assumed(The rate of interest was different for the different periods specified) Is my way of doing it correct? Please tell. :?:

[quote="skand"]I suppose question is more time consuming(taking in acc the 3 different situations)[/quote]
I think it's about some GP summation that was expected from the answerer, not just blunt calculation.

How many over here suck in commercial arithmetic?Maybe VAT?


Hello,

here is a code to illustrate the moon-earth-sun system with asymptote :

 [code]
import graph3;
import solids;
import three;
import animate;

// settings.render=2;
settings.tex="pdflatex";
settings.outformat="pdf";
currentprojection=orthographic(5,4,2);
currentlight=(epsilon,epsilon,epsilon);

size(15cm,0);

animation A;

real Rst=20, Rl=0.7, Rtl=5;
real ast=20, est=0.3, bst=ast*sqrt(1-est^2), cst=ast*est;
real atl=5, etl=0.8, btl=atl*sqrt(1-etl^2), ctl=atl*etl;

real xST(real t) {return ast*cos(t)+cst;}
real yST(real t) {return bst*sin(t);}
real zST(real t) {return 0;}

real xTL(real t) {return atl*cos(27*t);}
real yTL(real t) {return btl*sin(27*t);}
real zTL(real t) {return 0;}

real xLl(real t) {return Rl*cos(27*t);}
real yLl(real t) {return Rl*sin(27*t);}
real zLl(real t) {return 0;}

real xTt(real t) {return Rtl*cos(100*(t))/5;}
real yTt(real t) {return Rtl*sin(100*(t))/5;}
real zTt(real t) {return 0;}

real xl(real t) {return xST(t)+xTL(t)+xLl(t);}
real yl(real t) {return yST(t)+yTL(t)+yLl(t);}
real zl(real t) {return 0;}

real xt(real t) {return xST(t)+xTt(t);}
real yt(real t) {return yST(t)+yTt(t);}
real zt(real t) {return 0;}

real xL(real t) {return xST(t)+xTL(t);}
real yL(real t) {return yST(t)+yTL(t);}
real zL(real t) {return 0;}

path3 Pl=graph(xl,yl,zl,0,2*pi,1000), Pt=graph(xt,yt,zt,0,2*pi,3000),
Pts=graph(xST,yST,zST,0,2*pi,500);

picture pic;

draw(pic,Pl,lightgray);
draw(pic,Pt,lightblue);
draw(pic,Pts,blue+dashed);

draw(pic, shift(cst,0,0)*scale3(Rtl/2)*unitsphere, yellow, nolight);

surface terre=scale3(Rtl/5)*unitsphere;
surface lune=scale3(Rl)*unitsphere;

int n=20; // for short compilation
real step=2pi/n;
for(int i=0; i<n; ++i) {
  real k=i*step;
  add(pic);
  draw(shift(xL(k),yL(k),0)*lune,lightgray);
  draw(shift(xST(k),yST(k),0)*terre,lightblue);
  A.add();
       // shipout(prefix="terrelune"+string(i),view=false);
       erase();
       }

label(A.pdf(BBox(1mm,Fill(Black)), multipage=false));
[/code]

Unfortunately, lights on earth(terre) and moon(lune) don't work.
Uncommenting  settings.render=2; make lights right but compilation bug after 3 pdf-files creations.

Thanks for any help

--
Guillaume
[/code]

[asy]import graph3;
import solids;
import three;
import animate;

// settings.render=2;
settings.tex="pdflatex";
settings.outformat="pdf";
currentprojection=orthographic(5,4,2);
currentlight=(epsilon,epsilon,epsilon);

size(15cm,0);

animation A;

real Rst=20, Rl=0.7, Rtl=5;
real ast=20, est=0.3, bst=ast*sqrt(1-est^2), cst=ast*est;
real atl=5, etl=0.8, btl=atl*sqrt(1-etl^2), ctl=atl*etl;

real xST(real t) {return ast*cos(t)+cst;}
real yST(real t) {return bst*sin(t);}
real zST(real t) {return 0;}

real xTL(real t) {return atl*cos(27*t);}
real yTL(real t) {return btl*sin(27*t);}
real zTL(real t) {return 0;}

real xLl(real t) {return Rl*cos(27*t);}
real yLl(real t) {return Rl*sin(27*t);}
real zLl(real t) {return 0;}

real xTt(real t) {return Rtl*cos(100*(t))/5;}
real yTt(real t) {return Rtl*sin(100*(t))/5;}
real zTt(real t) {return 0;}

real xl(real t) {return xST(t)+xTL(t)+xLl(t);}
real yl(real t) {return yST(t)+yTL(t)+yLl(t);}
real zl(real t) {return 0;}

real xt(real t) {return xST(t)+xTt(t);}
real yt(real t) {return yST(t)+yTt(t);}
real zt(real t) {return 0;}

real xL(real t) {return xST(t)+xTL(t);}
real yL(real t) {return yST(t)+yTL(t);}
real zL(real t) {return 0;}

path3 Pl=graph(xl,yl,zl,0,2*pi,1000), Pt=graph(xt,yt,zt,0,2*pi,3000),
Pts=graph(xST,yST,zST,0,2*pi,500);

picture pic;

draw(pic,Pl,lightgray);
draw(pic,Pt,lightblue);
draw(pic,Pts,blue+dashed);

draw(pic, shift(cst,0,0)*scale3(Rtl/2)*unitsphere, yellow, nolight);

surface terre=scale3(Rtl/5)*unitsphere;
surface lune=scale3(Rl)*unitsphere;

int n=20; // for short compilation
real step=2pi/n;
for(int i=0; i<n; ++i) {
  real k=i*step;
  add(pic);
  draw(shift(xL(k),yL(k),0)*lune,lightgray);
  draw(shift(xST(k),yST(k),0)*terre,lightblue);
  A.add();
       // shipout(prefix="terrelune"+string(i),view=false);
       erase();
       }

label(A.pdf(BBox(1mm,Fill(Black)), multipage=false));[/asy]
It doesn't work on the forum...

[quote="AIME15"]It doesn't work on the forum...[/quote]
We don't care.
Asymptote is a real software, not exclusively a programming language for (this) forum(s).


In response to the attempted airline terrorism, some new rules have been implemented.

http://news.yahoo.com/s/ap/us_airline_attack_regulations

These rules are ridiculous.

"Some airlines were telling passengers on Saturday that new government security regulations prohibit them from leaving their seats beginning an hour before landing."

"Air Canada said in a statement that new rules imposed by the Transportation Security Administration limit on-board activities by passengers and crew in U.S. airspace. The airline said that during the final hour of flight passengers must remain seated. They won't be allowed access to carryon baggage or to have any items on their laps."

What if you need to use the restroom?  You are also barred from accessing any items at all during that hour.  Also, terrorists would just take action when the plane takes off.  Reasonable or foolish paranoia?  Discuss.

A bit of both.  Remember, they have to look like they're doing something in response to these threats.  It's not like they're (they=government) just going to sit there and twiddle their thumbs at such an event.

I hate the rules too...but if they did nothing, then people would be even angrier.

What would you propose in place of the current extra regulations (one hour is a bit harsh...)?

A terrorist could still take action before the hour of restriction.  Unless the restrictions are extended to cover the entire flight duration, there will still be attack opportunities.  Of course, doing so would be outrageous.

Instead of trying to stop terrorists from taking action on the plane, why not focus on keeping them off the plane in the first place?  If a terrorist successfully makes it onto the plane, they will likely carry through with their plan regardless of any rules.  Even though they aren't supposed to stand up/access items, they would anyway.  Perhaps the TSA should try screening improvements and implement/improve background checks when purchasing tickets.  After all, the suspected terrorist's father tried to warn officials that his son was radicalized.

And the chance that terrorists make it on the plane are and probably will remain really high unless the TSA inconvenience every single passenger by putting extra extra guidelines.

There was a study that showed that people carrying potentially explosive chemicals got by the TSA most of the time, so checking is clearly not quite foolproof.

There will be a tradeoff between convenience/cost to maintain such a stringent system and consumer happiness.  It won't be easy. ever.

These kinds of measures are not meant to stop terrorist attacks -- only to deter people from performing them. The man sewed an explosive into his underwear, if I'm not mistaken. What are you going to do about that? If someone [i]really[/i] wants to blow up a plane, they can.

And if he just didn't screw up, it would have happened, which would have been a great tragedy to the integrity of the TSA and to security in general.  I mean, unless you completely strip everyone down to their underwear, there is absolutely no way you'll catch 99% of potential terrorists.  (That is, unless our technology gets much better)

It is said the guy had enough explosives to take down the whole plane, so those passengers really lucked out from his inability to do so.

[quote="TZF"]If someone [i]really[/i] wants to blow up a plane, they can.[/quote]  This is obviously not true, and the current case is a clear counterexample.  Security measures designed to stop incompetent people who already are thwarted by some random other passengers and a fire extinguisher are pointless -- those people are already going to be prevented by random other passengers and fire extinguishers.  Similarly, the idea that everyone should have to specially have their shoes scanned after a demonstration that shoe bombs work very, very poorly is just ridiculous.  The main aggregate effect of these sorts of security theater is that they make flying a truly miserable experience.

The claim that the TSA needs to "do something" does not stand up to even mild scrutiny.  If this is your position, you should actually defend it, not just assert it and then resign yourself to utterly pointless regulation.  The true related claim is that some people somewhere [i]believe[/i] that it is necessary to "do something" in response to this non-event, but the fact that they believe it does not actually mean that they are right.  And in any case, the need to "do something" does not justify doing things that are self-evidently pointless.

Finally, if you're going to cite a study, you should do so in a way that allows other people to actually find out what you're talking about (e.g., a reference to an article about the study, or even better a link to the study itself).

Note also that the main goal of terrorists is not just to kill a few random people and destroy a few million dollar worth of equipment, which is next to nothing on the global scale (we've got much more people killed on the highways on an average day and more property damage from an average snowstorm) but to disrupt the normal life in the country and to plant fear in its citizens. From this point of view, this attack was quite successful (mainly due to the decisions of our wise security men): the new inconveniences will be felt by thousands of passengers for a long period of time.

Interestingly enough, in Israel, where the terrorist threat is much more real, they do not find it necessary to react to every occasion by introducing new senseless regulations. As a matter of fact, I do not remember a single case when the Israeli government would respond to a terrorist attack by making lives of their own citizens more complicated. True, they introduced security checks in many public places but, once you pass through the security, you are no longer "regulated" in any way beyond the common sense.

So, I would be OK with increased scrutiny at the check points and delays associated with that, but to treat passengers as suspects [b]after[/b] they cleared the security seems just ridiculous. The message sent by such regulations is "We trust neither you, nor our security checks". Is it what TSA wants to tell the USA citizens and visitors?

I acknowledge that the study may not have been up-to-date with the technology we have now to screen potential passengers.  (I read it somewhere, so I don't exactly remember it.)[url]
http://www.cbsnews.com/htdocs/pdf/gao_aviation_security.pdf[/url]

found it.

There's no guarantee that the TSA has to do anything.  However, it is likely that people would protest that the government is just a figurehead.  If they did absolutely nothing about this incident, there would have been an uproar.  In this case, it's better safe than sorry, because without some sort of extra restrictions, some terrorist somewhere might be able to slip on a plane headed for the U.S. and have the opportunity to detonate it.

Flying in this harried fashion is not fun, and I know it.  I was pretty annoyed when I got checked a few months ago.  However, at least I felt safe (until now...)

I also read that the full-body scanning machines that would have caught the culprit before he got on the airplane weren't in use/weren't at the airport where he board.  Investigators speculated that the Nigerian airport was the one without the scanning.

[url]http://news.yahoo.com/s/ap/20091228/ap_on_go_ot/us_airliner_attack_explosive[/url]

Whew.

[quote="JBL"][quote="TZF"]If someone [i]really[/i] wants to blow up a plane, they can.[/quote]  This is obviously not true, and the current case is a clear counterexample.[/quote] Of course, not anyone can blow up a plane. But the truth about terrorism is it is impossible to stop all of it. Everyday, new techniques/technologies are being made that can get past our current screening (at least our primary screening.)

As to the original question, I don't really think the government is being paranoid. As most people in this thread said, they are responding to some of their constituents who are being paranoid.

To say "better safe than sorry" [url=http://en.wikipedia.org/wiki/Begs_the_question]begs the question[/url] by presupposing that the security measures taken actually will increase safety.  [i]This[/i] is the proposition that needs defending.

Here's a thought experiment for you: how likely is it that in less than 24 hours, the people at the TSA were able to invent security procedures that will measurably increase safety and that no one had ever thought of or implemented before?  My [i]a priori[/i] estimate of that probability is 0.  Now, why should this estimate change because somebody lit his underwear on fire?  If anything, it makes me think these measures are [i]less[/i] useful: useful security measures require some nontrivial amount of careful thinking, planning and implementation.

Ihatepie: that's a much, much, much weaker claim.  But what does it have to do with anything?  I mean, if you're resigned to occasionally having something blown up, why therefore is it necessary to have people not get out of their seats for the last 60 minutes of every flight?  As far as I can tell, these two things (the inevitability of terrorist success and the implementation of a particular set of pointless and ridiculous security procedures) have nothing whatsoever to do with each other.*

* Actually, this reminds me of the old comment that the food at this restaurant is just truly awful -- and the portions are so small!  If security is inept and doomed to failure, why should we do more of it?  I happen to agree that occasionally things will get blown up regardless of the state of security, by the way, but if anything, this suggests to me that we should make security procedures as unmiserating as possible.

[quote="JBL"]Ihatepie: that's a much, much, much weaker claim.  But what does it have to do with anything?  I mean, if you're resigned to occasionally having something blown up, why therefore is it necessary to have people not get out of their seats for the last 60 minutes of every flight?  As far as I can tell, these two things (the inevitability of terrorist success and the implementation of a particular set of pointless and ridiculous security procedures) have nothing whatsoever to do with each other. [/quote][url=http://en.wikipedia.org/wiki/Straw_man]Straw man.[/url] I never said once that I thought the TSA's recent actions were justified. I was merely restating what I think TZF meant to say in his last sentence.

[quote="fedja"][quote="Ihatepie"]
I don't really think the government is being paranoid. As most people in this thread said, they are responding to some of their constituents who are being paranoid.[/quote]
Yeah, those who have the least common sense are the ones who are most likely to "uproar", so let's make our decisions based on their opinions :P. A friend of mine called all such measures "an imitation of vigorous activity".[/quote] I wish our society wasn't like this, and I agree that our government shouldn't govern by the whims of some crazy people or even sometimes the general populace (I have no hard data to back this up, but I assume that most people in America want the TSA to do something.) This actually reminds me of something I recently heard on a 60 minutes interview with Obama. It went something like this:

Interviewer: With recent polls showing that the majority of the American people disagree with sending more troops to Afghanistan, are you rethinking your position of doing so?
Obama: No.
Interviewer: Why is that?
Obama: Because I was elected to serve the American people. From all of the evidence I've been given, my advisors and I think that the best possible course in Afghanistan is to send more troops. What the polls say has no bearing on that.

As JBL said, I doubt the TSA could think of an effective security measure in 24 hours and I hope that they will think longer about any future measure before it's put in place.

Just wondering. How many of you guys who are actively posing views on this matter did/are doing speech/debate in school?  (Where you pull up a term like "straw man" I never heard used before  :| )

I'm not trying to say that the TSA absolutely has to do something about this terror act.

[quote="JBL"]To say "better safe than sorry" begs the question by presupposing that the security measures taken actually will increase safety[/quote]

Obviously, that's no guarantee.  However, in the line of events that have happened in the last decade or so, lowering that already infinitesimal probability may save a plane.  (I acknowledge it may not, because this is all hypothetical.)  There are many products that abide by that standard of prevention.  I mean, would you rather get vaccinated or get really sick?

[quote="JBL"]Incidentally, if you live in South Carolina, you might give your senator a call and ask him why he thinks union-busting is more important than having a TSA director:
http://www.mcclatchydc.com/227/story/81356.html[/quote]

It's these things that make our government screwy.  :(

[quote]I don't really think the government is being paranoid. As most people in this thread said, they are responding to some of their constituents who are being paranoid. [/quote]

word.

[quote="zapi2007"]Just wondering. How many of you guys who are actively posing views on this matter did/are doing speech/debate in school?  (Where you pull up a term like "straw man" I never heard used before  :| )[/quote]I actually don't do debate, but I frequent another forum that also has a similar debate section. People use words like that over there.

I just thought I'd try it out here. :)

It's going to boil down to how the policies are actually carried out, since the execution is in the hands of ordinary people who happen to have particular jobs. I remember visiting Florida about a year after 9/11 and getting a plastic model rocket from the Kennedy Space Center, and getting hell for it when I was trying to get on the plane. The fact that it looks like a rocket does not even remotely suggest that it is actually a rocket, but they confiscated it. Additionally, I was forced to break the tips the pencils in my backpack.


integration
calculus
calculus computations

[quote="levi"]:spider: $Prove\ that\ \int_{0}^{\frac{\pi}{2}}sinx\sqrt{sin2x}\ dx\ =\ \int_{-1}^{1}\sqrt{1-x^2}\ dx$
$....................only\ with\ the\ substitution\ t=\delta (x)=sinx-cosx\ !$[/quote]

Let I=$\int_{0}^{\frac{\pi}{2}}sinx\sqrt{sin2x}\ dx$
So, I=$\int_{0}^{\frac{\pi}{2}}cosx\sqrt{sin2x}\ dx$

adding we get

2I=$\int_{0}^{\frac{\pi}{2}}(sinx+cosx)\sqrt{sin2x}\ dx$

Now substitue $Sinx-Cosx=t$ and $(Cosx+Sinx)dx=dt$

and when x=0, then t= -1 and when ${x=\frac{\pi}{2}}$, then t=1
and use $Sin2x=1-(sinx-cosx)^2$

But this will give me $2I=\ \int_{-1}^{1}\sqrt{1-x^2}\ dx$

Levi...Pl check

It is O.K. I think that this solution is the shortest and the best solution !


limit
logarithms
trigonometry
algebra unsolved
algebra


Hey, can you show how you solved it. I'm really interested in the number theory formulas you used.

Turn it into $(1991-1)^{1991^{1992}}+(1991+1)^{1991^{1990}}$, use binomial expansion, and see (and prove without much difficulty) that the term divisible by the smallest power of $1991$ is $(1991^{1992}+1991^{1990})1991$


Does someone have the official solutions of BMO 2009? Please, I need them. Can I find them anywhere?

Click on the problems to see the solutions.

http://www.mathlinks.ro/resources.php?c=1&cid=19&year=2009

Video solutions to BMO1 from December 2009 are temporarily available here: http://www.ukmt.org.uk/


algebra polynomial geometry calculus integration function Putnam

If you know how to solve this already, don't post a solution.  Let other people play with it for a while first.

Give $\frac{a}{5}+\frac{b}{4}+\frac{c}{3}+\frac{d}{2}+e=0$, prove that the polynomial $ax^4+bx^3+cx^2+dx+e$ has at least one real zero.

If we integrate it and get the area under the curve from 0 to 1, we get it to be zero, so obviousally since this graph dosnt run along the x-axis (which would be right anyway) it must cross the x axis somewhere between 0 and 1. Am i close?

And actually since the graph dosnt go along the x-axis, it has to be above it and below it, therefore this isnt a double root, and since it is an equation of even degree that means it has two real roots.

Am i right? Or what other way is there to do it?

I think you've got the basic idea -- [hide]for the integral to be 0, either 1) the function is uniquely zero, in which case we're done, or the function is not uniquely zero in which case it must have both positive and negative values (in order to have 0 integral) and thus must have a zero.[/hide]

Nice question!

Ah i see, it can be the zero polynomial if $a=b=c=d=e=0$ whoops! :blush:

Yes, it's a lovely problem, and the solution above is probably the nicest one.  The problem appeared in the English Mathematical Tripos in 1929 (according to Hardy's [i]Course of Pure Mathematics[/i]), and the Putnam exam in February 1958.

You might wonder whether calculus is required.  Actually, there is a more elementary solution.  For simplicity, I'll give an elementary proof of the quadratic case: If $\frac{a}{3} + \frac{b}{2} + c = 0$, then $ax^2 + bx + c$ has a real zero.  The proof method can be extended to any degree.

[hide="Proof"]
Let $r = \frac{1}{2} + \frac{1}{\sqrt{12}}$ and $s = \frac{1}{2} - \frac{1}{\sqrt{12}}$.  If you evaluate the quadratic at r and s, and add up the two results, you get $2(\frac{a}{3} + \frac{b}{2} + c)$, which by hypothesis is 0.  Thus one of the values is nonnegative and one is nonpositive.  By continuity, there is a zero in between.

That solution gives the further information that there is a root of the quadratic in $[0.2, 0.8]$, not just $[0, 1]$.  The solution is a little messier than the calculus approach, but it's nice to see more than one approach.
[/hide]

This is a silly question perhaps, but where did you get your r and s from?

I worked backward.  To make the method work, I needed the equations
$\frac{r^2 + s^2}{2} = \frac{1}{3}$ and $\frac{r+s}{2} = \frac{1}{2}$.  I then solved those equations with a little algebra.

Here is a probabilistic way to look at what's going on.  The original proof relied on the fact that, if X is chosen with a uniform distribution from $[0, 1]$, then we have the expected values $E(X) = \frac{1}{2}$, $E(X^2) = \frac{1}{3}$, $E(X^3) = \frac{1}{4}$, and $E(X^4) = \frac{1}{5}$.  But that's not the only distribution with those properties.  There are discrete distributions with those properties.  For example, we can choose X to be equally likely one of four numbers p, q, r, and s to make those four expected values work out.  You can either take that fact for granted, or I give some details below in case you are interested.

[hide="Details"]We want to choose p, q, r, and s so that their average is $\frac{1}{2}$, the average of their squares is $\frac{1}{3}$, the average of their cubes is $\frac{1}{4}$, and the average of their fourth powers is $\frac{1}{5}$.  By using Newton's sums, we find that those four numbers are the real roots of the polynomial $45x^4 - 90 x^3 + 60 x^2 - 15x + 1$.  Those roots are in $[0.1, 0.9]$, so we have shown there is a root of Magnara's original problem in the range $[0.1, 0.9]$.

By the way, that quartic polynomial is symmetric around $x = \frac{1}{2}$, and thus we can solve it as a quadratic in a quadratic.  We find that the values of p, q, r, and s are
\[\frac{1}{2}\left(1 \pm \sqrt{\frac{5 \pm 2\sqrt{5}}{15}} \, \right).\]
[/hide]

A variant of Ravi B's argument:

the approximate integration formulae such as Simpson's rule, which sample a
function at some number of points in an interval with some positive
weights, are exact for polynomials of small degree (relative to the number of points).

So there is some explicit list of positive numbers $w_i > 0$ and $0 \leq p_i \leq 1$  for which the integral of
given polynomial on $[0,1]$ is $\Sigma w_i f(p_i)$.
This sum will be zero for the given polynomial, by direct calculation, so $f$ must change sign in the interval.

Oh yeah, that's a better way to think about it.

Here's a followup question.  Let's fix the degree of the polynomial.  What's the smallest value of c such that we can guarantee that the polynomial has a root in the range $[\frac{1-c}{2}, \frac{1+c}{2}]$?

For the quadratic case, the answer is $\frac{1}{\sqrt{3}}$.  My first post shows that this value of c guarantees a root, while the quadratic $6x^2 - 6x + 1$ shows that this value of c is optimal.

I don't the answer for cubic polynomials or higher.

[quote="Ravi B"]Here's a followup question. Let's fix the degree of the polynomial. What's the smallest value of c such that we can guarantee that the polynomial has a root in the range $\left[\frac{1-c}{2}, \frac{1+c}{2}\right]$?
[/quote]
Let the degree be $2k-1$ or $2k-2$. Define $p_k$ to be the $k-th$ derivative of $x^k(1-x)^k.$ The polynomials $1,p_1,p_2,\dots,p_k$ are an orthogonal basis for the polynomials with degree $\le k$ (the inner product here is the integral of the product on $[0,1]$), so by the theory of Gaussian quadrature, we can choose $x_1<x_2<\dots<x_k$ to be the roots of $p_k$ and find weights $w_j$ so that $\sum_i w_if(x_i)=\int_0^1f(t)\, dt$ for any polynomial $f$ of degree less than or equal to $2k-1.$
Then since some of the $f(x_i)$ must be nonnegative and some nonpositive, there is a root in $[x_1,x_k].$ To prove that this is the best we can do, let $g_k(x)=\frac{(p_k(x))^2}{(x-x_1)(x-x_k)}.$ $g$ is positive on $[0,x_1)$ and $(x_k,1]$ and nonpositive on $[x_1,x_k],$ so we should be able to add some small quantity to move it below the axis in the middle and above the axis at $x_1$ and $x_k$, thus moving the outer roots in by a tiny amount. The appropriate function to add, for $k\ge2$, is a small multiple of $g_{k-1}.$

On fleeting_guest's question: there are plenty of cheap answers. $f(a)=0$ is a linear equation in the coefficients.

[quote="Myth"][quote="fleeting_guest"]Another question.

Is there any other single linear equation on the coefficients that forces existence of a root in the interval [0,1]?[/quote]
Indeed, $\frac{a}{160}+\frac{b}{64}+\frac{c}{24}+\frac{d}{8}+\frac{e}{2}=0$.  :lol:  :lol:  :lol:[/quote]

Nice.  The functionals I had in mind were integrals of $f(x)W(x)$ with $W(x) > 0$.
Your example shows that $W(x) \geq 0$ is also good, and jmerry's observation shows that we can take take limits of these, or use distributional $W(x)$ as well.

[quote="jmerry"]In Stieltjes language, $\int f\, dY=0$ where $Y$ is a monotone function on $[0,1]$. That should cover everything, and all such functionals work..[/quote]

Yes, "everything" should be in the convex hull of Dirac delta functionals $f(a)$, in some appropriate sense of convex hull (finite linear combinations are not enough, I think). Stieltjes integrals can certainly reproduce those.

[quote="jmerry"]The polynomials $1,p_1,p_2,\dots,p_k$ are an orthogonal basis for the polynomials with degree $\le k$ (the inner product here is the integral of the product on $[0,1]$), so by the theory of Gaussian quadrature, we can choose $x_1<x_2<\dots<x_k$ to be the roots of $p_k$ and find weights $w_j$ so that $\sum_i w_if(x_i)=\int_0^1f(t)\, dt$ for any polynomial $f$ of degree less than or equal to $2k-1.$[/quote]
Thanks for your cool proof, jmerry.  Are the weights $w_i$ guaranteed to be nonnegative?  (I think we need nonnegativity to continue the rest of the proof.)  Sorry if my question is naive, but I don't know Gaussian quadrature very well.  Do you happen to know an online resource that describes the theorem you are using here?

[quote="jmerry"]To prove that this is the best we can do, let $g_k(x)=\frac{(p_k(x))^2}{(x-x_1)(x-x_k)}.$ $g$ is positive on $[0,x_1)$ and $(x_k,1]$ and nonnegative on $[x_1,x_k],$[/quote]
I think you mean nonpositive on $[x_1, x_k]$.

I realized we neede positive weights after I posted that. Here's the proof: Consider the nonnegative polynomial $g(x)=\left(\frac{p_k(x)}{x-x_j}\right)^2.$ Since its degree is only $2k-2,$ $\int_0^1g(x)\, dx=w_jg(x_j)>0$ and $w_j>0.$

I don't have a specific resource; this belongs to numerical analysis and is well known. The standard interval for an exposition of Gaussian quadrature is $[-1,1]$, which would make these the Legendre polynomials.
An outline of the proof:
First, we verify that the $p_k$ are orthogonal using integration by parts: ($k>j$) $\int_0^1 p_k(x)p_j(x)\, dx$
$=\int_0^1 ((x-x^2)^k)^{(k)}((x-x^2)^j)^{(j)}=\int_0^1 ((x-x^2)^k)^{(k-j-1)}((x-x^2)^j)^{(2j)}=0$
The boundary terms are all zero because of the zeros in the integrated polynomial.
Use the convention $p_0=1.$ The orthogonality still applies.
Next, we have a linear independence result: we can choose the $w_j$ such that $\sum_i w_ix_i^j=\frac1{j+1}$ for each $j$ between 0 and $k-1.$ This is in fact a $k\times k$ linear system, and the coefficient matrix is invertible by the Vandermonde determinant.
Now, for any polynomial $f$ of degree $\le 2k-1$, we can write $f(x)=p(x)g(x)+h(x)$ for some polynomials $g,h$ of degree $\le k-1.$ By orthogonality, $\int_0^1 p(x)g(x)\, dx=0$, and $p(x_i)g(x_i)=0.$ Then by our choice of $w_i,$ $\sum_i w_ih(x_i)=\int_0^1 h(x)\, dx$ and $\sum_i w_if(x_i)=\int_0^1 f(x)\, dx.$

I'll edit that "nonnegative" mistake.

Here's another approach:

Consider an indefinite integral of the quartic and apply Rolle's theorem.


vector
linear algebra
matrix
induction
linear algebra unsolved

This should be simple. Let $ T$ be a linear operator mapping a finite dimensional vector space $ V$ into itself.

Suppose $ T$ is nilpotent. Then there is a basis of $ V$ with respect to which the matrix represent $ T$ contains $ 0$'s on and below the main diagonal.

This is essentially Schur's Lemma. You usually see Schur's Lemma stated for complex scalars, but the only reason for that is that you need to know that eigenvectors exist. In the case of a nilpotent matrix, we do know that.

Let's walk through the proof, anyway. I'll phrase this in terms of matrices; a change of basis will be accounted for by a similarity of matrices. So suppose $ A$ is a nilpotent matrix. Find a vector $ v$ such that $ Av=0.$ (This must exist, because A is not invertible.) Now, extend $ v$ to a basis. That is, find a basis for $ F^{n}$ such that $ v$ is the first vector of that basis. Let $ P$ be the matrix whose columns are that basis; $ P$ is invertible.

$ AP$ is then a matrix whose first column is zero.

$ P^{-1}AP$ has the following $ [1+(n-1)]\times[1+(n-1)]$ block structure:

$ P^{-1}AP=\begin{bmatrix}0&*\\0&A'\end{bmatrix}=B.$

Now, $ B^{m}=\begin{bmatrix}0&*\\0&(A')^{m}\end{bmatrix}$ by direct calculation.

Since $ B$ is similar to $ A$, it is also nilpotent, so there is some $ m$ such that $ B^{m}=0.$ It follows that $ (A')^{m}=0$ and $ A'$ is also nilpotent.

Now the proof becomes a proof by induction on $ n,$ the size of the matrix. We assume that the theorem is true for $ (n-1)\times(n-1)$ matrices. The inductive hypothesis show that there exists $ Q$ such that $ Q^{-1}A'Q=T,$ where $ T$ has zeros on or below the main diagonal. Then

$ \begin{bmatrix}1&0\\0&Q^{-1}\end{bmatrix}\begin{bmatrix}0&*\\0&A'\end{bmatrix}\begin{bmatrix}1&0\\0&Q\end{bmatrix}=\begin{bmatrix}0&*\\0&T\end{bmatrix}$

And thus $ A$ is similar to that last matrix, which has zeros on and below the main diagonal.

Sorry. I posted here this prob, cos the solution in the notes that i was reading "seemed to be wrong".  but i realise it might be right and i do not comprehend its solution.

The proof is basically like this:

Pick any set of vectors that form a basis for Ker$ T$.  Extend this set to form a basis for Ker$ T^{2}$.  Continue this process until the set forms a basis for $ V$.  This will always happen since $ T$ is nilpotent.

I do not understand why this basis work.

If $ v \in \ker T$ then $ Tv = 0$.  For $ m > 1$, if $ v \in \ker (T^{m})$ then $ Tv \in \ker (T^{m-1})$.  Interpret that as a statement about nonzero entries in the matrix and you're done.

sorry. i think i'm quite dense.

but i rilli dun get the interpretation part.

You want $ U^{-1}T U = K$, where $ K$ is strictly upper triangular.

That's equivalent to $ TU = UK$, i.e. $ T U_{*i}= U K_{*i}, \, \forall i \in \overline{1,n}$ (here $ A_{*i}$ means the $ i$-th column of $ A$).
For each $ j$, we have that $ U_{*j}\in \ker \left( T^{p}\right)$, where $ p>0$. Hence, $ T U_{*j}\in \ker \left( T^{p-1}\right)$.
Now note that the span of $ U_{1}, \ldots, U_{j-1}$ includes $ \ker \left( T^{p-1}\right)$. Therefore, $ K_{*j}$ has $ 0$'s on the positions $ j, j+1, \ldots, n$.

Is it clear now?


geometry
circumcircle
perimeter
inequalities
inradius
geometry proposed

Let $ ABC$ be a triangle with circumradius $ R$, perimeter $ P$ and area $ K$. Determine the maximum value of: $ \frac{KP}{R^3}$.

Hope it right
We always have:
$ R\equal{}\frac{a}{2sinA}\equal{}\frac{b}{2sinB}\equal{}\frac{c}{2sinC}\equal{}\frac{P}{2(sinA\plus{}sinB\plus{}sinC)}$
$ K\equal{}\frac{abc}{4R}$
So:
$ \frac{KP}{R^3} \equal{} \frac{abcP}{4R^4} \equal{}$ $ \frac{abcP}{4\frac{P^4}{16(sinA\plus{}sinB\plus{}sinC)^4}}\equal{}$ $ \frac{4abc(sinA\plus{}sinB\plus{}sinC)^4}{P^3}$ $ \equal{} \frac{4abc}{(a\plus{}b\plus{}c)^3} . (sinA\plus{}sinB\plus{}sinC)^4$

By AM-GM inequality: $ \frac{16abc}{(a\plus{}b\plus{}c)^3} \le \frac{16abc}{27abc} \equal{} \frac{4}{27}$
Easily we get: $ sinA\plus{}sinB\plus{}sinC \le \frac{3\sqrt{3}}{2}$ so $ (sinA\plus{}sinB\plus{}sinC)^4 \le \frac{729}{16}$

Equality holds when $ ABC$ is an equilateral triangle
Finally, $ Max\frac{KP}{R^3}\equal{} \frac{4}{27} . \frac{729}{16} \equal{}$  [color=blue][size=200] [b]6.75[/b][/size][/color]
:lol: :lol:  :lol:

$K=\frac{1}{2}absinC$

$P=a+b+c$

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R$

$\frac{KP}{R^3}=\frac{1}{2}\frac{(2RsinA)(2RsinB)(sinC)2R(sinA+sinB+sinC)}{R^3}$

$=4sinAsinBsinC(sinA+sinB+sinC)$

now,$sinA,sinB,sinC\geq 0$

so, by AM-GM

$4sinAsinBsinC(sinA+sinB+sinC)\leq \frac{4}{27}(sinA+sinB+sinC)^4
\leq \frac{4}{27}(\frac{3\sqrt3}{2})^4=\frac{27}{4}$

so, $\frac{KP}{R^3}\leq\frac{27}{4}$ with equality in case of a equilateral triangle.

so, $max\frac{KP}{R^3}=\frac{27}{4}$

[b]sumanguha[/b]'s proof is the correct one, here is an easier way :
Denote $s$,$R$,$r$ the semi-perimeter, circumradius and inradius of $\triangle ABC$ respectively.
So : $\frac{KP}{R^3} = \frac{2s^2r}{R^3}$
According to Mitrinovic's Inequality :
$s \le \frac{3\sqrt{3}R}{2}$
$\Leftrightarrow s^2 \le \frac{27}{4} R^2$
And : $2r \le R$
Multiplying the two inequalities , we get : $2s^2r \le \frac{27}{4} R^3 \Leftrightarrow \frac{KP}{R^3} \le \frac{27}{4}$

Alternatively, $KP/R^3=abc(a+b+c)/4R^4$.  Now, this is then $4sin(a)sin(b)sin(c)(sin(a)+sin(b)+sin(c))$ which is at most $4(sin(a)+sin(b)+sin(c))((sin(a)+sin(b)+sin(c))/3)^3=(4/27)(sin(a)+sin(b)+sin(c))^4$.  By Jensen's, $sin(a)+sin(b)+sin(c)$ is at most $3sqrt(3)/2$, so this is at most $(4/27)(3sqrt(3)/2)^4=27/4$ and we're done.

[size=150]Answer:[/size]
\[\frac{KP}{R^3}\le \frac{27}{4}\]
[size=150]Proof:[/size]
First, we will show that \[\textcolor{blue}{3\sqrt 3 R\ge P.}\]
As $\sum_{cyc} \sin A=\sum_{cyc} \frac{a}{2R}=\frac{P}{2R}$, it's suffice to show that $\sum_{cyc} \sin A\le \frac{3\sqrt 3}{2}.$ This follows directly from Jensen's Inequality. $ \square$
Then, we have \[4\sqrt 3KP=\frac{\sqrt 3 abcP}{R}\le 9abc \implies KP\le \frac{9abc}{4\sqrt 3}.\] Meanwhile, $3\sqrt 3R \ge P\ge 3\sqrt[3]{abc} \implies \textcolor{blue}{\frac{1}{R^3}\le \frac{3\sqrt 3}{abc}}$, so \[\frac{KP}{R^3}\le \frac{9abc}{4\sqrt 3}\cdot \frac{3\sqrt 3}{abc}=\frac{27}{4}.\]


function
probability
integration
algebra
polynomial
expected value
geometry unsolved

We have n rods.
We place them end to end in a st line.
Earch is of length equally likely to be anything from 0 to 1 yard long.
How long, and with what probalility and uncertainty is the line of n rods, end to end?

I'm not sure what you mean by "uncertainty."  The expected value of the length of the sum is the expected value of the sum of the lengths, or $ \frac{n}{2}$.

[quote="t0rajir0u"]The expected value of the length of the sum is the expected value of the sum of the lengths, or $ \frac {n}{2}$.[/quote]

Yes t0r. That's the [b]most[/b] probable case. But I think he wants the [url=http://en.wikipedia.org/wiki/Probability_function]Probability density function[/url]. Then all information (i.e. What is the probability that the total length will be greater than $ \frac {2n}{3}$ etc.) will be available. You said that the maximum value of the function is achieved at $ \frac {n}{2}$.

Let $ f_i: \mathbb R\to \mathbb R\ \forall i\in \mathbb N$ be the probability density function when $ i$ rods  are placed end to end.

That means that $ \int_{a}^{b}f_i(x)\ dx$ is the probability of having the total length between $ a$ and $ b$ when $ i$ rods are placed end to end.

It is given that $ \begin {cases}f_1(x) \equal{} 0\ \forall x < 0 \\
f_1(x) \equal{} 1\ \forall x, 0\leq x\leq 1 \\
f_1(x) \equal{} 0\ \forall x > 1 \end {cases}$

We also have this: $ f_{n \plus{} 1}(x) \equal{} \int_{x \minus{} 1}^{x} f_n(t)\ dt$

For example $ \begin {cases}f_2(x) \equal{} 0\ \forall x < 0 \\
f_2(x) \equal{} x\ \forall x, 0\leq x\leq 1 \\
f_2(x) \equal{} 2 \minus{} x\ \forall x, 2\geq x > 1 \\
f_2(x) \equal{} 0\ \forall x > 2 \end {cases}$

I believe that between any two consecutive integers $ f_n(x)$ is a polynomial of degree $ n \minus{} 1$
Now we need a general formula for $ f_n(x)$.

P.S. This doesn't belong to this forum...


vector
abstract algebra
linear algebra
linear algebra unsolved

Let $ A,B\in \mathcal{M}_n(\mathbb{K})$ with same rank.
Denote $ \Gamma\equal{}\{M \in \mathcal{M}_n(\mathbb{K}) |  AM\equal{}BM \}$.
Find a sharp bound for $ \mathrm{dim} \Gamma$.

Let $ U\otimes{V}: X\rightarrow{UXV^T}$. Then $ \Gamma=ker((A-B)\otimes{I})$.
Let $ (u_i)_{i\leq{r}},(v_j)_{j\leq{s}}$ be the non zero singular values of $ U,V$ where $ r=rank(U),s=rank(V)$. Then the non zero singular values of $ U\otimes{V}$ are $ (u_iv_j)_{ij}$ and $ rank( U\otimes{V})=rank(U)rank(V)$.
Thus $ dim(\Gamma)=n^2-n\times{rank(A-B)}$

Rewrite $ AM \equal{} BM$ as $ \left(A \minus{} B\right)M \equal{} 0$, or, equivalently, as $ \mathrm{Im}M\subseteq\mathrm{Ker}\left(A \minus{} B\right)$. Thus, $ \Gamma$ is isomorphic to the vector space of all linear maps from $ \mathbb{K}^n$ to $ \mathrm{Ker}\left(A \minus{} B\right)$. Computing the dimension is obvious. No need for tensor products or singular values here, loup blanc.

By the way, I would denote the linear map $ X\mapsto UXV^T$ by $ U\otimes V^{\ast}$ rather than by $ U\otimes V$, since it corresponds to the multiplication by $ U\otimes V^{\ast}$ from the left if $ \mathrm{M}_n\left(\mathbb{K}\right)$ is canonically identified with $ \mathbb{K}^n\otimes\left(\mathbb{K}^n\right)^{\ast}$ (which is indeed canonically isomorphic to $ \mathrm{M}_n\left(\mathbb{K}\right)$, while $ \mathbb{K}^n\otimes\mathbb{K}^n$ is not). This viewpoint has the advantage that it guarantees $ \left(U_1\otimes V_1^{\ast}\right)\left(U_2\otimes V_2^{\ast}\right) \equal{} U_1V_2\otimes V_1^{\ast}V_2^{\ast}$, while with your notation $ \left(U_1\otimes V_1\right)\left(U_2\otimes V_2\right)$ is generally not $ U_1U_2\otimes V_1V_2$ (but $ U_1U_2\otimes V_2V_1$).

  darij

[quote]$ rank( U\otimes{V}) \equal{} rank(U)rank(V)$.[/quote]
I thinks: $ U\otimes{V}$ is equivalent $ \left[\begin{array}{cc}I_{r}\otimes{V}&O\\O&O\end{array}\right]$ use Gaussian method  :huh:

The real question should have been
Let $ A,B\in \mathcal{M}_n(\mathbb{K})$ with same rank.
Denote $ \Gamma \equal{} \{M \in \mathcal{M}_n(\mathbb{K}) | AM \equal{} MB \}$.
Find a sharp bound for $ \mathrm{dim} \Gamma$.
and not
[quote="JC_math"]... $ \Gamma \equal{} \{M \in \mathcal{M}_n(\mathbb{K}) | AM \equal{} BM \}$.
Find a sharp bound for $ \mathrm{dim} \Gamma$.[/quote]
Sorry for the typo.

The condition that $ A$ and $ B$ have the same rank is useless: for every $ \lambda\in\mathbb{K}$, the equation $ AM \equal{} MB$ is equivalent to $ \left(A \plus{} \lambda I_n\right)M \equal{} M\left(B \plus{} \lambda I_n\right)$, and (for infinite $ \mathbb{K}$) we can always find some $ \lambda\in\mathbb{K}$ such that the matrices $ A \plus{} \lambda I_n$ and $ B \plus{} \lambda I_n$ have the same rank (namely, full rank). I proved that $ \mathrm{dim}\Gamma > n\left(n \minus{} 1\right)\Longrightarrow\mathrm{dim}\Gamma \equal{} n^2$ in http://www.mathlinks.ro/viewtopic.php?t=256194 , and this cannot be improved. Whenever we have an integer $ k\in\mathbb{Z}$ such that there exist two arrays $ \left(\lambda_1,\lambda_2,...,\lambda_n\right)$ and $ \left(\mu_1,\mu_2,...,\mu_n\right)$ of elements of $ \mathbb{K}$ such that

$ k \equal{} \left|\left\{\left(i,j\right)\in\left\{1,2,...,n\right\}^2\mid \lambda_i \equal{} \mu_j\right\}\right|$,

we can construct two matrices $ A$ and $ B$ such that $ \dim\Gamma \equal{} k$ (in fact, just take $ A \equal{} \mathrm{diag}\left(\lambda_1,\lambda_2,...,\lambda_n\right)$ and $ B \equal{} \mathrm{diag}\left(\mu_1,\mu_2,...,\mu_n\right)$). In particular, $ \dim\Gamma$ can be any integer from $ 0$ to $ n$, but also any multiple of $ n$ (between $ 0$ and $ n^2$). So you are probably not looking for bounds, but for the exact set of all possible values of $ \dim\Gamma$. This seems to be a harder question...

[b]EDIT:[/b] According to loup blanc's idea in http://www.mathlinks.ro/viewtopic.php?t=256194 , our question is the following one: Which values can the dimension of the kernel (= the geometric multiplicity of the eigenvalue $ 0$) of the linear map $ A^T\otimes I \minus{} I\otimes B$ take? The values of the [i]algebraic[/i] multiplicity of $ 0$ are the values of $ k$ above.

  darij

Hi Darij,
Ok I used an axe to break a match.
About the Kronecker product ; it is defined as follows:
    $ U\otimes{V}: X\rightarrow{UXV^T}$or $ VXU^T$ in view of the following formula:
    $ (U\otimes{V})(U_1\otimes{V_1})\equal{}(UU_1)\otimes{(VV_1)}$.
One obtain $ UXV^T$ with $ U\otimes{V}\equal{}[u_{ij}V]_{ij}$ and by stacking the columns of $ X$ into a single column vector.
One obtain $ VXU^T$ with $ U\otimes{V}\equal{}[u_{ij}V]_{ij}$ by stacking the rows of $ X$ into a single column vector.

PS to JC_math:
before sending a post, please read and reread it.

You could have directed the PS to me as well - it seems I copypasted the $ X\mapsto UXV^T$ from your post but kept reading it as $ X\mapsto UXV$.

Let me add something to my latest post here. An integer $ k\in\mathbb{Z}$ will be called

- [i]$ n$-sylvestery[/i] if there exist two $ n\times n$ matrices $ A$ and $ B$ such that $ \dim\Gamma \equal{} k$ (where $ \Gamma \equal{} \left\{M \in \mathcal{M}_n(\mathbb{K}) | AM \equal{} MB \right\}$);
- [i]diagonally $ n$-sylvestery[/i] if there exist two diagonal $ n\times n$ matrices $ A$ and $ B$ such that $ \dim\Gamma \equal{} k$.

In my post above, I showed that:

- if an integer $ k$ is $ n$-sylvestery, we have $ k\in\left\{0,1,2,...,n\left(n \minus{} 1\right)\right\}\cup\left\{n^2\right\}$;
- an integer $ k$ is diagonally $ n$-sylvestery if and only if there exist two arrays $ \left(\lambda_1,\lambda_2,...,\lambda_n\right)$ and $ \left(\mu_1,\mu_2,...,\mu_n\right)$ of elements of $ \mathbb{K}$ such that

$ k \equal{} \left|\left\{\left(i,j\right)\in\left\{1,2,...,n\right\}^2\mid \lambda_i \equal{} \mu_j\right\}\right|$.

The latter property can be rewritten as follows: An integer $ k$ is diagonally $ n$-sylvestery if and only if there exist two sequences $ \left(\alpha_1,\alpha_2,...,\alpha_u\right)$ and $ \left(\beta_1,\beta_2,...,\beta_u\right)$ of positive integers such that $ \sum_{r \equal{} 1}^u\alpha_r\leq n$, $ \sum_{r \equal{} 1}^u\beta_r\leq n$ and $ \sum_{r \equal{} 1}^u\alpha_r\beta_r \equal{} k$.

Using this, it can be easily seen that any integer between $ 0$ and something approximately equal to $ \frac {n^2}{2}$ (which I am too lazy to compute more precisely) is diagonally $ n$-sylvestery.

Of course, every diagonally $ n$-sylvestery integer is $ n$-sylvestery as well. The interesting question is whether the converse holds.

According to [url=http://en.wikipedia.org/wiki/Sylvester_equation]this[/url], it is enough to consider the case when $ A$ and $ B$ are Jordan matrices. Besides, we also assume that $ A$ and $ B$ are nilpotent. Altogether, we can thus assume that $ A$ and $ B$ are nilpotent Jordan matrices. This turns everything into a combinatorial problem.
[b]EDIT:[/b] ... and that is solved. Yes, every $ n$-sylvestery integer is diagonally $ n$-sylvestery. This easily follows from chapter VIII.1 of
Felix R. Gantmacher, [i]The Theory of Matrices[/i], vol. 1.
I have absolutely no desire to post the proof here since it is ugly as hell.

  darij


For how many real numbers $a$ does the quadratic equation $x^2 + ax + 6a=0$ have only integer roots for $x$?

[hide]

Using quadratic, we get:

\[ x = \frac{-a \pm \sqrt{a^2-24a}}{2} \]

in which case we want to find all values of $a$ such that $a^2 - 24a = k^2$ for integers $k$

hence, we can use quadratic again to arrive at:

\[ a^2 - 24a - k^2 \implies a = \frac{24 \pm 4\sqrt{144 + k^2}}{2} = 12 \pm \sqrt{144 + k^2} \]

essentially we are trying to find all pythagorean triples with one of the sides of the triangle being 12, hence we have:

$\{ 5,12,13 \}$
$\{ 9,12,15 \}$
$\{ 12,16,20\}$
$\{ 12,35,37 \}$

and we must not forget when $a= 24$ or $0$, which make $k=0$

then, for each of the triples above, we see that they generate two valid values of $a$ each, thus our final answer is $\boxed{10}$

[/hide]

[quote="1234567890"]Now, becuase $s$ and $r$ are basically the same, we will look for only pairs of integers that satisfy this equation with $r \le s$.[/quote]

Just say "Without loss of generality, assume that $r \le s$." :)

Reminds me of...
http://artofproblemsolving.com/community/c5h623889p3734545

Are integer solutions in polynomials frequently tested?
Do you know any other good questions like this?

EDIT: Also, I don't quite understand:
[quote=1234567890][hide="From NT class"]Let $x^2+ax+6a=(x-s)(x-r)$
Using Viete's yields $s+r=-a, sr=6a$
$\Rightarrow sr+6s+6r=0$
$\Rightarrow sr+6s+6r+36=36$
$\Rightarrow (s+6)(r+6)=36$

Now, becuase $s$ and $r$ are basically the same, we will look for only pairs of integers that satisfy this equation with $r \le s$.

So possible values of $(r+6,s+6)$ are:
$(-36,-1),(-18,-2),(-12,-3),(-9,-4),(-6,-6),(1,36),(2,18),(3,12),(4,9),(6,6)$
$\Rightarrow \boxed{10} \text{values of } a$[/hide][/quote]

Can someone explain how he got to sr+6s+6r=0?

it is from vietas formulas:
We have that:
$$s +r=-a$$
$$sr=6a$$
Multiply the second equation by $6$:
$$6s+6r=-6a$$
Add this equation with the second equation:
$$sr + 6s + 6r = 6a  + (-6a) =0$$


Prove that : If $ \lim_{n\to \infty}a_n = a $ then
  $ \lim_{n\to \infty}\frac{a_1+a_2+...+a_n}{n} = a $ (do not use Toeplitz theorem )

I think it is not difficult.and I  will post the solution for you.(Althought it can be found in many books,I think)
first assume $a=0,then, \lim_{n\to \infty}a_n = 0$
for $\forall \epsilon >0$,we can find N,such that $a_n< \epsilon $
so we have  $ \lim{n\to \infty}\frac{a_1+a_2+...+a_n}{n}<\frac{(S_N+(n-N) \epsilon )}{n}$.
when $n \rightarrow \infty $ can be show  the $RHS  \rightarrow 0$
for every we can note $b_i=a_i-a$
we are done :)


calculus
integration
geometry
trapezoid
calculus computations

Problem #5 in this link: http://www.collegeboard.com/prod_downloads/ap/students/calculus/ap04_frq_calculus_ab.pdf

Can you be more specific? Most of the parts have you either think of the integral as an area, or make you use the fundamental theorem of calc. Do those hints help?

[quote="HilbertThm90"][...] use the fundamental theorem of calc. [...][/quote]

To explain a little further, recall that if $ g(x)\equal{}\int_a^x f(x)\ dt$, then $ g(b)$ is the area under the curve in the interval $ [a,b]$, and $ g'(b)\equal{}f(b)$.

[hide="To get you started..."]
In the given problem, we are asked to find $ g(0)$, where $ g(x)\equal{}\int_{\minus{}3}^x f(x)\ dt$, so we find the area under the curve from $ [\minus{}3,0]$. This can be done however you wish; for example, the area is a trapezoid with bases $ 2$ and $ 1$ and height $ 3$, so the area is $ A\equal{}g(0)\equal{}\frac{(2\plus{}1)(3)}{2}\equal{}\boxed{\frac{9}{2}}$.

Now to find $ g'(0)$, we simply note that $ g'(0)\equal{}f(0)\equal{}\boxed{1}$.

These few hints should be enough of a springboard to finish the problem.
[/hide]


trigonometry
calculus
calculus computations

a, b, c belongs to (0,pi/2)
   such that sin(a) +tan(b)+cos(c)=3
   then (3-sin(a))(3-tan(b))(3-cos(c)) is

      options are
1.	= to 8 sin(a)tan(b) cos(c)for some a ,b, c
2.	greater than 8 sin(a) tan(b)  cos(c)
3.	less than 8 sin(a) tan(b) cos(c)
4.	= to any real number

This doesn't seem to be calculus, or at least not necessarily calculus.  [hide="hint"]
\begin{eqnarray*}(3 - \sin a)(3 - \tan b)(3 - \cos c) & = & (3 - \sin a)(3 - \cos c)(\sin a + \cos c) \\
& = & (3 - x)(3 - y)(x + y)\end{eqnarray*}
for some $ x,y\in (0,1)$.  To prove that (2) is the answer it then suffices to show
\[ (3 - x)(3 - y)(x + y) > 8xy(3 - x - y) \ \ \ \ \forall x,y\in (0,1)
\]
[/hide]

[color=green]Redundant quote deleted.[/color]
thanks sir


inequalities
geometry
3D geometry
tetrahedron
geometry proposed

Given ortho tetrahedron $ABCD$ volume $V$ with area of triangle $BCD$ is $S_{a}$ similarly $S_{b},S_{c},S_{d}$ with any $P$ on space prove that
\[S_{a}PA+S_{b}PB+S_{c}PC+S_{d}PD\ge 6V \]

[color=purple]gemath, I think the$RHS$ must be $9V$ and the condition ortho tetrahedron is unnescessary because it's only a condition for equality occurs :) [/color]


abstract algebra
superior algebra
superior algebra unsolved

$R$ is a ring following property: $x^3=x$ for all $x$ in $R$.
Prove that $R$ is abelian.

dunno if it matters but do you assume it has unity?

did you assume that it is a group?
I'm sorry that can't be correct, there are many noncyclic groups with the $x^3=x$ property, like $\mathbb{Z}_2\times \mathbb{Z}_2$

Hehe this isn't such a hard problem at all. Anyway here it is a solution

 [b] Problem.[/b]

  Let $R$ be a ring such that $x^{3}=x\forall x\in R$. Prove that the ring $R$ is comutative.

 [i] Solution.[/i]

    From $x^{3}=x$ we get $x^{4}=x^{2}$. By doing some calculations one can easily get
    \[ (x^{2}yx^{2})^{2}=(x^{2}yx^{2}-yx^{2})^{2}=0.  \] Then we obtain
    \[ x^{2}yx^{2}-x^{2}y=(x^{2}yx^{2}-x^{2}y)^{3}=(x^{2}yx^{2}-x^{2}y)^{2}(x^{2}yx^{2}-x^{2}y)=0.  \]
     We get \[ x^{2}yx^{2}x^{2}y\forall x, y\in R.  \]
     Analogously, we have \[ x^{2}yx^{2}=yx^{2},  \] so we get \[ x^{2}y=yx^{2}.  \]
     Now, as final step we will get our conclusion like this:
     \[ xy=x^{3}y^{3}=x(x^{2}y^{3})=xy^{3}x^{2}=(xy^{2})(yx^{2})=(yx^{2})(xy^{2})=y(yx)(yx)x=(yx)^{2}x=(yx)^{3}=yx, \forall x, y\in R.  \]

i dont know why but i like this problem(maybe because its so simple to understand).
i found some relations wich are of not much help(i only used the operation of the
semigroup of the ring...namely (R,.) )

$(xy)^3=x^3 y^3$
$x^3 y= y x^3$

$xy=x^2 y$
$yx=y^2 x$

$x^2 y^2 = yx^2 y^2$
$yxy=y^2 x^2 y$
$xy^2=x^2 y^2$

cezar,your solution overall seems very nice.
but i would like to know how exactly that analogy is made(i didnt understand that).
also that equality you say of there is true...i checked it,it makes sense
what is the reason that particular one is chosen in the context of this problem ?
is this your solution or is this from a book ?
(i doubt that one person(like me for exampl...) would be able to find such an equation
 but with great skill and,intuition and experience probably one would be able to
 find something like this)
again,very nice solution   :)

Alternative proof:

1. $xy=0$ implies $yx=(yx)^3=yxyxyx=0$.
2. We have $0 = xy - x^3 y = x(y -x^2 y)$. By 1. we get $(y-x^2 y)x = 0$.
3. Then, 2. implies $y^2 (x^2 yx)=y^2 (yx) = y^3 x = yx =(yx)^3$, i.e. $y(yx-xy)xyx=0$.
Another application of 1. gives $(yx-xy)(xy)^2=0$, hence $0 = (yx-xy)(xy)^3=(yx-xy)xy$.
4. The final step is
\[ (xy-yx)^2 = (xy)^2 - xyyx - yxxy + (yx)^2 =(xy-yx)xy - (xy-yx)yx = 0 \]
by 3., which implies $xy - yx = 0$ after multiplication of $xy-yx$. This shows $xy=yx$.

Here's an additional exercise:

If for every element $x$ of $R$ there is an integer $n>1$ such that $x^n = x$, then $R$ is commutative.


limit
integration
real analysis
real analysis unsolved

$f(x)$ is increasing on $\mathbb{R}$ and it is bounded and differentiable on $\mathbb{R}$
$\lim_{x\to -\infty} f(x)=A, \quad \lim_{x\to+\infty}
f(x)=B$
show that $\int_\mathbb{R} f'(x)dx = B-A.$

i forget a condition again : $f$ is differentiable.  :oops:  Now I have modified the problem.


algebra
polynomial
induction
symmetry
Ring Theory
advanced fields
advanced fields unsolved

Let $ I$ be the monomial ideal in $ k[x_1,\cdots,x_n]$ with $ dim V(I)\equal{}0$.
a. What is $ V(I)$ in this case?
b. Prove that $ dimV(I)\equal{}0$ if and only if for every $ 1\leq i\leq n$, $ x_i^{l_i}\in I$ for some $ l_i\geq 1$.

Thanks in advance.

Here is something sketchy assuming that a monomial ideal is an ideal generated by monomials, i.e. by polynomials of the form $ X_1^{k_1}\cdot\ldots\cdot X_n^{k_n}$ with $ k_i\ge 0$.

Ad a: If $ I$ is a monomial ideal of $ k[X_1,\ldots,x_n]$ with $ \dim V(I)\equal{}0$, then $ R\equal{}k[X_1,\ldots,X_n]/I$ is a 0-dimensional ring, thus every prime ideal in $ R$ is maximal. This implies that $ Spec(R)$ is finite with closed points only (by quasi-compactness) and $ R$ is the finite product of Artinian rings.

Ad b: We need to show that given a monomial ideal $ I$ of $ k[X_1,\ldots,X_n]$ then $ \dim V(I)\equal{}0$ iff $ (X_1,\ldots,X_n)\subset\sqrt{I}$.

Note that in this case we have $ (X_1,\ldots,X_n)\equal{}\sqrt{I}$, since $ (X_1,\ldots,X_n)$ is a maximal ideal of $ k[X_1,\ldots,X_n]$. This implies a refinement of (a) that $ V(I)$ is not only finite, but it is a singleton. So this is already "$ \Leftarrow$". For the inverse implication "$ \Rightarrow$" we conduct induction on $ n$: For $ n\equal{}1$, $ I$ is generated by some monomials $ X_1^{l_i}$ with $ l_i>0$. This shows that $ I\equal{}(X_1^l)$ for some $ l>0$, thus the claim. For the induction step let $ n>1$ and $ I'\equal{}I\cap k[X_1,\ldots,X_{n\minus{}1}]$ which is monomial. Considering the projection $ \pi: \mathbb A^n\to\mathbb A^{n\minus{}1}$ dropping the $ n$-th component, then $ \pi(V(I))$ is a finite set of closed points of $ \mathbb A^{n\minus{}1}$, by Hilbert's Nullstellensatz, hence $ \pi(V(I))$ is closed and $ \dim \pi(V(I))\equal{}0$. Thus we get $ \pi(V(I))\equal{}V(I')$. By induction hypothesis, $ (X_1,\ldots,X_{n\minus{}1})\subset\sqrt{I'}$, in particular $ X_1,\ldots,X_{n\minus{}1}\in \sqrt{I}$. By symmetry we get $ (X_1,\ldots,X_n)\subset\sqrt{I}$ using the other projections dropping some component.


geometry
incenter
Pythagorean Theorem
geometry proposed

Let $ I$ be the incenter of triangle $ ABC.$ Let $ M,N$ be the midpoints of $ AB,AC,$ respectively. Points $ D,E$ lie on $ AB,AC$ respectively such that $ BD\equal{}CE\equal{}BC.$ The line perpendicular to $ IM$ through $ D$ intersects the line perpendicular to $ IN$ through $ E$ at $ P.$ Prove that $ AP\perp BC.$

Really nice one, but slightly easy to be in the Chinese TST. Just a simple application of a bit generalized Pythagorean theorem applied plenty of times.

In fact the point $ I$ does not play a great role here, the only thing matters is this point to lie on $ OI$.

[img]http://img230.imageshack.us/img230/6425/geofc7.jpg[/img]

Can you post a solution because I cannot solve it.Thanks.

Only calculate:
$ AP\perp BC\leftrightarrow \frac{1}{4}(AB^2\minus{}AC^2)\equal{}AM^2\minus{}AN^2\equal{}PM^2\minus{}PN^2$
$ \equal{}PK^2\plus{}KM^2\minus{}PL^2\minus{}NL^2\equal{}PI^2\minus{}IK^2\plus{}KM^2\minus{}(PI^2\minus{}IL^2)\minus{}NL^2$
$ \equal{}IL^2\minus{}IK^2\plus{}KM^2\minus{}NL^2\equal{}IE^2\minus{}EN^2\minus{}ID^2\plus{}DM^2\equal{}(IS^2\plus{}ES^2)\minus{}(IR^2\plus{}RD^2)\minus{}EN^2\plus{}DM^2$
$ \equal{}ES^2\minus{}RD^2\minus{}EN^2\plus{}DM^2\equal{}(a\minus{}(p\minus{}c))^2\minus{}(a\minus{}(p\minus{}b))^2\minus{}(a\minus{}b/2)^2\plus{}(a\minus{}c/2)^2$
It's true!
We end the proof

[quote="Erken"]
In fact the point $ I$ does not play a great role here, the only thing matters is this point to lie on $ OI$.
[/quote]
From [url=http://www.mathlinks.ro/viewtopic.php?t=281652]here[/url], we have $ OI\perp DE$ so for every point $ I'$ lies on $ OI$: $ I'D^2\minus{}I'E^2\equal{}ID^2\minus{}IE^2$ and we get the result.

My solution [b]:[/b]

From $ OI \perp DE $ (well-known) $ \Longrightarrow \triangle ADE $ and $ \triangle INM $ are orthogonal ,
so the perpendicular from $ A, D, E $ to $ NM, IM, IN $, resp are concurrent $ \Longrightarrow AP \perp MN $ .
i.e. $ AP \perp BC $

Q.E.D

Dear Mathlinkers,
see also
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=395892
Sincerely
Jean-Louis

Let $P, Q$ be the points of tangency of the incircle $\omega$ with sides $\overline{AB}, \overline{AC}$, respectively. Let $r$ denote the radius of $\omega$, and let $\Gamma_b, \Gamma_c, \gamma$ be the circle centered at $M$ with radius $MA$, the circle centered at $N$ with radius $NA$, and the circle centered at $I$ with radius $R := \sqrt{r^2 + s^2 - a^2 + bc}$, respectively. We claim that $P$ is the radical center of $\Gamma_b, \Gamma_c, \gamma.$ This will imply the desired result, because the radical axis of $\Gamma_b, \Gamma_c$ is just the $A$-altitude of $\triangle ABC.$

To see this, we apply the Pythagorean Theorem to $\triangle IPD$ to obtain \[ID^2 = IP^2 + PD^2 = r^2 + (BD - BP)^2 = r^2 + (s - c)^2.\] Meanwhile, $MA^2 = \tfrac{1}{4}c^2$ and $MD^2 = |BD - BM|^2 = |a - \tfrac{c}{2}|^2.$ Upon expansion, it is straightforward to see that $MD^2 - MA^2 = ID^2 - R^2.$ Therefore, $D$ has equal power w.r.t. $\Gamma_b$ and $\gamma.$ Then because $PD \perp MI$, it follows that $PD$ is the radical axis of $\Gamma_b$ and $\gamma.$ Similarly, $PE$ is the radical axis of $\Gamma_c$ and $\gamma$, and therefore $P$ is the radical center of these three circles. $\square$

[quote=TelvCohl]My solution [b]:[/b]

From $ OI \perp DE $ (well-known) $ \Longrightarrow \triangle ADE $ and $ \triangle INM $ are orthogonal [/quote]

Excuse me, but what does it for two triangles to be orthogonal to each other? Having orthic centre?

Another name for the relation is for the triangles to be [url=http://mathworld.wolfram.com/OrthologicTriangles.html]orthologic[/url]; two triangles $ABC$ and $DEF$ are orthologic if the perpendiculars from $A$ to $EF$, $B$ to $DF$, and $C$ to $DE$ concur. By [url=http://www.artofproblemsolving.com/wiki/index.php/Carnot's_Theorem]Carnot's Theorem[/url], this is equivalent to the reverse relation, that the perpendiculars from $D$ to $BC$, $E$ to $AC$, and $C$ to $AB$ concur, so the relation is symmetric.

Basically, we want to prove that $ADE$ and $IMN$ are orthologic triangles since we want perpendiculars from $E$ to $IM$, $D$ to $IN$ and $A$ to $MN$ concurrent. It is equivalent that perpendiculars from $M$ to $AE$ and from $N$ to $AD$ and $I$ to $DE$ are concurrent. Thus we want line joining mid point of $AH$ to $I$ perpendicular to $DE$.

By Linearity of power of a point applied to incircle and circumcircle we get that $OI \perp DE$ which is what we wanted.

Here is my bary solution.
[hide=My solution ]Since $BD=BC=BE$
$D=(a,c-a,0)$ and $E=(a,0,b-a)$
Since $M,N$ are midpoint we get $M=(\frac{1}{2},\frac{1}{2},0)$ and $N=(\frac{1}{2},0,\frac{1}{2})$
$I$ is incenter so $I=(a,b,c)$
Let $P=(x,y,1-x-y)$. Applying  EFFT to $EP\perp IN$ and $DP\perp IM$.We easily find coordinates of $P$.Now we apply EFFT to get the desired result.

[quote=nikolapavlovic]@above you left out bunch of nasty calculations i don't think you're done.

It suffice to prove that $\triangle IMN$ and $\triangle ADE$ are orthologic but as this is an symmetry relation it' sufficient to prove that  perpendicular bisectors of $AC,AB$ and the line thru $I$ perpendicular to $DE$ are concurrent.But $IO\perp DE$ and the result follows.[/quote]
Why?

[quote=Murad.Aghazade][quote=nikolapavlovic]Generally if you are bashing, bash right to the end.Solution like "we just do this and that " usually get 1 or zero .[/quote]
I dont post [b]My solution [/b] for asking How many points can I get.Also The last part can be done by everyone [b]who knows what computation is[/b].If you dont know how to do computations (last part)I can write for you and for everyone who doesnt know computations. You can be [b]sure[/b] that I will write completeley my solution [b]at exam[/b] if I can bash.[/quote]

Heyy
Don't get offended !
I am sure what he meant was...
That practice solving solutions to the end :)
That is all

Of course posting or not is your own choice

Let  incircle tangents $AB,CA at $$X,Y$ respectively. $AK \perp MN,EL \perp IL,DJ \perp MI$ and$ K \in MN,L \in IN,J \in MI $
Applying carnot theorem about$ \triangle MIN$
$AK,EL,JD$ are concurrent $\Leftrightarrow 0=(MK^2-KN^2)+(LN^2-LI^2)+(IJ^2-JM^2)$
 $\Leftrightarrow ((\frac{AB}{2})^2-(\frac{AC}{2})^2)+(EN^2-EI^2)+(DI^2-DM^2)=0$
 $\Leftrightarrow ((\frac{c}{2})^2-(\frac{b}{2})^2)+(EN^2-EY^2)+(XD^2-DM^2)=0$
 $\Leftrightarrow ((\frac{c}{2})^2-(\frac{b}{2})^2)+((a-\frac{b}{2})^2-(\frac{b+c-a}{2})^2)+((a-\frac{c}{2})^2-(\frac{a+b-c}{2})^2)=0 $We are done


[color=darkblue]A point $ D$ is chosen on the side $ AC$ of a triangle $ ABC$ with $ \angle C < \angle A < 90^0$  in such a way that $ BD \equal{} BA$. The incircle of $ \triangle ABC$ is tangent to $ AB$ and $ AC$ at points $ K$ and $ L$,respectively. Let $ J$ be the incentre of triangle $ BCD$. Prove that the line $ KL$ intersects the line segment $ AJ$ at it’s midpoint.[/color]

[color=darkred]Please tell me another solution. Thank you very much.[/color]

Thank you very much. Oh, it's France TST 2007 too.


algebra
polynomial
linear algebra
matrix
Rational Root Theorem

Solve for x
$ x^5 \minus{}x^4 \minus{} 5x^3 \minus{} x^2 \plus{}4x \plus{}3 \equal{} 0$

I'm not really sure how to do this, since I checked on a calculator, and there are only three real zeros, and all of them are irrational.

Yeah, I'm assuming my teacher is wanting a crazy "solution" that doesn't involve true mathematics.

[hide="solution"]From the Rational Root Theorem, the possible roots are 3, -3, 1, and -1. From the Even-Odd Test, there are no integer roots. Thus there are no rational roots either. Checking on Maple, it cannot be factored in the integers. Thus we plug it in into our TI-84 and get that -1.391946908 is a root, 1.0773943 is a root, and 2.7198782 is a root. If we zoom out, we can see that there are no other real roots.[/hide]

If there is a rigorous way to do this problem, I would like to know it.

Yeah, 1=2, I went through everything you had tried plus more, and I couldn't understand how to get those three real answers. It turns out that I was given this problem in order to review the Newton-Raphson Method. It was intentionally made so ugly. Sorry for dragging you into the same pain that I went through. Here's a nicer, prob, man, it's so cool, it's unreal!
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1231099#1231099

I'm not sure how to do this exactly (I have limited background in linear algebra) however this might help.

Given a polynomial $ f(x)\equal{}a_0\plus{}a_1x\plus{}...\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}a_nx^n$,

Let $ C(f)$ denote the companion matrix (see: [url]http://en.wikipedia.org/wiki/Companion_matrix[/url]) of $ f(x)$.

The eigenvalues (see: [url]http://www.sosmath.com/matrix/eigen1/eigen1.html[/url]) of $ C(f)$ are the zeros of $ f(x)$.

I hope this will lead to a solution, but I'm not sure about the extent to which this works. It might not work if certain solutions are complex or irrational (don't take my word for this). I hope this helps, good luck.[/hide]

[quote="Textangle"]What's the even-odd test and what's Maple? I can't find it anywhere :blush:.[/quote]
The Even-Odd test is when you take the polynomial mod 2 to find all possible roots mod 2, and Maple is a CAS.


Trying to solve the problem in the attached jpeg.
As I understand a linear system must be made up and the unknowns are $ \alpha_i$s
The only problem is that when I try to "unfold" the term in 3rd.jpeg(attached image) terms like $ \alpha_i\alpha_j$ appear
and this is not a linear system any more.

You have to solve $ \frac{\partial r}{\partial \alpha_i}\equal{}0$, not $ r\equal{}0$.

  darij

[quote="darij grinberg"]You have to solve $ \frac {\partial r}{\partial \alpha_i} \equal{} 0$, not $ r \equal{} 0$.

  darij[/quote]

oh yes,you're right,thanks Darij :)


logarithms
real analysis
real analysis unsolved

Let $ F,G\in\mathbb R$ be of measure 0. Is it true that the set $ H\equal{}\{f\cdot g|f\in F, g\in G\}$ is of measure 0?

Of course not. Sums are essentially the same as products (the logarithm and the exponential are absolutely continuous on the positive reals), and there are standard examples for sums. If $ C$ is the standard ternary Cantor set, $ C\plus{}C\equal{}[0,2]$. Transforming this, $ \exp(C)\cdot \exp(C)\equal{}[1,e^2]$.


trigonometry
real analysis
real analysis unsolved

Find the sum of $\sum_{k=1}^{n}\arctan\left(\frac{1}{k^{2}+k+1}\right)$.

Any ideas?

It's a fairly well-known problem. But rather than looking for it, let's just ask what we can do.

$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ can be rearranged to

$\arctan a+\arctan b=\arctan\left(\frac{a+b}{1-ab}\right).$

Try a few small partial sums, and see what you can get from this identity.

So are you suggesting it can be written as:

$\sum_{k=1}^{n}\arctan{\left(\frac{1}{k^{2}+k+1}\right)}=\sum_{k=1}^{n}\arctan{\left(k+1\right)}-\sum_{k=1}^{n}\arctan{\left(k\right)}$

?

Thanks a lot for the help.


geometry
angle bisector
geometry proposed

Let $AB$ and $CD$ be two chords in the circle $s$,with the midpoints $m$,$n$.(hmm let the chords intercept inside the circle).
Prove :
If $AB$ be the angle bisector of $\angle CMD$ then $CD$ is the angle bisector of $\angle ANB$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=50002 .

  darij

Actually it is not nessecery AB and CD intersect inside of circle .
We have another angle bisector with name of external angle bisector :D  :D


[quote="Lakshya"]There is a triangle AOB with angle O = 30 degrees. The length of the side AB is 1 unit. What is the maximum length of side AO?[/quote] angle B=b so using law of sin $ AO\equal{}sinb*\frac{1}{\frac{1}{2}}\equal{}2sinb\le 2$ equality when b=90 degrees


circumcircle
vector
trigonometry
inequalities
trig identities
Law of Sines

Suppose that $x,y,z$ are positive real numbers.  Prove that

\[\frac{1}{(x+y)^2(y+z)^2}+\frac{1}{(y+z)^2(z+x)^2}+\frac{1}{(z+x)^2(x+y)^2} \leq \frac{9}{16xyz(x+y+z)}\]

Here are three equivalent problems:

1.  Suppose that $x,y,z$ are positive real numbers such that $x+y+z=1$.  Prove that

\[\frac{1}{(x+yz)^2}+\frac{1}{(y+zx)^2}+\frac{1}{(z+xy)^2} \leq \frac{9}{16xyz}\]

2.  Suppose that $a,b,c$ are the sides of a triangle with area $K$.  Prove that

\[\left(\frac{1}{ab}\right)^2+\left(\frac{1}{bc}\right)^2+\left(\frac{1}{ca}\right)^2 \leq \left(\frac{3}{4K}\right)^2\]

3.  Suppose that $a,b,c$ are the sides of a triangle with circumradius $R$.  Prove that

\[a^2+b^2+c^2 \leq 9R^2\]

I'm pretty sure these are true, but I might have made a mistake in my proof.  Are these known, trivial, etc?

Indeed, these problems are all right and equivalent. However, problem 3 is well-known...  :(  See, for instance, http://www.mathlinks.ro/Forum/viewtopic.php?t=17068 for a generalization. Here is the simplest proof of problem 3:

If O is the circumcenter and G is the centroid of triangle ABC, then we have

$9\cdot GO^{2}=9R^{2}-\left( a^{2}+b^{2}+c^{2}\right)$.

In fact, using vectors, we have

$\overrightarrow{GO}=\frac{\overrightarrow{AO}+\overrightarrow{BO}+\overrightarrow{CO}}{3}$,

since G is the centroid of triangle ABC. We multiply this with 3, square it (using the scalar product) and get

$9\cdot \overrightarrow{GO}^{2}=\left( \overrightarrow{AO}+\overrightarrow{BO}+\overrightarrow{CO}\right)^{2}$
$=\overrightarrow{AO}^{2}+\overrightarrow{BO}^{2}+\overrightarrow{CO}^{2}+2\cdot \overrightarrow{BO}\cdot \overrightarrow{CO}+2\cdot \overrightarrow{CO}\cdot \overrightarrow{AO}+2\cdot \overrightarrow{AO}\cdot \overrightarrow{BO}$.

Since the scalar square of a vector equals the square of its length, and since the segments AO, BO, CO all have length R, we thus get

$9\cdot \overrightarrow{GO}^{2}=3R^{2}+2\cdot \overrightarrow{BO}\cdot \overrightarrow{CO}+2\cdot \overrightarrow{CO}\cdot \overrightarrow{AO}+2\cdot \overrightarrow{AO}\cdot \overrightarrow{BO}$.

But since

$a^{2}=BC^{2}=\overrightarrow{BC}^{2}=\left( \overrightarrow{BO}-\overrightarrow{CO}\right)^{2}$
$=\overrightarrow{BO}^{2}+\overrightarrow{CO}^{2}-2\cdot \overrightarrow{BO}\cdot \overrightarrow{CO}=2R^{2}-2\cdot \overrightarrow{BO}\cdot \overrightarrow{CO}$,

we have

$2\cdot \overrightarrow{BO}\cdot \overrightarrow{CO}=2R^{2}-a^{2}$,

and similarly

$2\cdot \overrightarrow{CO}\cdot \overrightarrow{AO}=2R^{2}-b^{2}$

and

$2\cdot \overrightarrow{AO}\cdot \overrightarrow{BO}=2R^{2}-c^{2}$.

Thus,

$9\cdot \overrightarrow{GO}^{2}=3R^{2}+2\cdot \overrightarrow{BO}\cdot \overrightarrow{CO}+2\cdot \overrightarrow{CO}\cdot \overrightarrow{AO}+2\cdot \overrightarrow{AO}\cdot \overrightarrow{BO}$
$=3R^{2}+\left( 2R^{2}-a^{2}\right)+\left( 2R^{2}-b^{2}\right)+\left( 2R^{2}-c^{2}\right) =9R^{2}-\left( a^{2}+b^{2}+c^{2}\right)$.

In other words, $9 \cdot GO^{2}=9R^{2}-\left( a^{2}+b^{2}+c^{2}\right)$, qed.. Now, since $9 \cdot GO^{2}\geq 0$, we thus have $9R^{2}-\left( a^{2}+b^{2}+c^{2}\right) \geq 0$, and $a^{2}+b^{2}+c^{2}\leq 9R^{2}$, completing our proof.

Note that the Extended Law of Sines yields $a=2R\sin A$, $b=2R\sin B$ and $c=2R\sin C$, so the inequality $a^{2}+b^{2}+c^{2}\leq 9R^{2}$ becomes $\left(2R\sin A\right)^{2}+\left(2R\sin B\right)^{2}+\left(2R\sin C\right)^{2}\leq 9R^{2}$. Upon division by $4R^{2}$, this simplifies to $\sin^{2}A+\sin^{2}B+\sin^{2}C\leq\frac94$.

  Darij

[\[ \frac {1}{(x \plus{} y)^2(y \plus{} z)^2} \plus{} \frac {1}{(y \plus{} z)^2(z \plus{} x)^2} \plus{} \frac {1}{(z \plus{} x)^2(x \plus{} y)^2} \leq \frac {9}{16xyz(x \plus{} y \plus{} z)}\]
]

Why,does not solution?
BQ

[quote=ThAzN1]Suppose that $x,y,z$ are positive real numbers.  Prove that
\[\frac{1}{(x+y)^2(y+z)^2}+\frac{1}{(y+z)^2(z+x)^2}+\frac{1}{(z+x)^2(x+y)^2} \leq \frac{9}{16xyz(x+y+z)}\][/quote]
Let $x+y=a,y+z=b,z+x=c,$ the inequality is equivalent to
$$(a+b+c)(a-b+c)(b-c+a)(c-a+b)(a^2+b^2+c^2) \le 9a^2b^2c^2$$
$$\Leftrightarrow \left[ 2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4]\right (a^2+b^2+c^2) \le 9a^2b^2c^2$$
$$a^6+b^6+c^6+3a^2b^2c^2 \ge a^2b^2(a^2+b^2)+b^2c^2(b^2+c^2)+c^2a^2(c^2+a^2)$$
This is just Schur's inequality.


calculus
derivative
logarithms
calculus computations

i have a problem to which i cant find a solution...iv been workin on it since hours...i get the answer till fractions and powers but i dont get the numerator rite....here r the questions:- FIND THE DERIVATIVE OF:-

1) y=  :sqrt: x+1/ :sqrt: x-x 2

2)p(x)= :sqrt: x/x 2 +1

1)$y=\frac{\sqrt{x+1}}{\sqrt{x-x^2}}\Longleftrightarrow y=\sqrt{\frac{x+1}{x-x^2}}$

Taking logarithm both sides of equation, $\ln y=\frac{1}{2}\ln \left|\frac{x+1}{x-x^2}\right|$

Differentiating both of sides,....

Good luck! :)

kunny

+infinity, your post is ambiguous without some parentheses to help us understand what you meant. kunny made one reasonable guess as to what you meant, but it was just a guess - we can't be sure.

Yeah dizzle, add a few $($ and $)$ dawg. For #2 I think he means:

$P(x) = \frac{\sqrt{x}}{x^{2}+1}$

And Kunny has the second one.

These are just straightforward quotient rizzles:

$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - g'(x)f(x)}{g^{2}(x)}$

But Kunny, I think, especially for a beginning calculus student, natural logarithms might not be the most friendly way to do it. I'd probably just do it the plug-and-chug way, but your way is obviously just as good. Plus it helps with creative thinking.


algebra
function
domain
set theory
advanced fields
advanced fields unsolved

edit: I apologize that this isn't TeXified.  I always use scientific word to generate TeX code and I'm away from home.  Later on tonight, I can TeXify this.

The idea I have is to, in the context of Aczel's presentation of non-well-founded set theory, add a universal set axiom within the context of having three truth values.  The third truth value makes standard contradictions possibly have the third truth value.

So I started by developing a semantics for ternary logic.  Next, I want to show relative consistency to ZFA (ZFC with anti-foundation axiom).

edit: I'm considering just dropping the foundation axiom rather than replace it with the anti-foundation axiom.  That way I'll have less to prove. :P

I want to show that if there is a model for ZFC- (which is ZFC without foundation axiom), then there is a model for TUZFC- (which is ternary-logical universal ZFC-).

Here's where I get stuck and would appreciate some feedback.  First of all, if any of the above sounds wrong, do tell me.  Now, let's let I=(A,a) be a model of ZFC-.  I need to find a model of TUZFC-.  Here's what I propose.  A does not contain all sets or else I'd be done.  Pick a set not in A and add it to A; call the result A*.  Denote this set by U, so A*=A∪{U}.  The symbol set may be enlarged from {∈} to {E, ∈, u}.  Extend the relation ∈ from sets in A to sets in A* by defining:
for a, b in A, aEb iff a∈b and
if b=U then for all a in A*, aEb.

Then I want to show that I*=(A*,a*) models TUZFC-.  Does this work or am I at a dead end?

edit (9-16/05)
It's actually good no one replied.  I figured it out for myself.   :tongue2:

I had to scrap that A*=A∪{U} idea.  (A*,a*) does not seem to model basically anything but the universal set axiom!

In the investigation, I found that given A (which is thought to be a domain for a model), there is a set containing A (as a subset) that is closed under pairing: for all x,y in the set, {x,y} is in the set.  Call this F.  Now I want F to satisfy extensionality but it doesn't clearly.  Imagine I adjusted F.  Then I'd lose the closed under pairing property.  I figure that if I could do what I was trying to do, I'd have a model for set theory which doesn't exist (yet).

So now what I'm doing is letting U be an arbitary set in A that is nonempty.  Then I define a relation, thinking of "is in", on A such that for all $x,y\in A$, $x\in _{2}y$ iff $x\in _{3}y$. Furthermore define $\in _{3}$so that for all $x\in A$, $x\in _{3}U$.  Here, the 2 subscript refers to "is in" in standard set theory and the new 3 subscript denotes "is in" in the new theory.

This setup will clearly not model the foundation axiom as U∈U.  Luckily, I dropped that in my theory.

Now I have to check that this models anything other than the universal set axiom!.

If you have any feedback, it would be appreciated.  If not, that's cool too.

http://home.comcast.net/~morlai/universal.pdf


The age of the father is 5.5 times  as that of the second daughter. Mom got married at 20; at that time grandfather was 57. The first son was born when mom was 22. At present, the first daughter is 19; her age differs from the second son by 5 and from the second daughter by 9. The last year, age of the third son was half of the first son. The sum of the age of the second daughetr and th ethird son equal the age of the second son.

What is the the age of the first son?

[hide]Let the first son be $ x$ years old.

We know that the second daughter must be $ 10$ years old and the third son's is $ \frac{x\plus{}1}{2}$ years old.

Also, $ 10\plus{}\frac{x\plus{}1}{2} \equal{} 14$ or $ 24$ since the second son is $ 5$ years older or younger than the first daughter.

If $ \frac{x\plus{}1}{2} \equal{} 14, x \equal{} 27$ and if $ \frac{x\plus{}1}{2} \equal{} 4, x \equal{} 7$. Since the first son must be older than the second, then $ \boxed{27}$.[/hide]

The first thing you need to do is list the information that you received; then elliminate that which is unneeded.
You do not need, for example, the fact that the father is 5-1/2 times the age of the second daughter. Nor that the mother was married at 20, had her first son at 22 (well, these two can rather be used to check the validity of the answers) or that Grandfather was 57 when she got married.

Now
S1 = first son
S2 = etc

D1 = first daughter
D2 = etc

D1 = 19.......... .(1)
D1-D2=9..... ....(2)
S3-1=1/2x(s1- 1) (-1 because last year: current age -1)........( 3)
D2+S3=S2.... ........( 4)

There are two case
S2>D1 S2-D1=5..... .....(5a)
or
S2<D1 D1-S2=5..... ......(5b)

From (2)
D1-D2=9
therefore
D2=D1-9
D2=19-9
D2=10....... (6)

If (5a)
S2-D1=5
then
S2=D1+5
S2=19+5
S2=24....... ..(7a)

If (5b)
D1-S2=5
Therefore
S2=D1-5
S2=19-5
S2=14....... ..(7b)

From (4)
D2+S3=S2
Therefore
S3=S2-D2

If (7a)
S3=24-10.... .from (7a) and (6)
S3=14......( 8a)

If (7b)
S3=14-10
S3=4.....(8b)

From (3)
S3-1=1/2x(S1- 1)
therefore
S1=2x(S3-1)+ 1
S1=2xS3-1

If 8a
S1=27

If 8b
S1=7

Now check, the mother gave birth to S1 at 22

Therefore
(8a) mother is currently 49, and she gave birth to D1 at 30 (possible)
(8b) mother is currently 29 and gave birth to D1 at 10 (highly unlikely)

Therefore S1 is 27 (age of first son). :lol:


Hello,

Could anyone please tell me the code for the plus or minus sign, if supported by LaTeX?

Thanks.

\pm as in $ \pm$. Guess what minus plus would be?


[i]Let a,b,c be positive real numbers
 Prove that[/i]

     $ \frac {a^2 \minus{} bc}{\sqrt { 8a^2 \plus{} (b \plus{} c)^2}} \plus{} \frac {b^2 \minus{} ca}{\sqrt { 8b^2 \plus{} (c \plus{} a)^2}} \plus{} \frac {c^2 \minus{} ab}{\sqrt { 8c^2 \plus{} (a \plus{} b)^2}} \geq 0$
       [i]  Love is  like Ineq
  Very beautiful but very hard[/i] :love:
      [i]  Y An Forever[/i]

[quote="nkht-tk14"][i]Let a,b,c be positive real numbers
 Prove that[/i]

     $ \frac {a^2 \minus{} bc}{\sqrt { 8a^2 \plus{} (b \plus{} c)^2}} \plus{} \frac {b^2 \minus{} cc}{\sqrt { 8b^2 \plus{} (c \plus{} a)^2}} \plus{} \frac {c^2 \minus{} ab}{\sqrt { 8c^2 \plus{} (a \plus{} b)^2}} \geq 0$
       [i]  Love is  like Ineq
  Very beautiful but very hard[/i] :love:
      [i]  Y An Forever[/i][/quote]
Yes, it's very easy!
$ \sum_{cyc}\frac {a^2 \minus{} bc}{\sqrt { 8a^2 \plus{} (b \plus{} c)^2}}\equal{}\frac{1}{2}\sum_{cyc}\left(\frac{(a\minus{}b)(a\plus{}c)}{\sqrt { 8a^2 \plus{} (b \plus{} c)^2}}\minus{}\frac{(c\minus{}a)(a\plus{}b)}{\sqrt { 8a^2 \plus{} (b \plus{} c)^2}}\right)\equal{}$
$ \equal{}\frac{1}{2}\sum_{cyc}\left(\frac{(a\minus{}b)(a\plus{}c)}{\sqrt { 8a^2 \plus{} (b \plus{} c)^2}}\minus{}\frac{(a\minus{}b)(b\plus{}c)}{\sqrt { 8b^2 \plus{} (a \plus{} c)^2}}\right)\equal{}$
$ \equal{}\sum_{cyc}\frac{(a\minus{}b)\left((a\plus{}c)^2(8b^2\plus{}(a\plus{}c)^2)\minus{}(b\plus{}c)^2( 8a^2 \plus{} (b \plus{} c)^2)\right)}{2\left((a\plus{}c)\sqrt{8b^2\plus{}(a\plus{}c)^2}\plus{}(b\plus{}c)\sqrt{8a^2 \plus{} (b \plus{} c)^2}\right)\sqrt { 8a^2 \plus{} (b \plus{} c)^2}\sqrt { 8b^2 \plus{} (a \plus{} c)^2}}\equal{}$
$ \equal{}\sum_{cyc}\frac{(a\minus{}b)^2\left(4c^3\minus{}2(a\plus{}b)c^2\plus{}4(a^2\minus{}3ab\plus{}b^2)c\plus{}a^3\plus{}a^2b\plus{}ab^2\plus{}b^3\right)}{2\left((a\plus{}c)\sqrt{8b^2\plus{}(a\plus{}c)^2}\plus{}(b\plus{}c)\sqrt{8a^2 \plus{} (b \plus{} c)^2}\right)\sqrt { 8a^2 \plus{} (b \plus{} c)^2}\sqrt { 8b^2 \plus{} (a \plus{} c)^2}}\geq$
$ \geq\sum_{cyc}\frac{(a\minus{}b)^2\left(4c^3\minus{}2(a\plus{}b)c^2\minus{}(a\plus{}b)^2c\plus{}\frac{(a\plus{}b)^3}{2}\right)}{2\left((a\plus{}c)\sqrt{8b^2\plus{}(a\plus{}c)^2}\plus{}(b\plus{}c)\sqrt{8a^2 \plus{} (b \plus{} c)^2}\right)\sqrt { 8a^2 \plus{} (b \plus{} c)^2}\sqrt { 8b^2 \plus{} (a \plus{} c)^2}}\equal{}$
$ \equal{}\sum_{cyc}\frac{(a\minus{}b)^2(2c\plus{}a\plus{}b)(2c\minus{}a\minus{}b)^2}{4\left((a\plus{}c)\sqrt{8b^2\plus{}(a\plus{}c)^2}\plus{}(b\plus{}c)\sqrt{8a^2 \plus{} (b \plus{} c)^2}\right)\sqrt { 8a^2 \plus{} (b \plus{} c)^2}\sqrt { 8b^2 \plus{} (a \plus{} c)^2}}.$ :)

[i]Very nice
  Out of my expectation                          :lol:
  Congratulation[/i] , [i]Argady[/i]


Show that there are no positive integers $x,y$ such that $x^{2}+y+2$ and $y^{2}+4x$ are both perfect squares.