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Questions 17
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 #6
avatar+26393 
+10

I called each of the four small angles at C, also  $$\alpha$$

$$\boxed {
\cos{(\alpha)} = \frac{h}{b} = \frac{h}{d} \quad | \quad h=\overline{EC} \quad \mbox{and} \quad d = \overline{FC}
}
\Rightarrow \boxed{d=b}$$

$$Areas: DCB = DCA = \frac{ \frac{c}{2} *h }{ 2 }$$

I.   Areas: ACF + FCD = DCB

$$b^2\sin{(2\alpha)} +be\sin{(\alpha)} =ae\sin{(\alpha)}$$

II.  Areas: FCD + DCB = FCB

$$be\sin{(\alpha)} +ae\sin{(\alpha)} =ab\sin{(2\alpha)}$$

I.+II.:

$$\begin{array}{rcl}
b^2\sin{(2\alpha)} +2be\sin{(\alpha)} &=&ab\sin{(2\alpha)} \quad | \quad :b\\
b\sin{(2\alpha)} +2e\sin{(\alpha)} &=& a\sin{(2\alpha)}
\end{array}$$

$$\Rightarrow
\boxed{
\frac { 2e\sin{( \alpha) } }
{ \sin{(2\alpha) } }
=a-b \qquad (1)
}$$

I.-II.:

$$\begin{array}{rcl}
b^2\sin{(2\alpha)} - ea\sin{(\alpha)} &=&ea\sin{(\alpha)} -ab\sin{(2\alpha)} \\
2ea\sin{(\alpha)} &=& ab\sin{(2\alpha)} + b^2\sin{(2\alpha)}
\end{array}$$

$$\Rightarrow
\boxed{
\frac { 2e\sin{( \alpha) } }
{ \sin{(2\alpha) } }
=\frac{ ab+b^2 } { a } \qquad (2)
}$$

(1)=(2):

$$a-b =\frac{ ab+b^2 } { a }$$

$$\Rightarrow
\boxed{
a^2 -2b*a -b^2 = 0
}$$

$$a_{1,2}={ 2b\pm\sqrt{4b^2-4*1*(-b^2)}\over2*1 }=b(1\pm\sqrt{2}) \quad | \quad \mbox{ b not negativ!}$$

$$\textcolor[rgb]{1,0,0}{
\boxed{
\textcolor[rgb]{0,0,0}{
a= b (1+\sqrt{2})
}
}
}$$

III. Areas: ACF + FCB = ACB

$$b^2\sin{(2\alpha)} + ba\sin{(2\alpha)} = ba\sin{(4\alpha)} \quad | \quad a=b(1+\sqrt{2})$$

 $$b^2\sin{(2\alpha)} + b^2(1+\sqrt{2})\sin{(2\alpha)} = b^2(1+\sqrt{2})\sin{(4\alpha)} \quad | \quad :b^2$$

$$\sin{(2\alpha)} + (1+\sqrt{2})\sin{(2\alpha)} = (1+\sqrt{2})\sin{(4\alpha)}$$

$$(2+\sqrt{2})\sin{(2\alpha)} = (1+\sqrt{2})\sin{(4\alpha)} \quad | \quad \leftrightarrow$$

$$(1+\sqrt{2})\sin{(4\alpha)} = (2+\sqrt{2})\sin{(2\alpha)} \quad | \quad \sin{(4\alpha)}=2\sin{(2\alpha)}\cos{2\alpha}$$

$$(1+\sqrt{2})2\sin{(2\alpha)}\cos{2\alpha}= (2+\sqrt{2})\sin{(2\alpha)} \quad | \quad : \sin{(2\alpha)}$$

$$(1+\sqrt{2})2\cos{2\alpha}= (2+\sqrt{2})$$

$$(2+2\sqrt{2})\cos{2\alpha}= (2+\sqrt{2})$$

$$\cos{2\alpha}=\frac{ (2+\sqrt{2}) }{ (2+2\sqrt{2}) } \quad | \quad *\frac{ \frac{\sqrt{2}}{2} }{ \frac{\sqrt{2}}{2} }$$

$$\cos{2\alpha}=\frac{ (2+\sqrt{2})\frac{\sqrt{2}}{2} }{ (\sqrt{2}+2) } = \frac{ \sqrt{2} }{ 2 } \quad | \quad \pm \cos^{-1}()$$

$$2\alpha= \pm \cos^{-1}( \frac{ \sqrt{2} } { 2 } ) \quad | \quad \textcolor[rgb]{1,0,0}{+} solution!$$

$$2\alpha= \textcolor[rgb]{1,0,0}{+} \cos^{-1}( \frac{ \sqrt{2} } { 2 } ) =45\ensurement{^{\circ}} \quad | \quad *2$$

$$\textcolor[rgb]{1,0,0}{
\boxed{
\textcolor[rgb]{0,0,0}{
4\alpha= 90\ensurement{^{\circ}}
}
}
}$$

.
2 sept. 2014
 #7
avatar+26393 
+10

Hi Melody,

 

François Viète (Latin: Franciscus Vieta; 1540 – 23 February 1603), Seigneur de la Bigotière, was a French mathematician whose work on new algebra was an important step towards modern algebra, due to its innovative use of letters as parameters in equations

The Laws

Basic formulas

Any general polynomial of degree n

P(x)=a_nx^n + a_{n-1}x^{n-1} +\cdots + a_1 x+ a_0 \,

(with the coefficients being real or complex numbers and an ≠ 0) is known by the fundamental theorem of algebra to have n (not necessarily distinct) complex roots x1, x2, ..., xn. Vieta's formulas relate the polynomial's coefficients { ak } to signed sums and products of its roots { xi } as follows:

\begin{cases} x_1 + x_2 + \dots + x_{n-1} + x_n = -\tfrac{a_{n-1}}{a_n} \\ (x_1 x_2 + x_1 x_3+\cdots + x_1x_n) + (x_2x_3+x_2x_4+\cdots + x_2x_n)+\cdots + x_{n-1}x_n = \frac{a_{n-2}}{a_n} \\ {} \quad \vdots \\ x_1 x_2 \dots x_n = (-1)^n \tfrac{a_0}{a_n}. \end{cases}

Equivalently stated, the (n − k)th coefficient an−k is related to a signed sum of all possible subproducts of roots, taken k-at-a-time:

\sum_{1\le i_1 < i_2 < \cdots < i_k\le n} x_{i_1}x_{i_2}\cdots x_{i_k}=(-1)^k\frac{a_{n-k}}{a_n}

for k = 1, 2, ..., n (where we wrote the indices ik in increasing order to ensure each subproduct of roots is used exactly once).

The left hand sides of Vieta's formulas are the elementary symmetric functions of the roots.

Example

Vieta's formulas applied to quadratic and cubic polynomial:

For the second degree polynomial (quadratic) P(x)=ax^2 + bx + c, roots x_1, x_2 of the equation P(x)=0 satisfy

x_1 + x_2 = - \frac{b}{a}, \quad x_1 x_2 = \frac{c}{a}.

The first of these equations can be used to find the minimum (or maximum) of P. See second order polynomial.

For the cubic polynomial P(x)=ax^3 + bx^2 + cx + d, roots x_1, x_2, x_3 of the equation P(x)=0 satisfy

x_1 + x_2 + x_3 = - \frac{b}{a}, \quad x_1 x_2 + x_1 x_3 + x_2 x_3 = \frac{c}{a}, \quad x_1 x_2 x_3 = - \frac{d}{a}.
28 août 2014