Ars Mathematica: Solución del problema 67

\frac{b}{c} = \frac{c}{D E} = \frac{a}{C D} \Rightarrow {\begin{matrix} D E = \frac{c^{2}}{b} \\ C D = \frac{a c}{b} \end{matrix} .

\begin{matrix} F E & = & k b \\ F K & = & k c \\ E K & = & k a \end{matrix}

\begin{matrix} F H & = & s b \\ F G & = & s a \\ G H & = & s c \end{matrix}

F K = F H

k c = s b \Rightarrow s = \frac{k c}{b}

F E + F G = c \Rightarrow k b + s a \Rightarrow k b + \frac{k c}{b} a = c \Rightarrow k (\frac{b^{2} + a c}{b}) = c \Rightarrow k = \frac{b c}{b^{2} + a c}

s = \frac{b c}{b^{2} + a c} \cdot \frac{c}{b} = \frac{c^{2}}{b^{2} + a c} \Rightarrow F K = F H = \frac{b c^{2}}{b^{2} + a c} (*)

\begin{matrix} H N & = & n b \\ H O & = & n a \\ O N & = & n c \end{matrix}

\begin{matrix} N L & = & p a \\ L M & = & p c \\ N M & = & p b \end{matrix}

O N = N M

n c = p b \Rightarrow n = \frac{p b}{c}

F H = H N + N L 0 n b + p a = \frac{b c^{2}}{b^{2} + a c}

\frac{p b^{2}}{c} + p a = \frac{b c^{2}}{b^{2} + a c} \Rightarrow p (\frac{b^{2} + a c}{c}) = \frac{b c^{2}}{b^{2} + a c} \Rightarrow p = \frac{b c^{3}}{(b^{2} + a c)^{2}}

n = \frac{b c^{3} b}{(b^{2} + a c)^{2} c} = \frac{b^{2} c^{2}}{(b^{2} + a c)^{2}} = k^{2}

k t = \frac{b c t}{(b^{2} + a c)^{2}}

k^{2} t

1 : k : k^{2}

k = \frac{b c}{b^{2} + a c}

r_{1}, r_{2}, r_{3}

r_{2}^{2} = r_{1} r_{3}

Ars Mathematica