Ars Mathematica: Solución al problema 60

A B C D

A C

M, N, R, P

r_{1}, r_{2}, r_{3}, r_{4}

C (N, r_{2})

\begin{matrix} (a - r_{2})^{2} & = & r_{2}^{2} + r_{2}^{2} \\ (a - r_{2})^{2} & = & 2 r_{2}^{2} \\ {(\frac{a - r_{2}}{\sqrt{2}})}^{2} & = & r_{2} \\ \frac{a - r_{2}}{\sqrt{2}} & = & r_{2} \\ a & = & r_{2} (1 + \sqrt{2}) \\ r_{2} & = & \frac{a}{1 + \sqrt{2}} \\ r_{2} & = & a (\sqrt{2} - 1) \end{matrix}

C (M, r_{1})

\begin{matrix} (a - 2 r_{2} - r_{1})^{2} & = & r_{1}^{2} + r_{1}^{2} \\ a - 2 r_{2} - r_{1} & = & r_{1} \sqrt{2} \\ a - 2 r_{2} & = & r_{1} (1 + \sqrt{2}) \\ r_{1} & = & \frac{a - 2 r_{2}}{\sqrt{2} + 1} \\ r_{1} = \frac{a - 2 (a (\sqrt{2} - 1))}{\sqrt{2} + 1} & = & a \frac{1 - 2 \sqrt{2} + 2}{\sqrt{2} + 1} = a \frac{3 - 2 \sqrt{2}}{\sqrt{2} + 1} \\ r_{1} = a (3 - 2 \sqrt{2}) (\sqrt{2} - 1) & = & a (3 \sqrt{2} - 4 - 3 + 2 \sqrt{2}) \\ r_{1} & = & a (5 \sqrt{2} - 7) \end{matrix}

C (R, r_{3})

\begin{matrix} (a + r_{3})^{2} & = & 2 (a - r_{3})^{2} \\ a + r_{3} & = & \sqrt{2} (a + r_{3}) \\ a + r_{3} & = & a \sqrt{2} - \sqrt{2} r_{3} \\ r_{3} (1 + \sqrt{2}) & = & a (\sqrt{2} - 1) \\ r_{3} & = & a \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = a (\sqrt{2} - 1)^{2} \\ r_{3} & = & a (3 - 2 \sqrt{2}) \end{matrix}

C (P, r_{4})

\begin{matrix} (a + 2 r_{3} + r_{4})^{2} & = & 2 (a - r_{4})^{2} \\ a + 2 r_{3} + r_{4} & = & \sqrt{2} (a - r_{4}) \\ a + 2 r_{3} + r_{4} & = & a \sqrt{2} - \sqrt{2} r_{4} \\ r_{4} (1 + \sqrt{2}) & = & a (\sqrt{2} - 1) - r_{3} = \\ = a (\sqrt{2} - 1 - 1 (3 - 2 \sqrt{2})) & = & = a (\sqrt{2} - 1 - 6 + 4 \sqrt{2}) = a (- 7 + 5 \sqrt{2}) \\ r_{4} = a \frac{- 7 + 5 \sqrt{2}}{\sqrt{2} + 1} & = & a (- 7 + 5 \sqrt{2}) (\sqrt{2} - 1) \\ r_{4} & = & a (17 - 12 \sqrt{2}) \end{matrix}

. \begin{matrix} \frac{r_{3}}{r_{4}} = \frac{3 - 2 \sqrt{2}}{17 - 12 \sqrt{2}} = 3 + 2 \sqrt{2} \\ \frac{r_{2}}{r_{1}} = \frac{\sqrt{2} - 1}{5 \sqrt{2} - 7} = 3 + 2 \sqrt{2} \end{matrix}} \Rightarrow \frac{r_{3}}{r_{4}} = \frac{r_{2}}{r_{1}} \Rightarrow r_{1} r_{3} = r_{2} r_{4}

Ars Mathematica