Inscribed circle

A circle is inscribed at the bottom wall of the cube with an edge (a = 1). What is the radius of the spherical surface that contains this circle and one of the vertex of the top cube base?

Correct result:

r =  0.8004

Solution:

$$\begin{gathered} a=1 \\ {r}_1=a\mathrm{/}2=1\mathrm{/}2={\displaystyle \frac{1}{2}}=0.5 \\ u=\sqrt{2}\cdot a=\sqrt{2}\cdot 1=\sqrt{2}\doteq 1.4142 \\ {r}_2=u\mathrm{/}2=1.4142\mathrm{/}2\doteq 0.7071 \\ {r}^2=x^2+r_1^2 \\ {r}^2=(a-x{)}^2+r_2^2 \\ {x}^2+r_1^2=(a-x{)}^2+r_2^2 \\ {r}_1^2=-2xa+a^2+r_2^2 \\ x={\displaystyle \frac{a^2+r_2^2-r_1^2}{2\cdot a}}={\displaystyle \frac{1^2+0.7071^2-0.5^2}{2\cdot 1}}={\displaystyle \frac{5}{8}}=0.625 \\ {r}_3=\sqrt{x^2+r_1^2}=\sqrt{0.625^2+0.5^2}\doteq 0.8004 \\ {r}_3=r \\ r=\sqrt{(x-a{)}^2+r_2^2}=\sqrt{(0.625-1{)}^2+0.7071^2}=0.8004 \end{gathered}$$

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