|ED|={\frac {1}{2}}\cdot a_{1}} (4) | E S 0 | = a 1 {\displaystyle \quad |ES_{0}|=a_{1}\quad } , since the inscribed hexagon is regular The height | C...
D F | = | E S 4 | = r 4 {\displaystyle |DF|=|ES_{4}|=r_{4}} , since F S 4 ¯ {\displaystyle {\overline {FS_{4}}}} is perpendicular to C D ¯ {\displaystyle...
D F | = | E S 4 | = r 4 {\displaystyle |DF|=|ES_{4}|=r_{4}} , since F S 4 ¯ {\displaystyle {\overline {FS_{4}}}} is perpendicular to C D ¯ {\displaystyle...
to the tangent (6) | F S 2 | = | E S 2 | = r 2 {\displaystyle \quad |FS_{2}|=|ES_{2}|=r_{2}\quad } , since B A → {\displaystyle {\overrightarrow {BA}}}...
{1}{2}}\cdot a_{0}} (3) | D S 0 | = | E S 0 | = r 1 {\displaystyle \quad |DS_{0}|=|ES_{0}|=r_{1}} (4) β = γ = π 6 {\displaystyle \quad \beta =\gamma ={\frac {\pi...
0 | = | F S 0 | = | E G | 2 = | F H | 2 = 1 2 ⋅ a 0 {\displaystyle |ES_{0}|=|FS_{0}|={\frac {|EG|}{2}}={\frac {|FH|}{2}}={\frac {1}{2}}\cdot a_{0}\quad...
E S 0 | = | F S 0 | = 1 2 ⋅ a 0 {\displaystyle |AE|=|BE|=|CF|=|DF|=|ES_{0}|=|FS_{0}|={\frac {1}{2}}\cdot a_{0}} (3) | A F | = | B F | = a 1 {\displaystyle...
| F S 0 | = a 0 {\displaystyle |AS_{0}|=|BS_{0}|=|CS_{0}|=|DS_{0}|=|ES_{0}|=|FS_{0}|=a_{0}} So the diameter between two corners is | A D | = | B E |...
1 = 2 ⋅ r 0 {\displaystyle a_{1}={\sqrt {2}}\cdot r_{0}\quad } , see CRC dia 2) The radius of the inscribed circle r 2 = 1 2 ⋅ r 0 {\displaystyle r_{2}={\frac...
{1}{2}}r_{0}} | E S 2 | = | H S 2 | = | F S 2 | = r 2 {\displaystyle \quad |ES_{2}|=|HS_{2}|=|FS_{2}|=r_{2}} (1) D H S 0 D S 1 ¯ {\displaystyle \quad {\overline {DHS_{0}DS_{1}}}}...
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