245
RÉSOLUES.
D’après la manière dont on vient de voir que se construisent les points
, on trouvera facilement pour leurs équations, savoir :
![{\displaystyle {\text{pour }}\mathrm {S_{1}} \left\{{\begin{array}{ll}x={\tfrac {1}{2}}\left\{1-\left({\tfrac {m-2n}{m}}\right)^{0}\right\}c=0,\\y={\tfrac {1}{2}}\left\{1-\left({\tfrac {m-2n}{m}}\right)^{0}\right\}d=0;\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d38431bd14b867993d59a6a7f4bfe2e7a872c6d)
![{\displaystyle {\text{pour }}\mathrm {S_{2}} \left\{{\begin{array}{ll}x={\tfrac {1}{2}}\left\{1-\left({\tfrac {m-2n}{m}}\right)^{1}\right\}c,\\y={\tfrac {1}{2}}\left\{1-\left({\tfrac {m-2n}{m}}\right)^{1}\right\}d;\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9111978402fb2d2ce5332ef00c3929d56f300a01)
![{\displaystyle {\text{pour }}\mathrm {S_{3}} \left\{{\begin{array}{ll}x={\tfrac {1}{2}}\left\{1-\left({\tfrac {m-2n}{m}}\right)^{2}\right\}c,\\y={\tfrac {1}{2}}\left\{1-\left({\tfrac {m-2n}{m}}\right)^{2}\right\}d;\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba242d9556f17f4fd84ca28af9b5905ec9840c66)
et en général
![{\displaystyle {\text{pour }}\mathrm {S_{k}} \left\{{\begin{array}{ll}x={\tfrac {1}{2}}\left\{1-\left({\tfrac {m-2n}{m}}\right)^{k-1}\right\}c,\\y={\tfrac {1}{2}}\left\{1-\left({\tfrac {m-2n}{m}}\right)^{k-1}\right\}d;\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cacd762dd61e9d02b97f60021c093f9a2d4dc3a9)
De même, d’après la manière dont a vu que se construisent les points
, …, on trouvera facilement pour leurs équations, savoir :
![{\displaystyle {\text{pour }}\mathrm {P_{1}} \left\{{\begin{array}{ll}x=\left({\tfrac {n}{m}}\right)^{0}a+{\tfrac {1}{2(m-3n)}}\left\{(m-3n)+2n\left({\tfrac {n}{m}}\right)^{0}-(m-n)\left({\tfrac {m-2n}{m}}\right)^{0}\right\}c,\\y=\left({\tfrac {n}{m}}\right)^{0}b+{\tfrac {1}{2(m-3n)}}\left\{(m-3n)+2n\left({\tfrac {n}{m}}\right)^{0}-(m-n)\left({\tfrac {m-2n}{m}}\right)^{0}\right\}d;\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/814baa982c9e774c439e5f444dcbabb8dec590ba)
![{\displaystyle {\text{pour }}\mathrm {P_{2}} \left\{{\begin{array}{ll}x=\left({\tfrac {n}{m}}\right)^{1}a+{\tfrac {1}{2(m-3n)}}\left\{(m-3n)+2n\left({\tfrac {n}{m}}\right)^{1}-(m-n)\left({\tfrac {m-2n}{m}}\right)^{1}\right\}c,\\y=\left({\tfrac {n}{m}}\right)^{1}b+{\tfrac {1}{2(m-3n)}}\left\{(m-3n)+2n\left({\tfrac {n}{m}}\right)^{1}-(m-n)\left({\tfrac {m-2n}{m}}\right)^{1}\right\}d;\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d6cf61ed67598bf956dfc7d4d87d28c2e3c3084)
![{\displaystyle {\text{pour }}\mathrm {P_{3}} \left\{{\begin{array}{ll}x=\left({\tfrac {n}{m}}\right)^{2}a+{\tfrac {1}{2(m-3n)}}\left\{(m-3n)+2n\left({\tfrac {n}{m}}\right)^{2}-(m-n)\left({\tfrac {m-2n}{m}}\right)^{2}\right\}c,\\y=\left({\tfrac {n}{m}}\right)^{2}b+{\tfrac {1}{2(m-3n)}}\left\{(m-3n)+2n\left({\tfrac {n}{m}}\right)^{2}-(m-n)\left({\tfrac {m-2n}{m}}\right)^{2}\right\}d;\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6601c449596afe241ff648af7530f2e303019341)