249
RÉSOLUES.
![{\displaystyle {\text{pour }}\mathrm {P_{k}} \left\{{\begin{array}{ll}x={\tfrac {1}{2}}\left\{c+\left[2a-(2k-1)c\right]\left({\tfrac {1}{3}}\right)^{k-1}\right\},\\y={\tfrac {1}{2}}\left\{d+\left[2b-(2k-1)d\right]\left({\tfrac {1}{3}}\right)^{k-1}\right\}.\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e89f0b22fc33225002e43f002c9674365747252)
En conséquence, les équations des points
, seront telles qu’il suit :
![{\displaystyle {\text{pour }}\mathrm {M_{1}} \left\{{\begin{array}{ll}x={\tfrac {1}{2}}\left\{c+(a-c)\left({\tfrac {1}{3}}\right)^{0}\right\},\\y={\tfrac {1}{2}}\left\{d+(b-d)\left({\tfrac {1}{3}}\right)^{0}\right\};\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/753c824add4f0ebeef1dd3991eb63ac97fd6c648)
![{\displaystyle {\text{pour }}\mathrm {M_{2}} \left\{{\begin{array}{ll}x={\tfrac {1}{2}}\left\{c+(a-2c)\left({\tfrac {1}{3}}\right)^{1}\right\},\\y={\tfrac {1}{2}}\left\{d+(b-2d)\left({\tfrac {1}{3}}\right)^{1}\right\};\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8871c52718021897b0541451f3749288114de79)
![{\displaystyle {\text{pour }}\mathrm {M_{3}} \left\{{\begin{array}{ll}x={\tfrac {1}{2}}\left\{c+(a-3c)\left({\tfrac {1}{3}}\right)^{2}\right\},\\y={\tfrac {1}{2}}\left\{d+(b-3d)\left({\tfrac {1}{3}}\right)^{2}\right\};\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9d5a0b9ce7a48efd4eccae694af2220190ae150)
. . . . . . . . . . . . . . . . . . . . . . . . .
![{\displaystyle {\text{pour }}\mathrm {M_{k}} \left\{{\begin{array}{ll}x={\tfrac {1}{2}}\left\{c+(a-kc)\left({\tfrac {1}{3}}\right)^{k-1}\right\},\\y={\tfrac {1}{2}}\left\{d+(b-kd)\left({\tfrac {1}{3}}\right)^{k-1}\right\};\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68d2e1d8d635006b25a5e26c4598060444b18207)
Il n’est donc plus question que d’éliminer
entre ces deux dernières équations pour obtenir celle de la courbe.
Pour cela, traitons-y d’abord
et
comme deux inconnues distinctes, il viendra ainsi ;
![{\displaystyle {\begin{aligned}k&={\tfrac {a(2y-d)-b(2x-c)}{2(cy-dx)}},\\\left({\tfrac {1}{3}}\right)^{k-1}&={\tfrac {2(cy-dx)}{bc-ad}}.\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee91cd147d49ab95917148e99905682b16d9e72a)
De la première de ces deux expressions on déduira
![{\displaystyle k-1={\tfrac {(a-c)(2y-b)-(b-d)(2x-a)}{2(cy-dx)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e6df7e95a3f40dbc23edf564393bdb65df6330c)
;
la seconde donnera, en passant aux logarithmes,
![{\displaystyle (k-1)\cdot \operatorname {log} .{\tfrac {1}{3}}=\operatorname {log} .{\tfrac {2(cy-dx)}{bc-ad}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b1b56125d817c5d78dc20e772fd81b37966aaa3)
;