Page:Annales de mathématiques pures et appliquées, 1811-1812, Tome 2.djvu/334

320

QUESTIONS

d’où résulte

${\begin{aligned}&\mathrm {A} a=\mathrm {AB} .{\tfrac {\operatorname {Sin} .(a+x)}{\operatorname {Sin} .a}},&&a\mathrm {B=AB} .{\tfrac {\operatorname {Sin} .x}{\operatorname {Sin} .a}},\\&\mathrm {B} b=\mathrm {BC} .{\tfrac {\operatorname {Sin} .(a+b-\mathrm {B} +x)}{\operatorname {Sin} .b}},&&b\mathrm {C=BC} .{\tfrac {\operatorname {Sin} .(a+b+x)}{\operatorname {Sin} .b}},\\&\mathrm {C} c=\mathrm {CD} .{\tfrac {\operatorname {Sin} .(a+b+c-\mathrm {B-C} +x)}{\operatorname {Sin} .c}},&&c\mathrm {D=CD} .{\tfrac {\operatorname {Sin} .(a+b-\mathrm {B-C} +x)}{\operatorname {Sin} .c}},\\&\mathrm {D} d=\mathrm {DA} .{\tfrac {\operatorname {Sin} .(a+b+c+d-\mathrm {B-C-D} +x)}{\operatorname {Sin} .d}},&&d\mathrm {A=DA} .{\tfrac {\operatorname {Sin} .(a+b+c-\mathrm {B-C-D} +x)}{\operatorname {Sin} .d}},\\\end{aligned}}$

PROBLÈME I. On donne le contour du polygone demandé.

D’après ce qui précède, on a

${\begin{aligned}\mathrm {A} a+a\mathrm {B} =&\mathrm {AB} .{\tfrac {\operatorname {Sin} .x+\operatorname {Sin} .(a+x)}{\operatorname {Sin} .a}}\\\mathrm {B} b+b\mathrm {C} =&\mathrm {BC} .{\tfrac {\operatorname {Sin} .(a-\mathrm {B} +x)+\operatorname {Sin} .(a+b-\mathrm {B} +x)}{\operatorname {Sin} .b}}\\\mathrm {C} c+c\mathrm {D} =&\mathrm {CD} .{\tfrac {\operatorname {Sin} .(a+b-\mathrm {B-C} +x)+\operatorname {Sin} .(a+b+c-\mathrm {B-C} +x)}{\operatorname {Sin} .c}}\\\mathrm {D} d+d\mathrm {A} =&\mathrm {DA} .{\tfrac {\operatorname {Sin} .(a+b+c-\mathrm {B-C-D} +x)+\operatorname {Sin} .(a+b+c+d-\mathrm {B-C-D} +x)}{\operatorname {Sin} .d}}\,;\\\end{aligned}}$

prenant la somme de ces équations, en remarquant qu’en général

${\frac {\operatorname {Sin} .z+\operatorname {Sin} .(k+z)}{\operatorname {Sin} .k}}=\operatorname {Cosec} .{\tfrac {1}{2}}k.\operatorname {Sin} .\left({\tfrac {1}{2}}k+z\right),$

il viendra

$ab+bc+cd+da=\left\{{\begin{aligned}&\mathrm {AB} .\operatorname {Cosec} .{\tfrac {1}{2}}a.\operatorname {Sin} .\left({\tfrac {1}{2}}a+x\right),\\+&\mathrm {BC} .\operatorname {Cosec} .{\tfrac {1}{2}}b.\operatorname {Sin} .\left(a+{\tfrac {1}{2}}b-\mathrm {B} +x\right),\\+&\mathrm {CD} .\operatorname {Cosec} .{\tfrac {1}{2}}c.\operatorname {Sin} .\left(a+b+{\tfrac {1}{2}}c-\mathrm {B-C} +x)\right),\\+&\mathrm {DA} .\operatorname {Cosec} .{\tfrac {1}{2}}d.\operatorname {Sin} .\left(a+b+c+{\tfrac {1}{2}}d-\mathrm {B-C-D} +x\right).\\\end{aligned}}\right.$

De là découle la construction suivante, fondée sur les propriétés du centre des moyennes distances :

Sur une droite $\mathrm {SE}$ (fig. 19), et en un de ses points $\mathrm {S,}$ soient faits les angles $\mathrm {ESA,ES} b,\mathrm {ES} c,\mathrm {ES} d,$ respectivement égaux aux angles