Question Number 222703 by gabthemathguy25 last updated on 05/Jul/25
Commented by gabthemathguy25 last updated on 06/Jul/25
$${Where}\:{the}\:{circles}\:{radius}\:{is}\:{R}\:=\:\mathrm{7} \\ $$
Answered by mr W last updated on 06/Jul/25
Commented by mr W last updated on 06/Jul/25
$${R}={radius}\:{of}\:{circle} \\ $$$${a}_{\mathrm{5}} ={side}\:{length}\:{of}\:{regular}\:{pentagon} \\ $$$$\:\:\:\:\:\:\:\:\:{inscribed}\:{in}\:{the}\:{circle} \\ $$$${a}_{\mathrm{4}} ={side}\:{length}\:{of}\:{largest}\:{square}\: \\ $$$$\:\:\:\:\:\:\:\:{inscribed}\:{in}\:{the}\:{pentagon} \\ $$$${a}_{\mathrm{3}} ={side}\:{length}\:{of}\:{largest}\:{equilateral}\: \\ $$$$\:\:\:\:\:\:\:\:\:{inscribed}\:{in}\:{the}\:{square} \\ $$$${a}_{\mathrm{5}} =\mathrm{2}{R}\:\mathrm{sin}\:\mathrm{36}°\approx\mathrm{1}.\mathrm{176}{R} \\ $$$$\frac{{a}_{\mathrm{4}} }{\mathrm{2}\:\mathrm{cos}\:\mathrm{36}°}+\frac{{a}_{\mathrm{4}} \:\mathrm{sin}\:\mathrm{18}°}{\mathrm{sin}\:\mathrm{72}°}={a}_{\mathrm{5}} \\ $$$${a}_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{36}°}+\frac{\mathrm{sin}\:\mathrm{18}°}{\mathrm{sin}\:\mathrm{72}°}\right)=\mathrm{2}{R}\:\mathrm{sin}\:\mathrm{36}° \\ $$$$\Rightarrow{a}_{\mathrm{4}} =\frac{\sqrt{\mathrm{5}}{R}}{\mathrm{1}+\sqrt{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}}}\approx\mathrm{0}.\mathrm{984}{R} \\ $$$${a}_{\mathrm{3}} \:\mathrm{sin}\:\mathrm{15}°+\frac{{a}_{\mathrm{3}} }{\:\sqrt{\mathrm{2}}}={a}_{\mathrm{4}} \\ $$$${a}_{\mathrm{3}} \left(\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\:\mathrm{2}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\frac{\sqrt{\mathrm{5}}{R}}{\mathrm{1}+\sqrt{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}}} \\ $$$${a}_{\mathrm{3}} =\frac{\sqrt{\mathrm{10}}{R}}{\left(\mathrm{1}+\sqrt{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)\left(\mathrm{1}+\sqrt{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}}\right)}\approx\mathrm{1}.\mathrm{019}{R} \\ $$
Commented by gabthemathguy25 last updated on 06/Jul/25
perfection. ��