Question-218733

Question Number 218733 by Spillover last updated on 14/Apr/25

Commented by A5T last updated on 15/Apr/25

This leads to a contradiction(if the other chord were also a diameter).

$$\mathrm{This}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{contradiction}\left(\mathrm{if}\:\mathrm{the}\:\mathrm{other}\:\mathrm{chord}\right. \\ $$$$\left.\mathrm{were}\:\mathrm{also}\:\mathrm{a}\:\mathrm{diameter}\right). \\ $$

Answered by mr W last updated on 15/Apr/25

Commented by mr W last updated on 15/Apr/25

R=(√(3^2 +((3/2))^2 ))=((3(√5))/2) a=R−(3/2)=((3((√5)−1))/2) b=2R−3a=2×((3(√5))/2)−3×((3((√5)−1))/2)=((3(3−(√5)))/2) c^2 =3^2 +b^2 =((27(3−(√5)))/2) cd=2a(a+b) (h/3)=(d/c)=((2a(a+b))/c^2 ) =((2×2)/(27(3−(√5))))×((3((√5)−1))/2)×(((3((√5)−1))/2)+((3(3−(√5)))/2)) =((2((√5)−1))/(3(3−(√5))))=(((√5)+1)/3) ⇒h=(√5)+1 A_(pink) =((2ah)/2)=ah=((3((√5)−1)((√5)+1))/2)=6 ✓

$${R}=\sqrt{\mathrm{3}^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${a}={R}−\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${b}=\mathrm{2}{R}−\mathrm{3}{a}=\mathrm{2}×\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{3}×\frac{\mathrm{3}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}}=\frac{\mathrm{3}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{\mathrm{27}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}{\mathrm{2}} \\ $$$${cd}=\mathrm{2}{a}\left({a}+{b}\right) \\ $$$$\frac{{h}}{\mathrm{3}}=\frac{{d}}{{c}}=\frac{\mathrm{2}{a}\left({a}+{b}\right)}{{c}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}×\mathrm{2}}{\mathrm{27}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}×\frac{\mathrm{3}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}}×\left(\frac{\mathrm{3}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}}+\frac{\mathrm{3}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{h}=\sqrt{\mathrm{5}}+\mathrm{1} \\ $$$${A}_{{pink}} =\frac{\mathrm{2}{ah}}{\mathrm{2}}={ah}=\frac{\mathrm{3}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{6}\:\checkmark \\ $$

Commented by Spillover last updated on 15/Apr/25