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Question Number 54045 by ajfour last updated on 28/Jan/19

Commented by ajfour last updated on 28/Jan/19

ABCD is a square, and its diagonal as a side of the rectangle. Find R/r.

$${ABCD}\:{is}\:{a}\:{square},\:{and}\:{its}\:{diagonal} \\ $$$${as}\:{a}\:{side}\:{of}\:{the}\:{rectangle}.\:{Find}\:{R}/{r}. \\ $$

Answered by ajfour last updated on 28/Jan/19

(R/r) = 2((((((√5)+1)/2)+(√2)+1)/((√5)+(√2)+3))) = (((√5)+2(√2)+3)/((√5)+(√2)+3)) .

$$\:\:\frac{{R}}{{r}}\:=\:\mathrm{2}\left(\frac{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{2}}+\mathrm{1}}{\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}+\mathrm{3}}\right)\: \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}}{\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}+\mathrm{3}}\:. \\ $$

Answered by mr W last updated on 28/Jan/19

let AB=1 tan α=(1/2)⇒tan (α/2)=(√5)−2 tan β=2⇒tan (β/2)=(((√5)−1)/2) tan ((45°)/2)=(√2)−1 (R/(tan (α/2)))+(R/(tan ((45°)/2)))=(√2) (R/((√5)−2))+(R/((√2)−1))=(√2) ⇒R=(((2−(√2))((√5)−2))/((√5)+(√2)−3)) (r/(tan ((45°)/2)))+(r/(tan (β/2)))=((√2)/2) (r/((√2)−1))+((2r)/((√5)−1))=((√2)/2) ⇒r=(((2−(√2))((√5)−1))/(2((√5)+2(√2)−3))) ⇒(R/r)=(((3−(√5))((√5)+2(√2)−3))/(2((√5)+(√2)−3)))≈1.21265

$${let}\:{AB}=\mathrm{1} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\sqrt{\mathrm{5}}−\mathrm{2} \\ $$$$\mathrm{tan}\:\beta=\mathrm{2}\Rightarrow\mathrm{tan}\:\frac{\beta}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\frac{\mathrm{45}°}{\mathrm{2}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$ \\ $$$$\frac{{R}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}+\frac{{R}}{\mathrm{tan}\:\frac{\mathrm{45}°}{\mathrm{2}}}=\sqrt{\mathrm{2}} \\ $$$$\frac{{R}}{\sqrt{\mathrm{5}}−\mathrm{2}}+\frac{{R}}{\sqrt{\mathrm{2}}−\mathrm{1}}=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)}{\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}−\mathrm{3}} \\ $$$$\frac{{r}}{\mathrm{tan}\:\frac{\mathrm{45}°}{\mathrm{2}}}+\frac{{r}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\frac{{r}}{\sqrt{\mathrm{2}}−\mathrm{1}}+\frac{\mathrm{2}{r}}{\sqrt{\mathrm{5}}−\mathrm{1}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}\right)} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)\left(\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}\right)}{\mathrm{2}\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}−\mathrm{3}\right)}\approx\mathrm{1}.\mathrm{21265} \\ $$

Commented by ajfour last updated on 29/Jan/19

thanks Sir.

$${thanks}\:{Sir}.\: \\ $$

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