Question-219117

Question Number 219117 by Spillover last updated on 19/Apr/25

Answered by A5T last updated on 20/Apr/25

(r+1)^2 =1^2 +(3−r−(√3))^2 ⇒4r=6+r(√3)−3(√3) r=((6−3(√3))/(4−(√3)))

$$\left(\mathrm{r}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{r}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4r}=\mathrm{6}+\mathrm{r}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{r}=\frac{\mathrm{6}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}−\sqrt{\mathrm{3}}} \\ $$

Commented by Spillover last updated on 20/Apr/25

$${thank}\:{you} \\ $$

Answered by Nicholas666 last updated on 20/Apr/25

$${r}\:=\:\mathrm{3}−\mathrm{2}\:\sqrt{\mathrm{2}} \\ $$$$ \\ $$

Commented by Nicholas666 last updated on 20/Apr/25

using the system Circles Theorem with curvatures k = (1/r), and nothing the large circle′s internal tangency (k_1 = −1), the two medium circles (k_2 = k_3 = (√2) + 1), and the small circle (k_4 = (1/r)), we have; (−1+((√2)+1)+((√2)+1)+(1/r))^2 = 2((−1)^2 +((√2)+1)^2 +((√2)+1)^2 +((1/r))^2 ) solving the equatio for r yields to possible values. By visual inspection of the diagram (the smallest circle).. r = 3−2(√(2 )) ✓

$${using}\:{the}\:{system}\:{Circles}\:{Theorem}\:{with}\:{curvatures}\:{k}\:=\:\frac{\mathrm{1}}{{r}},\:\:\:\:\:\:\:\:\: \\ $$$${and}\:{nothing}\:{the}\:{large}\:{circle}'{s}\:{internal}\:{tangency}\:\left({k}_{\mathrm{1}} \:=\:−\mathrm{1}\right),\:\:\:\: \\ $$$${the}\:{two}\:{medium}\:{circles}\:\left({k}_{\mathrm{2}} \:=\:{k}_{\mathrm{3}} \:=\:\sqrt{\mathrm{2}}\:+\:\mathrm{1}\right),\:\:\:{and}\:{the}\:{small}\:{circle}\:\left({k}_{\mathrm{4}} \:=\:\frac{\mathrm{1}}{{r}}\right),\:{we}\:{have};\:\:\:\:\: \\ $$$$\left(−\mathrm{1}+\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}} \:=\:\mathrm{2}\left(\left(−\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}} \right)\:\:\:\:\:\: \\ $$$${solving}\:{the}\:{equatio}\:{for}\:{r}\:{yields}\:{to}\:{possible}\:{values}.\:\:\:\:\: \\ $$$${By}\:{visual}\:{inspection}\:{of}\:{the}\:{diagram}\:\left({the}\:{smallest}\:{circle}\right).. \\ $$$${r}\:=\:\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}\:}\:\:\checkmark \\ $$$$ \\ $$

Answered by Spillover last updated on 20/Apr/25