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Question Number 61559 by Tawa1 last updated on 04/Jun/19

Commented by Tawa1 last updated on 04/Jun/19

Find the length of the pink

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pink} \\ $$

Answered by ajfour last updated on 04/Jun/19

x^2 +y^2 =2    (right circle)  (x+(√2)+1)^2 +(y−1)^2 =2   (left one)  ⇒ 2((√2)+1)x+3+2(√2)−2y+1=0  ⇒ y=((√2)+1)x+(√2)+2  ⇒ x^2 +[((√2)+1)x+(√2)+2]^2 =2  ⇒ (4+2(√2))x^2 +2(√2)(3+2(√2))x+4(√2)=0  one root is x=−(√2)  ⇒ other root =−((2(√2)((√2)+1)^2 )/(2(√2)((√2)+1)))+(√2)      = −(√2)−1+(√2)= −1  y=1  pink segment length=2

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}\:\:\:\:\left({right}\:{circle}\right) \\ $$$$\left({x}+\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}\:\:\:\left({left}\:{one}\right) \\ $$$$\Rightarrow\:\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){x}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}{y}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:{y}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){x}+\sqrt{\mathrm{2}}+\mathrm{2} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +\left[\left(\sqrt{\mathrm{2}}+\mathrm{1}\right){x}+\sqrt{\mathrm{2}}+\mathrm{2}\right]^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow\:\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\right){x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right){x}+\mathrm{4}\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$${one}\:{root}\:{is}\:{x}=−\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{other}\:{root}\:=−\frac{\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}+\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:=\:−\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{2}}=\:−\mathrm{1} \\ $$$${y}=\mathrm{1} \\ $$$${pink}\:{segment}\:{length}=\mathrm{2} \\ $$

Commented by Tawa1 last updated on 04/Jun/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 04/Jun/19

Commented by mr W last updated on 04/Jun/19

∠DCA=∠DCB  α=((135°)/2)=67.5°  β=180°−2α=45°  γ=135°−β=90°  DE=(√2)AE=(√2)×(√2)=2    or  BCAD is lozenge, ⇒AD//BC  ⇒β=180−135=45°  ......

$$\angle{DCA}=\angle{DCB} \\ $$$$\alpha=\frac{\mathrm{135}°}{\mathrm{2}}=\mathrm{67}.\mathrm{5}° \\ $$$$\beta=\mathrm{180}°−\mathrm{2}\alpha=\mathrm{45}° \\ $$$$\gamma=\mathrm{135}°−\beta=\mathrm{90}° \\ $$$${DE}=\sqrt{\mathrm{2}}{AE}=\sqrt{\mathrm{2}}×\sqrt{\mathrm{2}}=\mathrm{2} \\ $$$$ \\ $$$${or} \\ $$$${BCAD}\:{is}\:{lozenge},\:\Rightarrow{AD}//{BC} \\ $$$$\Rightarrow\beta=\mathrm{180}−\mathrm{135}=\mathrm{45}° \\ $$$$...... \\ $$

Commented by Tawa1 last updated on 04/Jun/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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