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Question Number 183773 by HeferH last updated on 30/Dec/22

In a square (ABCD) there is a quarter of a circle ADC (AD = DC), put a point N in the arc AC such that AN = 1 and NC = 2(√2) find BN.

$$\:{In}\:{a}\:{square}\:\left({ABCD}\right)\:{there}\:{is}\:{a}\:{quarter}\:{of} \\ $$$$\:{a}\:{circle}\:{ADC}\:\left({AD}\:=\:{DC}\right),\:{put}\:{a}\:{point}\:{N} \\ $$$$\:{in}\:{the}\:{arc}\:{AC}\:{such}\:{that}\:{AN}\:=\:\mathrm{1}\:{and}\:{NC}\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\:{find}\:{BN}.\: \\ $$$$\: \\ $$

Answered by mr W last updated on 30/Dec/22

Commented by mr W last updated on 30/Dec/22

AC=(√2)a AC^2 =1^2 +(2(√2))^2 −2×1×2(√2)×cos 135° 2a^2 =13 ⇒a=(√((13)/2)) cos α=((2(√2))/(2a))=(2/( (√(13)))) ⇒sin α=(3/( (√(13))))=cos β BN^2 =(2(√2))^2 +((√((13)/2)))^2 −2(2(√2))×(√(((13)/2) ))×(3/( (√(13))))=(5/2) ⇒BN=((√(10))/2) ✓

$${AC}=\sqrt{\mathrm{2}}{a} \\ $$$${AC}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\mathrm{2}\sqrt{\mathrm{2}}×\mathrm{cos}\:\mathrm{135}° \\ $$$$\mathrm{2}{a}^{\mathrm{2}} =\mathrm{13} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{13}}{\mathrm{2}}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}{a}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}=\mathrm{cos}\:\beta \\ $$$${BN}^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{13}}{\mathrm{2}}}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)×\sqrt{\frac{\mathrm{13}}{\mathrm{2}}\:}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow{BN}=\frac{\sqrt{\mathrm{10}}}{\mathrm{2}}\:\checkmark \\ $$

Answered by mr W last updated on 30/Dec/22

Method II D as origin (0,0) say N(x,y) x^2 +y^2 =a^2 x^2 +(a−y)^2 =1^2 ⇒a^2 −ay=(1/2) (a−x)^2 +y^2 =(2(√2))^2 ⇒a^2 −ax=4 (((a^2 −4)/a))^2 +(((a^2 −(1/2))/a))^2 =a^2 a^4 −9a^2 +((65)/4)=0 ⇒a^2 =((13)/2) ⇒a=(√((13)/2)) ⇒x=(5/( (√(26)))) ⇒y=((12)/( (√(26)))) BN=(√(((√((13)/2))−(5/( (√(26)))))^2 +((√((13)/2))−((12)/( (√(26)))))^2 )) =((√(10))/2) ✓

$${Method}\:{II} \\ $$$${D}\:{as}\:{origin}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$$${say}\:{N}\left({x},{y}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left({a}−{y}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \:\Rightarrow{a}^{\mathrm{2}} −{ay}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({a}−{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:\Rightarrow{a}^{\mathrm{2}} −{ax}=\mathrm{4} \\ $$$$\left(\frac{{a}^{\mathrm{2}} −\mathrm{4}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}}{{a}}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{4}} −\mathrm{9}{a}^{\mathrm{2}} +\frac{\mathrm{65}}{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{13}}{\mathrm{2}}\:\Rightarrow{a}=\sqrt{\frac{\mathrm{13}}{\mathrm{2}}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{5}}{\:\sqrt{\mathrm{26}}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{12}}{\:\sqrt{\mathrm{26}}} \\ $$$${BN}=\sqrt{\left(\sqrt{\frac{\mathrm{13}}{\mathrm{2}}}−\frac{\mathrm{5}}{\:\sqrt{\mathrm{26}}}\right)^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{13}}{\mathrm{2}}}−\frac{\mathrm{12}}{\:\sqrt{\mathrm{26}}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{10}}}{\mathrm{2}}\:\checkmark \\ $$

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