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Question Number 39089 by ajfour last updated on 02/Jul/18

Commented by ajfour last updated on 02/Jul/18

Find (a/b) and 𝛉 . The curve is a semicircle and the two red segments are equal.

$${Find}\:\:\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}\:\:{and}\:\boldsymbol{\theta}\:. \\ $$$${The}\:{curve}\:{is}\:{a}\:{semicircle}\:{and} \\ $$$${the}\:{two}\:{red}\:{segments}\:{are}\:{equal}. \\ $$

Answered by MJS last updated on 02/Jul/18

circle: (x−(a/2))^2 +y^2 =((a/2))^2 upper semicircle y=(√(ax−x^2 )) tangent in P= ((p),((√(ap−p^2 ))) ) y=−((2p−a)/(2(√(ap−p^2 ))))x+((ap)/(2(√(ap−p^2 )))) (b=((ap)/(2(√(ap−p^2 ))))) Q= ((a),(b) ) line PQ: y=((p(2p−a))/(2(a−p)(√(ap−p^2 ))))x+((ap(2a−3p))/(2(a−p)(√(ap−p^2 )))) y=0 ⇒ x=((ap(2a−3p))/(a−2p)) x=a−p ⇒ ⇒ p=((a(√2))/2) ⇒ tangent: y=−((√(−2+2(√2)))/2)x+((√(1+(√2)))/2)a tan θ=−((√(−2+2(√2)))/2) ⇒ θ≈−24.47°

$$\mathrm{circle}: \\ $$$$\left({x}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{upper}\:\mathrm{semicircle} \\ $$$${y}=\sqrt{{ax}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{tangent}\:\mathrm{in}\:{P}=\begin{pmatrix}{{p}}\\{\sqrt{{ap}−{p}^{\mathrm{2}} }}\end{pmatrix} \\ $$$${y}=−\frac{\mathrm{2}{p}−{a}}{\mathrm{2}\sqrt{{ap}−{p}^{\mathrm{2}} }}{x}+\frac{{ap}}{\mathrm{2}\sqrt{{ap}−{p}^{\mathrm{2}} }} \\ $$$$\left({b}=\frac{{ap}}{\mathrm{2}\sqrt{{ap}−{p}^{\mathrm{2}} }}\right) \\ $$$${Q}=\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix} \\ $$$$\mathrm{line}\:{PQ}: \\ $$$${y}=\frac{{p}\left(\mathrm{2}{p}−{a}\right)}{\mathrm{2}\left({a}−{p}\right)\sqrt{{ap}−{p}^{\mathrm{2}} }}{x}+\frac{{ap}\left(\mathrm{2}{a}−\mathrm{3}{p}\right)}{\mathrm{2}\left({a}−{p}\right)\sqrt{{ap}−{p}^{\mathrm{2}} }} \\ $$$${y}=\mathrm{0}\:\Rightarrow\:{x}=\frac{{ap}\left(\mathrm{2}{a}−\mathrm{3}{p}\right)}{{a}−\mathrm{2}{p}} \\ $$$${x}={a}−{p}\:\Rightarrow \\ $$$$\Rightarrow\:{p}=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{tangent}: \\ $$$${y}=−\frac{\sqrt{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}{x}+\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}}{\mathrm{2}}{a} \\ $$$$\mathrm{tan}\:\theta=−\frac{\sqrt{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\Rightarrow\:\theta\approx−\mathrm{24}.\mathrm{47}° \\ $$

Commented by ajfour last updated on 02/Jul/18

No Sir, tan θ =+((√(−2+2(√2)))/2) .

$${No}\:{Sir},\:\mathrm{tan}\:\theta\:=+\frac{\sqrt{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}\:. \\ $$

Commented by MJS last updated on 02/Jul/18

yes of course, if you measure θ as a positive angle. my solution comes from the tangent which is decreasing...

$$\mathrm{yes}\:\mathrm{of}\:\mathrm{course},\:\mathrm{if}\:\mathrm{you}\:\mathrm{measure}\:\theta\:\mathrm{as}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{angle}.\:\mathrm{my}\:\mathrm{solution}\:\mathrm{comes}\:\mathrm{from}\:\mathrm{the}\:\mathrm{tangent} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{decreasing}... \\ $$

Commented by ajfour last updated on 02/Jul/18

Thanks Sir, i also solved (and of course created the question myself).

$${Thanks}\:{Sir},\:{i}\:{also}\:{solved}\:\left({and}\right. \\ $$$${of}\:{course}\:{created}\:{the}\:{question} \\ $$$$\left.{myself}\right). \\ $$

Commented by MJS last updated on 02/Jul/18

I like your geometric questions and it′s very interesting how different people differently solve the same problems

$$\mathrm{I}\:\mathrm{like}\:\mathrm{your}\:\mathrm{geometric}\:\mathrm{questions}\:\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{very} \\ $$$$\mathrm{interesting}\:\mathrm{how}\:\mathrm{different}\:\mathrm{people}\:\mathrm{differently} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{same}\:\mathrm{problems} \\ $$

Answered by ajfour last updated on 02/Jul/18

Let origin be center of circle. Point of tangency P≡((a/2)sin θ,(a/2)cos θ) The dashed line makes angle θ with vertical ; so tan θ =((sin θ)/(cos θ))= ((b−(a/2)cos θ)/((a/2)+(a/2)sin θ)) ⇒ (a/2)sin θ+(a/2)sin^2 θ=bcos θ−(a/2)cos^2 θ bcos θ−(a/2) =(a/2)sin θ ...(i) let the red segment be x =(a/2)−(a/2)sin θ Then ((a−x)/b) = (x/(b−y)) ⇒ (((a/2)+(a/2)sin θ)/b) = (((a/2)−(a/2)sin θ)/(b−(a/2)cos θ)) b+bsin θ−(a/2)cos θ−(a/2)cos θsin θ = b−bsin θ 2bsin θ−(a/2)cos θ = (a/2)sin θcos θ ....(ii) Now (i)×cos θ is bcos^2 θ−(a/2)cos θ = (a/2)sin θcos θ equating this with (ii) 2bsin θ = bcos^2 θ ⇒ sin^2 θ +2sin θ−1=0 (sin θ+1)^2 =2 sin θ = ±(√2)−1 considering the +ve value sin θ = (√2)−1 tan θ = (((√2)−1)/(√(1−((√2)−1)^2 ))) θ=tan^(−1) ((((√2)−1)/(√(2(√2)−2)))) = tan^(−1) (((√(2(√2)−2))/2) ) (a/b) = ((2cos θ)/(1+sin θ)) = ((2(√(2(√2)−2)))/(√2)) hence (a/b) = 2(√((√2)−1)) .

$${Let}\:{origin}\:{be}\:{center}\:{of}\:{circle}. \\ $$$${Point}\:{of}\:{tangency}\:{P}\equiv\left(\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\theta,\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\theta\right) \\ $$$${The}\:{dashed}\:{line}\:{makes}\:{angle}\:\theta \\ $$$${with}\:{vertical}\:;\:{so} \\ $$$$\mathrm{tan}\:\theta\:=\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}=\:\frac{{b}−\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\theta}{\frac{{a}}{\mathrm{2}}+\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\:\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\theta+\frac{{a}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \theta={b}\mathrm{cos}\:\theta−\frac{{a}}{\mathrm{2}}\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:{b}\mathrm{cos}\:\theta−\frac{{a}}{\mathrm{2}}\:=\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\theta\:\:\:\:...\left({i}\right) \\ $$$${let}\:{the}\:{red}\:{segment}\:{be} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:=\frac{{a}}{\mathrm{2}}−\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\theta \\ $$$${Then}\:\:\:\:\:\frac{{a}−{x}}{{b}}\:=\:\frac{{x}}{{b}−{y}} \\ $$$$\:\Rightarrow\:\:\frac{\frac{{a}}{\mathrm{2}}+\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\theta}{{b}}\:=\:\frac{\frac{{a}}{\mathrm{2}}−\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\theta}{{b}−\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\theta} \\ $$$$\:{b}+{b}\mathrm{sin}\:\theta−\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\theta\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{b}−{b}\mathrm{sin}\:\theta \\ $$$$\mathrm{2}{b}\mathrm{sin}\:\theta−\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\theta\:=\:\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({ii}\right) \\ $$$${Now}\:\left({i}\right)×\mathrm{cos}\:\theta\:\:{is} \\ $$$$\:\:\:\:{b}\mathrm{cos}\:^{\mathrm{2}} \theta−\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\theta\:=\:\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$${equating}\:{this}\:{with}\:\left({ii}\right) \\ $$$$\:\:\:\:\mathrm{2}{b}\mathrm{sin}\:\theta\:=\:{b}\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\:\:\mathrm{sin}\:^{\mathrm{2}} \theta\:+\mathrm{2sin}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{sin}\:\theta+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\theta\:=\:\pm\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\:\:\:{considering}\:{the}\:+{ve}\:{value} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\theta\:=\:\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{tan}\:\theta\:=\:\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{1}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}}{\mathrm{2}}\:\right)\: \\ $$$$\:\:\:\frac{{a}}{{b}}\:=\:\frac{\mathrm{2cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}}{\sqrt{\mathrm{2}}} \\ $$$${hence}\:\:\:\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}\:=\:\mathrm{2}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\:\:. \\ $$

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