Question Number 43023 by ajfour last updated on 06/Sep/18
Commented by ajfour last updated on 06/Sep/18
$${Find}\:{location}\:{of}\:{centre}\left(\mathrm{0},{h}\right)\:{and} \\ $$$${radius}\:{R}\:{of}\:{circle}\:{such}\:{that} \\ $$$${all}\:{four}\:{bounded}\:{areas}\: \\ $$$${O}\overset{\frown} {{AD}}\:,\:\overset{\frown} {{APB}A}\:,\:{A}\overset{\frown} {{BC}}\overset{\frown} {{DA}}\:,\:\&\:\overset{\frown} {{CQD}C}\: \\ $$$${are}\:{equal}. \\ $$$${Determine}\:\boldsymbol{{R}},\:\boldsymbol{{h}}\:{in}\:{terms}\:{of}\:\boldsymbol{\theta}\:. \\ $$
Answered by MrW3 last updated on 07/Sep/18
Commented by MrW3 last updated on 07/Sep/18
![R sin γ=h sin θ ⇒sin γ=(h/R) sin θ with λ=(h/R) ⇒γ=sin^(−1) (λ sin θ) α=γ−θ β=γ+θ Δ_(OMD) =(h/2)×R sin α Δ_(OMC) =(h/2)×R sin β A_(OAD) =2×(h/2)×R sin α−αR^2 =((πR^2 )/3) ⇒(h/R) sin α−α=(π/3) ⇒λ sin [sin^(−1) (λ sin θ)−θ]−[sin^(−1) (λ sin θ)−θ]=(π/3) ...(i) A_(OBC) =2×(h/2)×R sin β+βR^2 =((2πR^2 )/3) ⇒(h/R) sin β+β=((2π)/3) ⇒λ sin [sin^(−1) (λ sin θ)+θ]+[sin^(−1) (λ sin θ)+θ]=((2π)/3) ...(ii) from (i) and (ii) we get λ and θ. I found a solution: λ=(h/R)≈5.763 θ≈2.64°](Q43099.png)
$${R}\:\mathrm{sin}\:\gamma={h}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{sin}\:\gamma=\frac{{h}}{{R}}\:\mathrm{sin}\:\theta \\ $$$${with}\:\lambda=\frac{{h}}{{R}} \\ $$$$\Rightarrow\gamma=\mathrm{sin}^{−\mathrm{1}} \left(\lambda\:\mathrm{sin}\:\theta\right) \\ $$$$\alpha=\gamma−\theta \\ $$$$\beta=\gamma+\theta \\ $$$$ \\ $$$$\Delta_{{OMD}} =\frac{{h}}{\mathrm{2}}×{R}\:\mathrm{sin}\:\alpha \\ $$$$\Delta_{{OMC}} =\frac{{h}}{\mathrm{2}}×{R}\:\mathrm{sin}\:\beta \\ $$$${A}_{{OAD}} =\mathrm{2}×\frac{{h}}{\mathrm{2}}×{R}\:\mathrm{sin}\:\alpha−\alpha{R}^{\mathrm{2}} =\frac{\pi{R}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\frac{{h}}{{R}}\:\mathrm{sin}\:\alpha−\alpha=\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\lambda\:\mathrm{sin}\:\left[\mathrm{sin}^{−\mathrm{1}} \left(\lambda\:\mathrm{sin}\:\theta\right)−\theta\right]−\left[\mathrm{sin}^{−\mathrm{1}} \left(\lambda\:\mathrm{sin}\:\theta\right)−\theta\right]=\frac{\pi}{\mathrm{3}}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$${A}_{{OBC}} =\mathrm{2}×\frac{{h}}{\mathrm{2}}×{R}\:\mathrm{sin}\:\beta+\beta{R}^{\mathrm{2}} =\frac{\mathrm{2}\pi{R}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\frac{{h}}{{R}}\:\mathrm{sin}\:\beta+\beta=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\lambda\:\mathrm{sin}\:\left[\mathrm{sin}^{−\mathrm{1}} \left(\lambda\:\mathrm{sin}\:\theta\right)+\theta\right]+\left[\mathrm{sin}^{−\mathrm{1}} \left(\lambda\:\mathrm{sin}\:\theta\right)+\theta\right]=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get}\:\lambda\:{and}\:\theta. \\ $$$$ \\ $$$${I}\:{found}\:{a}\:{solution}: \\ $$$$\lambda=\frac{{h}}{{R}}\approx\mathrm{5}.\mathrm{763} \\ $$$$\theta\approx\mathrm{2}.\mathrm{64}° \\ $$
Commented by ajfour last updated on 07/Sep/18
$${Very}\:{Nice}\:{Sir},\:{half}\:{the}\:{task}\:{is} \\ $$$${done}.\:{Rene}\:{Descartes}\:{said},\:'\mathrm{G}{ive} \\ $$$${names}\:{to}\:{all}\:{the}\:{lines}'\:\left({in}\:{Geometry}\right). \\ $$
Commented by ajfour last updated on 07/Sep/18
$${Wonderful}\:\&\:{Beautiful}\:{Sir}. \\ $$$$\left({I}\:{had}\:{suspected}\:{what}\:{i}\:{had}\:{asked}\right. \\ $$$$\left.{to}\:{determine},\:{i}\:{see}\:{it}\:{clearer}\:{now}\right)! \\ $$$${So}\:\theta\:{accepts}\:{just}\:{one}\:{value},\:{under} \\ $$$${the}\:{imposed}\:{condition}..{it}\:{means}. \\ $$
Commented by MrW3 last updated on 07/Sep/18
$${thank}\:{you}\:{sir}\:{for}\:{this}\:{nice}\:{question}. \\ $$
Commented by MrW3 last updated on 07/Sep/18
$${for}\:{a}\:{moment}\:{I}\:{was}\:{unsure}\:{if}\:{such}\:{a} \\ $$$${configuration}\:{is}\:{possible}\:{at}\:{all}.\:{now} \\ $$$${I}\:{know}\:{there}\:{is}\:{one}\:{and}\:{only}\:{one}\:{solution}. \\ $$
Commented by ajfour last updated on 07/Sep/18
$${A}\:{Very}\:{wise}\:{and}\:{practical}\:{solution} \\ $$$${too},\:{Sir}. \\ $$