Question Number 203055 by mr W last updated on 08/Jan/24
Commented by mr W last updated on 08/Jan/24
$${is}\:{it}\:{possible}\:{to}\:{determine}\:{the}\: \\ $$$${unknown}\:{area}\:{in}\:{terms}\:{of}\:{other} \\ $$$${partial}\:{areas}\:{A},{B},{C},{D},{E}? \\ $$
Commented by ajfour last updated on 09/Jan/24
$${i}\:{think},\:{yes}. \\ $$$${We}\:{assume}\:{four}\:{angles}\:{and}\:{one}\:{side} \\ $$$${as}\:{variables}. \\ $$$${System}\:{is}\:{unique}.\:{Five}\:{equations} \\ $$$${can}\:{be}\:{written}\:{using}\:{these}\:{given}\: \\ $$$${areas}\:{to}\:{find}\:{them}. \\ $$
Commented by mr W last updated on 09/Jan/24
$${but}\:{the}\:{equations}\:{are}\:{non}−{linear}, \\ $$$${see}\:{below}.\:{so}\:{i}\:{don}'{t}\:{think}\:{they}\:{are} \\ $$$${solvable}. \\ $$
Commented by mr W last updated on 12/Jan/24
$${four}\:{angles}\:{and}\:{one}\:{side}\:{don}'{t}\:{make} \\ $$$${the}\:{system}\:{unique}\:{as}\:{following} \\ $$$${figure}\:{shows}.\:{though}\:{the}\:{blue}\:{star} \\ $$$${and}\:{the}\:{green}\:{star}\:{have}\:{equal}\:{angles} \\ $$$${and}\:{one}\:{common}\:{equal}\:{side}\:\left({red}\right),\: \\ $$$${but}\:{they}\:{are}\:{different}. \\ $$
Commented by mr W last updated on 12/Jan/24
Answered by mr W last updated on 09/Jan/24
Commented by mr W last updated on 09/Jan/24
$${at}\:{first}\:{let}'{s}\:{find}\:{the}\:{side}\:{lengthes}\: \\ $$$${of}\:{a}\:{triangle}\:{if}\:{its}\:{area}\:{is}\:{S}\:{and}\:{its} \\ $$$${interior}\:{angles}\:{are}\:\alpha,\:\beta\:{and}\:\gamma\: \\ $$$${respectively}\:{with}\:\alpha+\beta+\gamma=\mathrm{180}°. \\ $$$${a}=\mathrm{2}{R}\:\mathrm{sin}\:\alpha \\ $$$${b}=\mathrm{2}{R}\:\mathrm{sin}\:\beta \\ $$$${c}=\mathrm{2}{R}\:\mathrm{sin}\:\gamma \\ $$$${S}=\frac{{abc}}{\mathrm{4}{R}}=\frac{\mathrm{8}{R}^{\mathrm{3}} \:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}{\mathrm{4}{R}}=\mathrm{2}{R}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma \\ $$$$\Rightarrow{R}=\sqrt{\frac{{S}}{\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{2}{S}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma}} \\ $$$$\Rightarrow{b}=\sqrt{\frac{\mathrm{2}{S}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\gamma\:\mathrm{sin}\:\alpha}} \\ $$$$\Rightarrow{c}=\sqrt{\frac{\mathrm{2}{S}\:\mathrm{sin}\:\gamma}{\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta}} \\ $$
Commented by mr W last updated on 10/Jan/24
Commented by mr W last updated on 12/Jan/24
$${say}\:{the}\:{angles}\:{of}\:{the}\:\mathrm{5}−{point}\:{star}\: \\ $$$${are}\:\theta_{\mathrm{1}} ,\theta_{\mathrm{2}} ,\theta_{\mathrm{3}} ,\theta_{\mathrm{4}} ,\theta_{\mathrm{5}} \:{respectively}. \\ $$$$\theta_{\mathrm{1}} +\theta_{\mathrm{2}} +\theta_{\mathrm{3}} +\theta_{\mathrm{4}} +\theta_{\mathrm{5}} =\mathrm{180}°\:\:\:…\left({vi}\right) \\ $$$${so}\:{in}\:{fact}\:{we}\:{have}\:{only}\:{four}\: \\ $$$${unknown}\:{angles}. \\ $$$$\varphi_{\mathrm{1}} =\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \\ $$$$\varphi_{\mathrm{2}} =\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \\ $$$$\varphi_{\mathrm{3}} =\theta_{\mathrm{4}} +\theta_{\mathrm{1}} \\ $$$$\varphi_{\mathrm{4}} =\theta_{\mathrm{5}} +\theta_{\mathrm{2}} \\ $$$$\varphi_{\mathrm{5}} =\theta_{\mathrm{1}} +\theta_{\mathrm{3}} \\ $$$$ \\ $$$${say}\:{the}\:{unknown}\:{area}\:{in}\:{the}\:{middle} \\ $$$${is}\:{X}. \\ $$$${according}\:{to}\:{above}\:{we}\:{have} \\ $$$${a}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}{S}_{\mathrm{1}} \:\mathrm{sin}\:\varphi_{\mathrm{2}} }{\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\varphi_{\mathrm{1}} }}=\sqrt{\frac{\mathrm{2}{S}_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)}{\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)}} \\ $$$${b}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}{S}_{\mathrm{1}} \:\mathrm{sin}\:\varphi_{\mathrm{1}} }{\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\varphi_{\mathrm{2}} }}=\sqrt{\frac{\mathrm{2}{S}_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)}{\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)}} \\ $$$${c}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}{S}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} }{\mathrm{sin}\:\varphi_{\mathrm{1}} \:\mathrm{sin}\:\varphi_{\mathrm{2}} }}=\sqrt{\frac{\mathrm{2}{S}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} }{\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)}} \\ $$$${P}_{\mathrm{1}} {Q}_{\mathrm{3}} ={b}_{\mathrm{1}} +{c}_{\mathrm{2}} =\sqrt{\frac{\mathrm{2}{S}_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)}{\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)}}+\sqrt{\frac{\mathrm{2}{S}_{\mathrm{2}} \:\mathrm{sin}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{4}} +\theta_{\mathrm{1}} \right)}} \\ $$$${P}_{\mathrm{1}} {Q}_{\mathrm{3}} =\sqrt{\frac{\mathrm{2}\left({S}_{\mathrm{1}} +{S}_{\mathrm{4}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{4}} }{\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{4}} \right)}} \\ $$$$\sqrt{\frac{{S}_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)}{\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)}}+\sqrt{\frac{{S}_{\mathrm{2}} \:\mathrm{sin}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{4}} +\theta_{\mathrm{1}} \right)}}=\sqrt{\frac{\left({S}_{\mathrm{1}} +{S}_{\mathrm{4}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{4}} }{\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{4}} \right)}}\:\:\: \\ $$$$\Rightarrow\sqrt{{S}_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{4}} +\theta_{\mathrm{1}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)}+\sqrt{{S}_{\mathrm{2}} \:\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{2}} }=\sqrt{\left({S}_{\mathrm{1}} +{S}_{\mathrm{4}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{4}} \:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)}\:\:\:…\left({i}\right) \\ $$$${similarly} \\ $$$$\sqrt{\frac{{S}_{\mathrm{2}} \:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)}{\mathrm{sin}\:\theta_{\mathrm{2}} \:\mathrm{sin}\:\left(\theta_{\mathrm{4}} +\theta_{\mathrm{1}} \right)}}+\sqrt{\frac{{S}_{\mathrm{3}} \:\mathrm{sin}\:\theta_{\mathrm{3}} }{\mathrm{sin}\:\left(\theta_{\mathrm{4}} +\theta_{\mathrm{1}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{5}} +\theta_{\mathrm{2}} \right)}}=\sqrt{\frac{\left({S}_{\mathrm{2}} +{S}_{\mathrm{5}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{5}} }{\mathrm{sin}\:\theta_{\mathrm{2}} \:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{5}} \right)}}\: \\ $$$$\Rightarrow\sqrt{{S}_{\mathrm{2}} \:\mathrm{sin}\:\left(\theta_{\mathrm{5}} +\theta_{\mathrm{2}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)}+\sqrt{{S}_{\mathrm{3}} \:\mathrm{sin}\:\theta_{\mathrm{2}} \:\mathrm{sin}\:\theta_{\mathrm{3}} }=\sqrt{\left({S}_{\mathrm{2}} +{S}_{\mathrm{5}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{5}} \:\mathrm{sin}\:\left(\theta_{\mathrm{4}} +\theta_{\mathrm{1}} \right)}\:\:\:…\left({ii}\right) \\ $$$$\sqrt{\frac{{S}_{\mathrm{3}} \:\mathrm{sin}\:\left(\theta_{\mathrm{4}} +\theta_{\mathrm{1}} \right)}{\mathrm{sin}\:\theta_{\mathrm{3}} \:\mathrm{sin}\:\left(\theta_{\mathrm{5}} +\theta_{\mathrm{2}} \right)}}+\sqrt{\frac{{S}_{\mathrm{4}} \:\mathrm{sin}\:\theta_{\mathrm{4}} }{\mathrm{sin}\:\left(\theta_{\mathrm{5}} +\theta_{\mathrm{2}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{3}} \right)}}=\sqrt{\frac{\left({S}_{\mathrm{3}} +{S}_{\mathrm{1}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{1}} }{\mathrm{sin}\:\theta_{\mathrm{3}} \:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{1}} \right)}}\:\:\: \\ $$$$\Rightarrow\sqrt{{S}_{\mathrm{3}} \:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{3}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{4}} +\theta_{\mathrm{1}} \right)}+\sqrt{{S}_{\mathrm{4}} \:\mathrm{sin}\:\theta_{\mathrm{3}} \:\mathrm{sin}\:\theta_{\mathrm{4}} }=\sqrt{\left({S}_{\mathrm{3}} +{S}_{\mathrm{1}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{1}} \:\mathrm{sin}\:\left(\theta_{\mathrm{5}} +\theta_{\mathrm{2}} \right)}\:\:\:…\left({iii}\right) \\ $$$$\sqrt{\frac{{S}_{\mathrm{4}} \:\mathrm{sin}\:\left(\theta_{\mathrm{5}} +\theta_{\mathrm{2}} \right)}{\mathrm{sin}\:\theta_{\mathrm{4}} \:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{3}} \right)}}+\sqrt{\frac{{S}_{\mathrm{5}} \:\mathrm{sin}\:\theta_{\mathrm{5}} }{\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{3}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)}}=\sqrt{\frac{\left({S}_{\mathrm{4}} +{S}_{\mathrm{2}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{2}} }{\mathrm{sin}\:\theta_{\mathrm{4}} \:\mathrm{sin}\:\left(\theta_{\mathrm{4}} +\theta_{\mathrm{2}} \right)}}\:\: \\ $$$$\Rightarrow\sqrt{{S}_{\mathrm{4}} \:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{5}} +\theta_{\mathrm{2}} \right)}+\sqrt{{S}_{\mathrm{5}} \:\mathrm{sin}\:\theta_{\mathrm{4}} \:\mathrm{sin}\:\theta_{\mathrm{5}} }=\sqrt{\left({S}_{\mathrm{4}} +{S}_{\mathrm{2}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{2}} \:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{3}} \right)}\:\:\:…\left({iv}\right) \\ $$$$\sqrt{\frac{{S}_{\mathrm{5}} \:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{3}} \right)}{\mathrm{sin}\:\theta_{\mathrm{5}} \:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)}}+\sqrt{\frac{{S}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} }{\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{4}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)}}=\sqrt{\frac{\left({S}_{\mathrm{5}} +{S}_{\mathrm{3}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{3}} }{\mathrm{sin}\:\theta_{\mathrm{5}} \:\mathrm{sin}\:\left(\theta_{\mathrm{5}} +\theta_{\mathrm{3}} \right)}}\:\: \\ $$$$\Rightarrow\sqrt{{S}_{\mathrm{5}} \:\mathrm{sin}\:\left(\theta_{\mathrm{3}} +\theta_{\mathrm{5}} \right)\:\mathrm{sin}\:\left(\theta_{\mathrm{1}} +\theta_{\mathrm{3}} \right)}+\sqrt{{S}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{5}} \:\mathrm{sin}\:\theta_{\mathrm{1}} }=\sqrt{\left({S}_{\mathrm{5}} +{S}_{\mathrm{3}} +{X}\right)\:\mathrm{sin}\:\theta_{\mathrm{3}} \:\mathrm{sin}\:\left(\theta_{\mathrm{2}} +\theta_{\mathrm{3}} \right)}\:\:\:…\left({v}\right) \\ $$$$ \\ $$$${we}\:{have}\:\mathrm{6}\:{equations}\:{for}\:\mathrm{6}\:{unknowns}: \\ $$$$\theta_{\mathrm{1}} ,\theta_{\mathrm{2}} ,\theta_{\mathrm{3}} ,\theta_{\mathrm{4}} ,\theta_{\mathrm{5}} \:{and}\:{X}. \\ $$
Commented by lazyboy last updated on 11/Jan/24
Commented by MathematicalUser2357 last updated on 20/Jan/24
$${unsolvable}\:{integral} \\ $$