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Question Number 195484 by Shlock last updated on 03/Aug/23

Answered by mr W last updated on 03/Aug/23

Commented by mr W last updated on 03/Aug/23

$$c=\frac{s}{2\sin \frac{270°}{7}}$$$${(c+b)}^2+{(s-b)}^2={(c+a)}^2+s^2-2s(c+a)\cos \frac{450°}{7}$$$${(\frac{s}{2\sin \frac{270°}{7}}+b)}^2+{(s-b)}^2={(\frac{s}{2\sin \frac{270°}{7}}+a)}^2+s^2-2s(\frac{s}{2\sin \frac{270°}{7}}+a)\cos \frac{450°}{7}$$$$letp=\frac{a}{s},q=\frac{b}{s}$$$$\Rightarrow {(\frac{1}{2\sin \frac{270°}{7}}+q)}^2+{(1-q)}^2={(\frac{1}{2\sin \frac{270°}{7}}+p)}^2+1-2(\frac{1}{2\sin \frac{270°}{7}}+p)\cos \frac{450°}{7}$$$${(c+b)}^2+{(c+a)}^2-2(c+b)(c+a)\cos \frac{540°}{7}={(s-b)}^2+s^2-2(s-b)s\cos \frac{900°}{7}$$$${(\frac{s}{2\sin \frac{270°}{7}}+b)}^2+{(\frac{s}{2\sin \frac{270°}{7}}+a)}^2-2(\frac{s}{2\sin \frac{270°}{7}}+b)(\frac{s}{2\sin \frac{270°}{7}}+a)\cos \frac{540°}{7}={(s-b)}^2+s^2+2(s-b)s\cos \frac{360°}{7}$$$$\Rightarrow {(\frac{1}{2\sin \frac{270°}{7}}+q)}^2+{(\frac{1}{2\sin \frac{270°}{7}}+p)}^2-2(\frac{1}{2\sin \frac{270°}{7}}+q)(\frac{1}{2\sin \frac{270°}{7}}+p)\cos \frac{540°}{7}={(1-q)}^2+1+2(1-q)\cos \frac{360°}{7}$$$$\Rightarrow p\approx 0.57064,q\approx 0.28532$$$$\Rightarrow \frac{a}{b}=\frac{p}{q}=2✓$$

Commented by Shlock last updated on 03/Aug/23

Very great solution, Prof!

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