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Question Number 195178 by Shlock last updated on 26/Jul/23

Answered by mr W last updated on 26/Jul/23

$$DC=BC=2,say$$$$KD=1$$$$\mathrm{\angle }KCD=\alpha ,say$$$$(1+2\cos 60°)\tan (60°-\alpha )=2\sin 60°$$$$\Rightarrow \tan (60°-\alpha )=\frac{\sqrt{3}}{2}$$$$\Rightarrow \alpha =60°-{\tan }^{-1}\frac{\sqrt{3}}{2}$$$$\mathrm{\angle }BCI=\beta ,say$$$$\beta =120°-90°-\alpha =30°-\alpha ={\tan }^{-1}\frac{\sqrt{3}}{2}-30°$$$$BI=x,say$$$$(x+2\cos 60°)\tan (60°-\beta )=2\sin 60°$$$$(x+1)\tan (90°-{\tan }^{-1}\frac{\sqrt{3}}{2})=\sqrt{3}$$$$\Rightarrow x=\sqrt{3}\times \frac{\sqrt{3}}{2}-1=\frac{1}{2}$$$$\frac{{\mathrm{\Delta }}_{BIC}}{{\mathrm{\Delta }}_{CKD}}=\frac{BI}{KD}=\frac{x}{1}=\frac{1}{2}✓$$

Commented by Shlock last updated on 26/Jul/23

Nice work, Prof!

Answered by mr W last updated on 27/Jul/23

Commented by mr W last updated on 27/Jul/23

$$\tan \alpha =\frac{1+1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$$$$\alpha +\beta =180°-90°=90°$$$$\tan \beta =\frac{1}{\tan \alpha }=\frac{\sqrt{3}}{2}=\frac{x+1}{\sqrt{3}}$$$$\Rightarrow x=\frac{3}{2}-1=\frac{1}{2}$$$$\frac{\mathrm{\Delta }BIC}{\mathrm{\Delta }KDC}=\frac{IB}{KD}=\frac{1}{2}$$

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