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Question Number 186737 by Rupesh123 last updated on 09/Feb/23

Commented by Frix last updated on 09/Feb/23

$$\mathrm{No}.$$$${\mathrm{There}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{area}\frac{1}{17425}\mathrm{of}\mathrm{the}\mathrm{circle}.$$

Answered by mr W last updated on 09/Feb/23

Commented by mr W last updated on 09/Feb/23

$$57+32+25+101+98+47=360✓$$$$\frac{q}{p}=\frac{\sin \frac{25+101}{2}}{\sin \frac{57+47}{2}}=\frac{\sin 63}{\sin 52}$$$$\frac{p}{r}=\frac{\sin \frac{57}{2}}{\sin \frac{101}{2}}=\frac{\sin 28.5}{\sin 50.5}$$$$\frac{r}{q}=\frac{\sin \frac{101+98}{2}}{\sin \frac{32+57}{2}}=\frac{\sin 99.5}{\sin 44.5}$$$$\frac{q}{p}\times \frac{p}{r}\times \frac{r}{q}=\frac{\sin 63}{\sin 52}\times \frac{\sin 28.5}{\sin 50.5}\times \frac{\sin 99.5}{\sin 44.5}$$$$1\stackrel{?}{=}\frac{\sin 63}{\sin 52}\frac{\sin 28.5}{\sin 50.5}\frac{\sin 99.5}{\sin 44.5}\approx 0.98389\ne 1$$$$\Rightarrow thethreelines{don}^{\prime }tintersectat$$$$thesamepoint!$$

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