Question Number 187774 by normans last updated on 21/Feb/23 | ||
Answered by mr W last updated on 21/Feb/23 | ||
Commented by mr W last updated on 21/Feb/23 | ||
$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$${a}={side}\:{length}\:{of}\:{regular}\:{hexagon} \\ $$$${let}\:{OP}={d} \\ $$$$\mathrm{sin}\:\alpha=\frac{{h}_{\mathrm{1}} }{{d}} \\ $$$$\mathrm{sin}\:\beta=\frac{{h}_{\mathrm{3}} }{{d}} \\ $$$$\alpha+\beta=\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\mathrm{cos}\:\frac{\pi}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\sqrt{\left({d}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} \right)\left({d}^{\mathrm{2}} −{h}_{\mathrm{3}} ^{\mathrm{2}} \right)}}{{d}^{\mathrm{2}} }−\frac{{h}_{\mathrm{1}} {h}_{\mathrm{3}} }{{d}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({d}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} \right)\left({d}^{\mathrm{2}} −{h}_{\mathrm{3}} ^{\mathrm{2}} \right)=\left(\frac{{d}^{\mathrm{2}} }{\mathrm{2}}+{h}_{\mathrm{1}} {h}_{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{d}^{\mathrm{2}} =\mathrm{4}\left({h}_{\mathrm{1}} ^{\mathrm{2}} +{h}_{\mathrm{3}} ^{\mathrm{2}} +{h}_{\mathrm{1}} {h}_{\mathrm{3}} \right) \\ $$$$\mathrm{cos}\:\left(\alpha−\frac{\pi}{\mathrm{6}}\right)=\frac{{h}_{\mathrm{2}} +\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}}{{d}} \\ $$$$\sqrt{\mathrm{3}}\:\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha=\frac{\mathrm{2}{h}_{\mathrm{2}} +\sqrt{\mathrm{3}}{a}}{{d}} \\ $$$$\frac{\sqrt{\mathrm{3}\left({d}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} \right)}}{{d}}+\frac{{h}_{\mathrm{1}} }{{d}}=\frac{\mathrm{2}{h}_{\mathrm{2}} +\sqrt{\mathrm{3}}{a}}{{d}} \\ $$$$\sqrt{\mathrm{3}\left({d}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} \right)}+{h}_{\mathrm{1}} =\mathrm{2}{h}_{\mathrm{2}} +\sqrt{\mathrm{3}}{a} \\ $$$$\sqrt{\mathrm{4}\left({h}_{\mathrm{1}} ^{\mathrm{2}} +{h}_{\mathrm{3}} ^{\mathrm{2}} +{h}_{\mathrm{1}} {h}_{\mathrm{3}} \right)−\mathrm{3}{h}_{\mathrm{1}} ^{\mathrm{2}} }+{h}_{\mathrm{1}} =\mathrm{2}{h}_{\mathrm{2}} +\sqrt{\mathrm{3}}{a} \\ $$$${h}_{\mathrm{1}} +\mathrm{2}{h}_{\mathrm{3}} +{h}_{\mathrm{1}} =\mathrm{2}{h}_{\mathrm{2}} +\sqrt{\mathrm{3}}{a} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}={h}_{\mathrm{1}} +{h}_{\mathrm{3}} −{h}_{\mathrm{2}} \\ $$$${similarly} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}={h}_{\mathrm{4}} +{h}_{\mathrm{2}} −{h}_{\mathrm{3}} \\ $$$$\Rightarrow{h}_{\mathrm{4}} +{h}_{\mathrm{2}} −{h}_{\mathrm{3}} ={h}_{\mathrm{1}} +{h}_{\mathrm{3}} −{h}_{\mathrm{2}} \\ $$$$\Rightarrow{h}_{\mathrm{4}} ={h}_{\mathrm{1}} +\mathrm{2}\left({h}_{\mathrm{3}} −{h}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{h}={h}_{\mathrm{1}} +{h}_{\mathrm{4}} =\mathrm{2}\left({h}_{\mathrm{1}} +{h}_{\mathrm{3}} −{h}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{X}+{A}_{\mathrm{2}} =\mathrm{2}\left({A}_{\mathrm{1}} +{A}_{\mathrm{3}} −{A}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{X}+\mathrm{7}=\mathrm{2}\left(\mathrm{10}+\mathrm{3}−\mathrm{7}\right)=\mathrm{12} \\ $$$$\Rightarrow{X}=\mathrm{5} \\ $$ | ||
Answered by mr W last updated on 21/Feb/23 | ||
Commented by mr W last updated on 21/Feb/23 | ||
$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$${say}\:{P}\left({k},{h}_{\mathrm{3}} \right) \\ $$$${eqn}.\:{line}\:{L}\mathrm{1}: \\ $$$${y}=\sqrt{\mathrm{3}}{x} \\ $$$${h}_{\mathrm{1}} =\frac{\sqrt{\mathrm{3}}{k}−{h}_{\mathrm{3}} }{\:\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}{k}=\mathrm{2}{h}_{\mathrm{1}} +{h}_{\mathrm{3}} \\ $$$${eqn}.\:{line}\:{L}\mathrm{2}: \\ $$$${y}=−\sqrt{\mathrm{3}}\left({x}−{a}\right) \\ $$$${h}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}{k}+{h}_{\mathrm{3}} −\sqrt{\mathrm{3}}{a}}{\mathrm{2}} \\ $$$$\mathrm{2}{h}_{\mathrm{2}} =\sqrt{\mathrm{3}}{k}+{h}_{\mathrm{3}} −\sqrt{\mathrm{3}}{a} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}{a}=\mathrm{2}\left({h}_{\mathrm{1}} +{h}_{\mathrm{3}} −{h}_{\mathrm{2}} \right) \\ $$$${eqn}.\:{of}\:{line}\:{L}\mathrm{4}: \\ $$$${y}=\sqrt{\mathrm{3}}\left({x}−\mathrm{2}{a}\right) \\ $$$${h}_{\mathrm{4}} =−\frac{\sqrt{\mathrm{3}}{k}−{h}_{\mathrm{3}} −\mathrm{2}{a}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${h}_{\mathrm{4}} =−\frac{\mathrm{2}{h}_{\mathrm{1}} +{h}_{\mathrm{3}} −{h}_{\mathrm{3}} −\mathrm{4}\left({h}_{\mathrm{1}} +{h}_{\mathrm{3}} −{h}_{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$${h}_{\mathrm{4}} ={h}_{\mathrm{1}} +\mathrm{2}\left({h}_{\mathrm{3}} −{h}_{\mathrm{2}} \right) \\ $$$${h}={h}_{\mathrm{1}} +{h}_{\mathrm{4}} =\mathrm{2}\left({h}_{\mathrm{1}} +{h}_{\mathrm{3}} −{h}_{\mathrm{2}} \right) \\ $$$${X}+{A}_{\mathrm{2}} =\mathrm{2}\left({A}_{\mathrm{1}} +{A}_{\mathrm{3}} −{A}_{\mathrm{2}} \right) \\ $$$${X}+\mathrm{7}=\mathrm{2}\left(\mathrm{10}+\mathrm{3}−\mathrm{7}\right)=\mathrm{12} \\ $$$${X}=\mathrm{5} \\ $$ | ||
Commented by mr W last updated on 21/Feb/23 | ||