Processing math: 100%

Thursday, December 13, 2018

Vector: Regular octagon

Solution

\displaystyle \overrightarrow{{AB}}=\overrightarrow{p}\ \ \ ,\ \ \overrightarrow{{BC}}=\overrightarrow{q}

\displaystyle \overrightarrow{{AH}}=\overrightarrow{{AP}}+\overrightarrow{{PH}}

\displaystyle \ \ \ \ \ \ =-\frac{1}{{\sqrt{2}}}\overrightarrow{p}+\overrightarrow{{QC}}

\displaystyle \ \ \ \ \ \ =-\frac{1}{{\sqrt{2}}}\overrightarrow{p}+\overrightarrow{{QB}}+\overrightarrow{{BC}}

\displaystyle \ \ \ \ \ \ =-\frac{1}{{\sqrt{2}}}\overrightarrow{p}-\frac{1}{{\sqrt{2}}}\overrightarrow{p}+\overrightarrow{q}

\displaystyle \ \ \ \ \ \ =\overrightarrow{q}-\sqrt{2}\ \overrightarrow{p}

\displaystyle \overrightarrow{{AE}}+\overrightarrow{{BH}}+\overrightarrow{{CG}}+\overrightarrow{{DF}}

\displaystyle =(\overrightarrow{{AH}}+\overrightarrow{{HE}})+\overrightarrow{{BH}}+(\overrightarrow{{CF}}+\overrightarrow{{FG}})+\overrightarrow{{BH}}

\displaystyle =\overrightarrow{{AH}}+\overrightarrow{{HE}}+2\overrightarrow{{BH}}+\overrightarrow{{CF}}+\overrightarrow{{FG}}

\displaystyle =\overrightarrow{{AH}}+(1+\sqrt{2})\overrightarrow{q}+2(\overrightarrow{{BA}}+\overrightarrow{{AH}})+(1+\sqrt{2})\overrightarrow{{DE}}-\overrightarrow{q}

\displaystyle =\overrightarrow{{AH}}+\sqrt{2}\ \overrightarrow{q}-2\overrightarrow{p}+2\overrightarrow{{AH}}+(1+\sqrt{2})\overrightarrow{{AH}}

\displaystyle =(4+\sqrt{2})\overrightarrow{{AH}}+\sqrt{2}\ \overrightarrow{q}-2\overrightarrow{p}

\displaystyle =(4+\sqrt{2})(\overrightarrow{q}-\sqrt{2}\ \overrightarrow{p})+\sqrt{2}(\overrightarrow{q}-\sqrt{2}\ \overrightarrow{p})

\displaystyle =(4+2\sqrt{2})(\overrightarrow{q}-\sqrt{2}\ \overrightarrow{p})

\displaystyle =2(2+\sqrt{2})(\overrightarrow{q}-\sqrt{2}\ \overrightarrow{p})

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