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})$