（4分）设点$P$在单位圆的内接正八边形$A_{1}A_{2}\ldots A_{8}$的边$A_{1}A_{2}$上，则$\overrightarrow{P{A}_{1}}^{2}+\overrightarrow{P{A}_{2}}^{2}+\ldots +\overrightarrow{P{A}_{8}}^{2}$的取值范围是 　$[12+2\sqrt{2}$，$16]$　．
分析：以圆心为原点，$A_{7}A_{3}$所在直线为$x$轴，$A_{5}A_{1}$所在直线为$y$轴，建立平面直角坐标系，求出正八边形各个顶点坐标，设$P(x,y)$，进而得到$\overrightarrow{P{A}_{1}}^{2}+\overrightarrow{P{A}_{2}}^{2}+\ldots +\overrightarrow{P{A}_{8}}^{2}=8(x^{2}+y^{2})+8$，根据点$P$的位置可求出$x^{2}+y^{2}$的范围，从而得到$\overrightarrow{P{A}_{1}}^{2}+\overrightarrow{P{A}_{2}}^{2}+\ldots +\overrightarrow{P{A}_{8}}^{2}$的取值范围．
解：以圆心为原点，$A_{7}A_{3}$所在直线为$x$轴，$A_{5}A_{1}$所在直线为$y$轴，建立平面直角坐标系，如图所示，

则$A_{1}(0,1)$，${A}_{2}(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2})$，$A_{3}(1,0)$，${A}_{4}(\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2})$，$A_{5}(0,-1)$，${A}_{6}(-\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2})$，$A_{7}(-1,0)$，${A}_{8}(-\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2})$，
设$P(x,y)$，
则$\overrightarrow{P{A}_{1}}^{2}+\overrightarrow{P{A}_{2}}^{2}+\ldots +\overrightarrow{P{A}_{8}}^{2}=\vert PA_{1}\vert ^{2}+\vert PA_{2}\vert ^{2}+\vert PA_{3}\vert ^{2}+\vert PA_{4}\vert ^{2}+\vert PA_{5}\vert ^{2}+\vert PA_{6}\vert ^{2}+\vert PA_{7}\vert ^{2}+\vert PA_{8}\vert ^{2}=8(x^{2}+y^{2})+8$，
$\because \cos 22.5^\circ \leqslant \vert OP\vert \leqslant 1$，$\therefore$$\dfrac{1+\cos 45^\circ }{2}\leqslant {x}^{2}+{y}^{2}\leqslant 1$，
$\therefore$$\dfrac{2+\sqrt{2}}{4}\leqslant {x}^{2}+{y}^{2}\leqslant 1$，
$\therefore 12+2\sqrt{2}\leqslant 8(x^{2}+y^{2})+8\leqslant 16$，
即$\overrightarrow{P{A}_{1}}^{2}+\overrightarrow{P{A}_{2}}^{2}+\ldots +\overrightarrow{P{A}_{8}}^{2}$的取值范围是$[12+2\sqrt{2}$，$16]$，
故答案为：$[12+2\sqrt{2}$，$16]$．

点评：本题主要考查了平面向量数量积的运算和性质，考查了学生分析问题和转化问题的能力，属于中档题．