Question

正八边形 $ABCDEFGH$ 的面积为 $n$ 。设四边形 $ACEG$ 的面积为 $m$ 。 $\dfrac{m}{n}$ 是多少？

Regular octagon $ABCDEFGH$ has area $n.$ Let $m$ be the area of quadrilateral $ACEG.$ What is $\dfrac{m}{n}?$

$\dfrac{\sqrt{2}}{4}$

$\dfrac{\sqrt{2}}{2}$

$\dfrac{3}{4}$

$\dfrac{3\sqrt{2}}{5}$

$\dfrac{2\sqrt{2}}{3}$

$ACEG$ 是内接于八边形的正方形

$ACEG$ is a square inscribed in the octagon

对称放置八边形并计算两个面积；四个角上的三角形组成了差额

Place the octagon symmetrically and compute both areas; the four corner triangles make up the difference

四个顶点 $A, C, E, G$ 构成一个正方形，因为它们是正八边形中每隔一个顶点取出的点。

取外接圆半径为一，八边形面积为 $2\sqrt2$ ，而正方形 $ACEG$ 的对角线等于该圆的直径，因此面积为 $2$ 。

比值为 $\dfrac{2}{2\sqrt2} = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}$ 。

因此，正确答案是 B。

The four vertices $A, C, E, G$ form a square, since they are every other vertex of the regular octagon.

Taking a unit circumradius, the octagon’s area is $2\sqrt2$ and the square $ACEG$ has diagonal equal to the circle’s diameter, giving area $2.$

The ratio is $\dfrac{2}{2\sqrt2} = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}.$

Thus, B is the correct answer.