AMC12 2020 A

AMC12 2020 A · Q14

AMC12 2020 A · Q14. It mainly tests Area & perimeter, Polygons.

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

正八边形 $ABCDEFGH$ 的面积为 $n$。设 $m$ 为四边形 $ACEG$ 的面积。$\frac{m}{n}$ 是多少？

(A) \frac{\sqrt{2}}{4} \frac{\sqrt{2}}{4}

(B) \frac{\sqrt{2}}{2} \frac{\sqrt{2}}{2}

(D) \frac{3\sqrt{2}}{5} \frac{3\sqrt{2}}{5}

(E) \frac{2\sqrt{2}}{3} \frac{2\sqrt{2}}{3}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Without loss of generality, let the side length of the regular octagon be 2. Extend sides $AB$ and $DC$ to meet at point $P$. Then $BP=CP=\sqrt{2}$. By symmetry, $ACEG$ is a square. Let its side length be $s$. Then by the Pythagorean Theorem applied to $\triangle APC$, the area of $ACEG$ is \[ m=s^2=(2+\sqrt{2})^2+(\sqrt{2})^2=8+4\sqrt{2}. \] The area of the octagon can be found by subtracting the areas of 4 triangles congruent to $\triangle BPC$ from the area of the square circumscribing the octagon shown below. Thus \[ n=(2+2\sqrt{2})^2-4\cdot\frac12\cdot(\sqrt{2})^2=8+8\sqrt{2}. \] Then \[ \frac{m}{n}=\frac{8+4\sqrt{2}}{8+8\sqrt{2}}=\frac{2+\sqrt{2}}{2+2\sqrt{2}}=\frac{\sqrt{2}}{2}. \]

答案（B）：不失一般性，设正八边形的边长为 2。延长边 $AB$ 和 $DC$ 相交于点 $P$。则 $BP=CP=\sqrt{2}$。由对称性，$ACEG$ 是一个正方形。设其边长为 $s$。对 $\triangle APC$ 应用勾股定理，$ACEG$ 的面积为 \[ m=s^2=(2+\sqrt{2})^2+(\sqrt{2})^2=8+4\sqrt{2}. \] 八边形的面积可由下图所示的外接正方形面积减去 4 个全等于 $\triangle BPC$ 的三角形面积得到。因此 \[ n=(2+2\sqrt{2})^2-4\cdot\frac12\cdot(\sqrt{2})^2=8+8\sqrt{2}. \] 于是 \[ \frac{m}{n}=\frac{8+4\sqrt{2}}{8+8\sqrt{2}}=\frac{2+\sqrt{2}}{2+2\sqrt{2}}=\frac{\sqrt{2}}{2}. \]

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